ZICA T3 - Business Mathematics & Statistics

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CHAPTER 1

ARITHMETIC This Chapter provides the students with general awareness and understanding of basic arithmetic used in the daily transactions. The Chapter looks at: • Operations on rational and real numbers • Proportions • Foreign currency conversion • Weights and measures. Pre – reading Knowledge of O-level Mathematics is required. 1.0 Introduction

1.1 Operations On Rational And Real Numbers Most of the algebra that we will study in this text is referred to as the

algebra of real numbers. This simply means that the variables represent real numbers. A set is a collection of objects and the objects are its members or elements. Hence a set of real numbers consists of rational numbers and irrational numbers.

A rational number is a number which can be written in the form ,b

a where

a and b are integers and b is not zero. Example 1

The numbers 8,0,4

16,4.,5,

4

3,

5

2 −− are rational numbers. This is

because they can be written in the form

.1

88,

4

25

4

16,

1

00,

10

44.,

1

55 ====−=−

A rational number can be defined as a terminating or repeating decimal.

2

Example 2

75.4

3

min5.22

5

3.10

3

=

=

=

decimalsatingter

...0454545.022

1

...166.06

1

...333.3

1

=

=

=

decimalsrepeating

A number which cannot be represented in the form b

a as above is called an irrational

number where a and b are integers and b is not zero. Furthermore, an irrational number has a nonrepeating decimal representation. Example 3 Some of the examples of irrational numbers are:

...058823529.017

1

...076923076.013

1

...142857142.07

1

=

=

=

decimalsrepeatingnon

Other examples are:

.,3,2 π

3

The entire set of real numbers is composed of the rational numbers along with the irrational numbers. It is necessary for us to be familiar with the various terminology used to classify different types of real numbers. { }...4,3,2,1 Natural numbers, counting numbers, positive integers. { }...,3,2,1,0 Whole numbers, nonnegative integers { }1,2,3,... −−− Negative integers { },1,2,3,... −−− Non positive integers { }...,2,1,0,1,2... −− Integers Some Definitions The result obtained by adding numbers is called the sum. The sum of 10, 7 and 15 is 10 + 7 + 15 = 32. The order in which numbers are added is not important. 10 + 7 + 15 = 7 + 15 + 10 = 15 + 10 + 7 = 32. The difference of two numbers is the larger number minus the smaller number. The difference of 23 and 5 is 23 – 5 = 18. The order of subtraction is very important. 23 – 5 is not the same as 5 – 23. The result obtained by multiplying numbers is called the product. The product of 5 and 6 is 5 x 6 = 30. The order in which multiplication is performed is not important. 5 x 6 = 6 x 5 = 30. The result obtained by division is called quotient. The quotient of 315÷ is 5. The order of division is important. .153315 ÷≠÷ Sequence of Arithmetical Operations Numbers are often combined in a series of arithmetical operations. When this happens a definite sequence must be observed.

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1) Brackets are used if there is any danger of ambiguity. The contents of the bracket must be evaluated before performing any other operation. Thus

16925)514(25

6488)35(8

=−=−−=×=+×

2) Multiplication and division must be done before addition and subtraction.

Thus

.445395342526314

.3252525

.83117211126

=+=+−=+÷−×=−=−÷

=+=+×

Exercise 1

Evaluate the following expressions

1) 386 ×+ 2) 3102 −×

3) 23263 ×+−× 4) 436 +÷ 5) 6318335 +÷−× 6) )53(6 +×

7) )46(325 −×− 8) )38(314 +×−

9) )58(21420 −×÷− 10) 521025 +÷− 11) 325732835 ÷−×−÷+× ` 12) )29)(35( ++

13) )53(2(86 ++ 14) 2523

5486

×−×−÷×

15) )3237)(2532( ×−××+×

1.2 Proportions

A ratio is a comparison between two similar quantities. If the length of a certain road is 34km and a model of it is 1 centimeter long, then the length of the model

is 34

1 of the length of the road. This can also be written as 1 to 34 and usually

represented as 1:34. The ratio has no units, it is dimensionless.

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Example 1 Express the ratio 50n to K4 000 in its simplest form. K4 000 × 100n = 400 000n

0008

1

000400

50000400:50 == .

Example 2

Express the ratio 5

1:25 in its lowest terms

1

125

1

525

5

125

5

1:25 =×=÷=

Example 3 The lengths are in the ratio 4:3. If the first length is 34km, what is the second length?

The second length = 4

3 of the first length

= 4

10234

4

3 =× km.

Example 4 Two amounts of money are in the ratio of 5:2. If the second amount is K189 000, what is the first amount?

First amount = .5004720001892

5K=×

Example 5 Divide K5.5m into two parts in the ration 3:7. Total number of parts = 3 + 7 = 10

Amount of each part = =10

0005005 K 550 000

Amount of the first part = 0005503× = K1 650 000 Amount of second part = =× 0005507 K3 850 000.

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Exercise 2

1) Express the following ratios as fractions in their lowest terms:

a) 16:8 b) 3

1:

5

2 c) 16:24 d) 27:9

e) K10 000 to K500

2) Express the ratio of K500 to K250 000 as a fraction in its lowest terms. 3) Two lengths are in the ratio 5:8. If the first length is 260 metres, what is the

second length? 4) Two amounts of money are in the ratio 7:5. If the second amount is K240 000,

what is the first amount? 5) At a sale, prices of children’s wear were reduced in the ratio 5:4. Find the sale

price of a girls suit picked at K108 000. Direct Proportion Two quantities are said to be in direct proportion if they increase or decrease at the same rate. If we buy 2kg packets of sugar at K7 000 then we suggest to pay K10 500 for 3kg sugar and K3 500 for 1 kilogram packet. That is if we double the amount bought, we double the costs; if we halve the amount bought, we halve the cost. In solving problem on direct proportion we can use either the unitary method or the fractional method. We show this in the following examples. Example 6 If 5 litres of semi-sweet wine cost K75 000 , how much does 50 litres cost? 1) Using the unitary method:

5 litres cost K75 000 or 7 500 000 ngwee

1 litre costs =5

00075 K15 000

50 litres cost 0001550× = K750 000.

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2) Using the fractional method:

Cost of 50 litres = 5

000755000075

5

50 ×=×

= K750 000 Example 7 15 litres of water have a mass of 30 000grams. Find the mass of 5 litres of water. Here the number of litres of water decreases, so the mass must decrease also. The mass and the volume of water are in direct proportion. 15 litres of water weigh 30 grams

1 litre weighs =15

00030 2 000 grams

5 litres cost ×5 2000 grams = 10 000 grams Example 8 A tourist changed $180 (180 United States dollars) for K810 000. How many Zambian kwacha would she get for $50? Here the amount is reduced $ 50 is worth

.000225180

00081050

000810180

50

K=

×=

×

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Example 9 If 6 packets of salt cost K21 000, how much do 25 packets of salt cost? 6 packet of salt cost K21 000

1 packet of salt cost =6

00021K 3 500

25 packet of salt cost .50087500325 K=× Exercise 3 1) If 9 kilograms of oranges cost K25 650, how much do 32 kilograms cost? 2) If 85 exercise books cost K112 500, how much do 17 cost? 3) Eggs cost K550 per unit. How much will 15 eggs cost? 4) If 6 metres of chitenge material cost K75 000, how much will 108 metres cost? 5) A car travels 250km on 15.625 litres of petrol. Ho far can it go on 25 litres? 6) 2.5kg of beef cost K150 000. Find the cost of 7.5kg. 7) If a tourist gets K9 500 for £1, how many Zambian kwacha could he get for £75? 8) K2 000 000 is exchanged for $364 (United States dollars). How much is

K500 000 worth in US dollars? 9) A train travels 250 kilometres in 5 hours. How long will it take to complete a

journey of 750 kilometres? 10) If 15 metres of carpet cost K1 800 000, how much will 50 metres cost? Inverse Proportion If 10 men can dig a trench in 8 hours, how long would 20 men take to dig the same trench, working at the same rate? If we double the number of men then we should halve the time taken. If we halve the number of men, then the job will probably take twice as long. This is an example of inverse proportion.

The number of men is increased in the ratio .1

2

10

20 =

9

Since this is an example of inverse proportion, the number of hours required must be

decreased in the ratio .2

1

Number of hours required = 482

1 =× hours.

Example 10 A packet of sweets is shared among 15 children and each gets 6 sweets. How many sweets will each child get if the same packet of sweets is shared among 10 children? Here the number of children decreases, so the number of sweets each gets increases.

The number of children decreases in the ratio .3

2

15

10 = Since is an example of inverse

proportion, the number of sweets increases in the ratio .2

3

Number of sweet each gets = 962

3 =× sweets.

Example 11 A car takes 8 hours to cover a journey at an average speed of 100 kilometres per hour. What average speed would be necessary to cover the same journey in 5 hours. Here, the time has been reduced, so the speed must be increased; the time and speed are

inversely proportioned. The ratio of decrease is .8

5 The average speed will increase in

the ratio .5

8

The average speed = hkm /1005

8 ×

= hkmkm /160208 =× .

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Exercise 4

1) If 20 men can dig in 5 hours, how long would it take 25 men to dig the same trench?

2) A farmer employs 24 men to harvest her maize crop. They take 12 days to

do the job. If she had employed 9 men, how long would it have taken them?

3) A bag contains sweets. When divided amongst 12 children, each receives

5 sweets. If the sweets were divided amongst 15 children, how many sweets would each receive?

4) In how many days could 25 men do a piece of work which 15 men can do

in 40 days? 5) 8 men produce 350 articles in 4 working days. How long would it take 14

men to produce the same amount? 1.3 Foreign Currency Conversion

Every country has its own monetary system. If there is to be trade and travel between any two countries, there must be a rate at which the money of one country can be converted into money of the other country. This rate is called the rate of exchange

Foreign Monetary Systems and Exchange rates 08/12/05

Country Currency Rate of exchange United States Dollar K4,368 = $1 United Kingdom British Pound (Sterling) K7,900 = $1 South Africa Rand K686 = 1 SAR European union Euro K5,414 = 1£ The methods used for direct proportion are applicable to problems in foreign exchange.

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Example 12 If K686 = 1 SAR, find the nearest SAR the value in South African money of 1 500 000 kwachas. 686 kwacha = 1 SAR

1 kwacha = SAR686

1

1 500 00 kwachas = 0005001686

1 × SAR

= 2 187 SAR.

Example 13 A tourist changes traveller’s cheques for $500 into South African rands at 6.37 rands to the dollars. How many rands does she get? $ 500 = 37.6500× rands = 3 185 rands.

Exercise 5 Where necessary give the answer to 2 places of decimals. Using the exchange rate given above, find; 1) The number of British pounds equivalent to 2,350 SAR. 2) The number of Euros equivalent to 800 000 kwachas. 3) The number of South African rands equivalent to $350 000 (United States

dollars). 4) A stereo system costs £350 in the United Kingdom. A Zambian visitor wants to

purchase a set but wishes to pay in Zambian kwachas. What is the equivalent price in kwachas?

5) A tourist changes cheques for £150 into South African rands at 9.50 rands to the

pound (£1). She spends 250 rands and changes the remainder back into Sterling at the same rate. How much did the tourist receive?

6) Calculate the rate of exchange if a bank exchanges 2 520 rands for K1 890 000.

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7) A person on holiday in United states of America changed K5 000 000 into

dollar at a rate of K4 500 to the $1. Her hotel expenses were $55 per day for 5 days and his expenses were $250. On returning home she changed the dollars she had into kwacha at a rate of K4 000 to the dollar. Calculate:

a) The number of dollars received for the K5 000 000.

b) The total expenses in kwachas. c) The number of dollars left after paying these expenses.

d) The amount in kwachas obtained from the dollars she had left.

(Give your answer to the nearest ngwee).

1.4 Weights and Measures We shall consider four of the original basic units used in the metric system. The meter (m) used to measure length The litre (l) used to measure volume The gram (g) used to measure *”weight” The Celsius degree (oc) used to measure temperature (also called centigrade

degree) The Meter The basic unit for measuring length in the metric system is the meter (m). It

corresponds roughly to the yard in the English system of measurement. 1 meter = 39.37 inches 1 yard = 36 inches * To be exact, the gram measures mass and not weight. However, in

everyday, non scientific use, the gram is used for ‘weight’. The Litre The basic unit for measuring volume (capacity) in the metric system is the litre

(l). It corresponds roughly to the quart in the English system of measurement. 1 litre = 1.06 qt

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The Gram The original basic unit for measuring “weight” in the metric system is the gram (g). One gram is a small quantity of weight. 454 grams = 1 pound The Celsius Degree The basic unit for measuring temperature in the metric system is the Celsius (often called centigrade degree).

Boiling point of water 100oC = 212 oF Normal body temperature 37 oC = 98.6 oF Freezing point of water 0 oC = 32 oF

Changing units within the Metric System Basic units can be changed to larger or smaller units by means of prefixes. We consider the three commonly used prefixes.

1) Kilo means 1 000 Therefore, 1 kilometer (km) = 1 000 meters (m) 1 kilo liter (kl) = 1 000 litres (l) 1 kilogram (kg) = 1 000 grams (g)

2) Centi means 100

1. (remember, 1 ngwee =

100

1 kwacha)

Therefore, 1 centimeter (cm) = 100

1 meter or 100 cm = 1m

1 centilitre (cl) = 100

1 litre or 100 cl = 1l

1 centigram (cg) = 100

1gram or 100cg = 1g

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3) Milli means 1000

1.

Therefore, 1 millimeter (mm) = 1000

1 meter or 1 000mm = 1m

1 millilitre (ml) = 1000

1.litre or 1 000ml = 1l

1 milligram (mg) = 1000

1 gram or 1 000mg = 1g

Changing Units All the prefixes involve either a multiplication or a division by a power of ten. Multiplying or dividing a number by a power of ten can be carried out just by moving the decimal point. Therefore, to change to larger or smaller units in the metric system, it is only necessary to move the decimal point. This is one of the main advantages to using the metric system. Example 14 Changing from larger units to small units involving kilo a) 0.84km = 0 8 4 0 . m = 840m +3 b) 7.125kl = 7 1 2 5l = 7 125l + 3 c) 10.5kg = 10 5 0 0 .g = 10 500g + 3 * +3 move decimal three places to the right.

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Example 15 Changing from small units to large units involving kilo a) 3 020m = 3 . 0 2 0m = 3 . 0 2 0 km = 3.02km -3 b) 550l = . 5 5 0 kl = 0.550kl -3 c) 52530g = 52. 5 3 0 kg = 52.55kg -3 Move decimal three places to the left. Note that the –3 means a movement of 3 places to the left. Example 16 Changing from small units to large units involving centi a) 245cm = 2 . 4 5 m = 2.45m -2 b) 25cl = . 2 5 l = 0.25l -2 c) 5.4cg = .0 5 4 g = 0.054g -2 * -2 move decimal two places to the left.

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Example 17 Changing from large units to small units involving centi a) 7.5m = 7 5 0 . cm = 750cm

2

b) 0.45l = 0 4 5 . cl = 45cl 2 c) 8.532g = 8 5 3 2 cg = 8.53.2cg 2 Move decimal two places to the right. Example 18 Changing from small units to large units involving milli. a) 75mm = . 7 5 m = 0.075m -3 b) 5 400 ml = 5 . 4 0 0 l = 5.4l -3 c) 530g = . 5 3 0 g = 0.53g -3 Move decimal three palaces to the left.

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Example 19 Changing from large units to small units involving milli a) 1.5m = 1 5 0 0. mm = 1 500mm b) 0.78l = 0 7 8 0. ml = 780ml c) 0.359g = 0 3 5 9 . mg = 359m Note that another commonly used metric unit is the cubic centimeter. 1 cubic centimeter (cc) = 1 millilitre (ml). Exercise 6 1) 1.5 km = …………m 2) 550g …………….kg 3) 55650g = ……………kg 4) 0.513kg …………….g 5) 3.5m …………………..cm 6) 4 560cm …………… m 7) A dairy produce board houses 135 litres of milk per day. How many kilo litres is

this per day? 8) A rectangular football ground measures 0.95km by 0.25km. Find the area of this

field in square meters (m2). 9) Chileshe’s height is 1.67 metres. Express his height in centimeters. 10) Chola’s gift weighed 2 520 grams. Express the weight of the gift in kilograms 11) Find the volume in cubic centimeters of a container that holds 3.735 litres of

water. (1cm3 = 1ml).

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Temperature When planning what clothing to wear, or what activity to engage in, we usually check the temperature first. Changing Celsius to Fahrenheit. This can be done by using the following formula.

325

9 += CF

where F is the number of oF and C = number of oC. Example 20 25 oC = ………... oF

773245

32)25(5

9

=+=

+=F

Therefore 25 oC = 77oF Example 21 75 oF = …………. oC (Round answer to nearest degree)

89.239

215

9215

1609375

325

975

==

=+==

+=

C

C

c

C

Therefore, 75 oF = 23.89 oC

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Perimeter The perimeter is the total distance round a given figure. Example 22 Find the perimeter of the figure below. 10cm 3cm 8cm 5cm 6cm 4cm Perimeter = 8cm + 10cm + 3cm + 6cm + 5cm + 4cm = 36cm Example 23 A rectangular piece of land is 50m by 30m. A farmer intends to cultivate this land and fence it at the cost of K10.50 per meter of fence wire. How much does it cost her? 50m 30m Perimeter = 50 + 30 + 30 + 50 or 2(l + b) where l = length and b = breadth =

160m. Total Cost = 10.50 (160) = K1680.00

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CHAPTER 2

ALGEBRA

2.0 ALGEBRA 2.1 Numbers Numbers were discussed in Chapter 1. Here the emphasis is on the following

rules which are important in the manipulation of equations. Rule 1. To add two numbers having like signs, add the numerical values and prefix their

common sign. Example 1

a) +15 + (+6) = + (15 + 6) = +21 b) –5 + (−25) = −(25 + 5) = −30

Rule 2

To add two numbers having unlike signs, subtract the smaller numerical value from the

larger, and prefix the sign of the number having the larger numerical value.

Example 2 a) 9)716()7(16 +=−+=−++ b) 18)523()23(5 −=−−=−=+ Rule 3 To subtract a number, change the sign and add. Example 3

11)6(5)6(5

7103)10(3

25520)5(20

−=−+−=−−=+−=−−−

=+=−−

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Note that .,,, −=−+−=−++=−−+=++ Therefore, the product of the two numbers can be stated in the following rule.

Rule 4

To multiply two numbers or to divide one number by another (note that division by zero is not allowed), multiply or divide the numerical values and prefix a + sign if the two numbers have like signs and a - sign if the two numbers have unlike signs.

Exercise 1

1) Perform the indicated operations

a) )2(6 −+ b) )15(16 −− c) )4(5 −+− d) )3(15 −−− e) )4(3 − f) )5)(3( −− g) )2(10 −÷− h) )6(90÷− 2.2 Exponents

Powers, indices all mean exponents, when ....... aaaaaa is abbreviated to 5a , a is called the base and 5 is called the exponent. An exponent is then a positive integer, written to the right and slightly above the base, which indicates the number of times the base is to appear as a factor.

Example 4

a) aaaa ...3 = b) 422.2.2.216 ==

c) 33 3.2216=

Laws of Exponents

If m and n are positive integers and 0≠a , we have nmnm aaa +=. (1)

22

Thus

62424

85353

.

.

bbbb

aaaa

====

+

+

nmn

m

aa

a −= (2)

where nm > Thus:

2353

5

aaa

a == −

If mna

nm −< 1, (3)

mnnm aa =)( (4) Thus 62323)( aaa == × nnn baab =)( (5) Thus:

n

nn

b

a

b

a

baab

=

= 333)(

(6)

Thus

3

33

b

a

b

a =

.

Zero, Negative and Fractional Exponents The extension of notion of an exponent to include any rational number (i.e, zero, positive and negative integers and common fractions) is made by the additional definitions. 0,10 ≠= aa (7)

23

nandaa

an

n 0,1 ≠=− a positive (8)

naa nn ,1

= a positive integer (9)

The exponent in na1

, for example, has nothing to do with the number of time the base is to appear as a factor. This is the nth root of the number ""a Example 5

a) 0666

6

333

31 === − e) 1046

4

6

. aaaa

a ==−

b) 8

1

2

12

33 ==− f)

9

12

12

93

4

3

a

b

b

a

b

a ==

−

−−

c) 3222

1 55

==− g) 33)81( 4 441

==

d) 416)16( 21

== h) 6

933

2

3 33

d

a

d

a =

Note that the above rules apply to products and quotients only. The rules are not for simplifying sums or differences of numbers. These rules cannot be used to simplify or reduce .33 yx+ For example

1968324381

333 954

≠+≠+

Logarithms The logarithm, base b, of a positive number N (written )log Nb is the exponent x such

that .Nbx = Example 6 a) 416log2 = b) 481log3 =

Since 1624 = Since 8134 =

24

Logs to Base 10 and logs to Base e If logs of all numbers were tabulated for every possible base, there would be endless sets of tables, therefore for convenience, two sets of tables are available, one for logs to base 10, and one for logs to base e. On your calculator, these may be found on the log key (for logs to base 10) and on the In key (for logs to base e). Rules for Logs Rule 1 For logarithm of the product of two or more positive numbers is the sum of the logarithms of the numbers.

MNNM bbb logloglog ⇔+

Rule 2 The logarithm of the quotient of two positive numbers is the logarithm of the numerator minus the logarithm of the denominator.

⇔−N

MNM bbb logloglog

Rule 3 The logarithm of a power of a positive number is the exponent of the power times the logarithm of the power.

)(log)(log MrM br

b =

Example 7 Given log2 = 0.3010 and log 3 = 0.477121; then a) )23log()43log(12log 2×=×=

079121.1

)3010.0(2477121.0

2log23log

2log3log 2

=+=

+=+=

25

b) )1023log()1043log(120log 2 ××=××=

079121.2

00000.1)3010.0(2477121.0

10log2log23log

=++=

++=

c) 100log12log100

12log12.0log −=

=

920819.0

2079181.1

−=−=

d) 31

)81.0log(81.0log3 =

[ ]

[ ] [ ]03051.0

2)477121.0(43

110log23log4

3

1

10log3log3

1

81.0log3

1

24

−=

−=−=

−=

=

Example 8 Solve the following equations a) 5.3)3log( =+x b) 0)2ln(ln2 =+− xx c) 86.1)3ln( =+x d) 83.3 1 =+xx a) 5.3)3log( =+x

to isolate x it is necessary to go from log form to index form 310 5.3 += x .

Using a calculator to evaluate ,10 5.3 we have

27766.3159

327766.3162

327766.3162

=−=+=

x

x

x

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b) 0)2ln(ln2 =+= xx

,0)2ln(ln 2 =+− xx rule 3 bring the 2 in as a power

,02

ln2

=

+x

x rule 2, log of a quotient

,12

02

==

+e

x

x going from log form to index form

22 += xx multiplying both sides by 2+x 022 =−− xx is a quadratic equation. So to solve the quadratic using completing square method or formula

.12 −= orx c) 386.1 += xe going from log form to index form using a calculator, we find

42374.686.1 =e ( 5 decimal places) then

!4237.3

342374.6

342374.6

solutionthecheck

x

x

=−===

d) 83.3 1 =+xx 83 12 =+x Taking logs of both sides, we have

4464.0

14771.0

9031.0

2

1

13log

8log

2

1

13log

8log2

3log

8log12

8log3log)12(

=

−=

−=

−=

=+

=+

x

x

x

x

27

Check the solution! Exercise 2 1) Simplify a) 65.aa b) 78.aa c) aaa ... 3

d) 5

8

a

a e)

3

6

a

a f)

6

39

1

g) 12

4

3

9

9

h)

4

32

63

.

.

bb

aa

2) Solve for n

a) 5)06.1( =n b) 8536.2)01.1( =n c) 67585.0)0325.1( =−n

d) 36.3505.0

1)05.1( =−n

3. Solve for x a) 52.0ln =x b) 2.2ln −=x c) 5.0)1ln( =+ x

d) 2

12

2

1ln

−=

−x e) 5.2ln =x f) 5)3ln( =+ x

g) 02.0ln −=x h) 85.0)2.01ln( −=− x

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2.3 Equations

Equations are statements of equality- a way of expressing a balanced relationship between two expression. These statements may be either true (for example 5 + 2 = 7) or false for (for example 13 – 6 = 9).

The equations of interest in algebra are those that involve at least one variable or

unknown on either (or both) sides of the equal sign. There are two types:

1) An Identity is an equation that is true for every permissible value of the variable. For example,

yy

andxxxx 25.04

,2882,55 =+=++=+ are identities because

regardless of which real number replaces the variable, each equation is true.

2) A Conditional equation is an equation that is not true for every possible real value replacement of the variable. For example,

246,7354,82 =−=+=+ zandyyx are conditional equations.

In this Manual, our main concern is to solve conditional equations – that is, to find the value of the variable that satisfies the equation (makes it true). A value of the variable that makes the equation true is called a solution and the collection of all solutions is called the solution set. Rules for Solving Conditional Equations 1) Any real number can be added to (or subtracted from) both sides of

an equation. In symbols this rule says; a = b if and only if a + c = b + c.

2) Both sides of an equation can be multiplied (or divided) by any

non-zero real number. In symbols, this rule says: if bac =≠ ,0 if and only if ac = bc.

Not all equations have solutions. In fact equations may have no solution at all or may have infinitely many solutions.

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Example 9 Solve for x in the following equations a) 145 =+x . b) 0)4)(3( =+− xx c) xxx 6834 +−=+ a) 145 =+x .

9

5514

=−=

x

sidesbothongsubtractinx

b) 0)4)(3( =+− xx This equation has the product of two terms (x-3) and (x+4) on the left-hand side

(LHS). A product is equal to 0, if either terms in the product is 0.

43

0403

−===+=−

xorx

xorx

Here there are two solutions. 43 −== xandx . Each solution can be confirmed

by checking that they satisfy the original equation. c) .6834 xxx +−=+ Grouping the like terms together, we have

.2

3

128

4863

=

−=−−−=−−

x

x

xxx

Quadratic Equations A quadratic equation has the general form, 02 =++ cbxax . Where a, b, c are numbers (constants). Note that .0≠a If a quadratic equation can be written as the product of two factors, then the solution can be written down immediately as in Example 9(b)

30

If the factors are not obvious or there are no easy factors the following method may be used. The solution to a quadratic equation 02 =++ cbxax is given by the formula.

a

acbbx

2

42 −±−= (2.1)

Example 10 Given the equation, .010133 2 =−+ xx Solve for x . Using formula (2.1), a = 3, b = 13 and c = -10.

3

25

6

1713

6

1713

6

1713

16912013

)3(2

)10)(3(4)13(13 2

orx

orx

bx

x

−=

+−−−=

±−=

±−=

−−±−=

Simultaneous Equations In many applications, both practical and theoretical, there will be several equations with several variables or unknowns. These are referred to generally as Simultaneous equations. The solution of a set of simultaneous equations is a set of values for the variable, which satisfy all the equations.

31

Two Equations in Two Unknowns A standard method for solving two linear equations in two unknowns is outlined in the following example. Example 11 Given the simultaneous equations

72

165

=+=+

yx

yx

a) Solve for x andy algebraically b) Solve for x and y graphically

a) Method: Eliminate x from the systems of equation by subtracting equation (1) and (2). The two equations reduce to a single equation in which the only unknown isy . Solve for y , then substitute the value of y into either the original equation and solve for x .

Step 1: )1(165 =+ yx

gsubtractiny

yx )2(

930

)72(

=+=+

3=y solving for y .

Step 2: Solve for x by substituting y =3 into either equation (1) or equation (2).

X + 5(3) = 16 substituting y =3 into equation (1)

x = 1

Step 3: Check the solution x = 1, y =3 by substituting these values into equations (1) and (2) and confirm that both equations balance.

• Substitute x = 1, y =3 into equation (1)

165 =+ yx (1) + 5(3) = 16 Substituting x = 1, y =3

32

1 + 15 = 16 16 = 16 So equation (1) balances and x = 1 and y =3 is a solution.

• Substitute x = 1 and y =3 into equation (2). 72 =+ yx

(1) + 2(3) = 7 substituting in x = 1 and y =3 1 + 6 = 7 7 = 7

So equation (2) balances and x = 1 and y =3 is a solution. Since the point (1, 3) satisfies equations (1) and (2), then this point is at the point

of intersection of the lines represented by equations (9)1 and (2) as shown in (b). y 3.5 72 =+ yx 3.2 • 3 • (1, 3) 2 1 • -6 -4 -2 0 2 4 6 7 8 10 16 x -1

165 == yx Figure 2.1 Unique Solution

33

b) The two lines are plotted in Figure 2.1. The point of intersect is the solution. The coordinates of this point are 1=x and 3=y . In this case, it is a unique solution, that is, the lines intersect at only one point. This point is on the first line so it satisfies equation (1) and also on the second line, so it satisfies equation (2).

A set of simultaneous equations may have • A unique solution as in the above example where a set of equations has one set of

values which satisfy all equations. • No solution. This occurs when a set of equations has no set of values, which

satisfy all equations. Example 12 Given the simultaneous equations

)2(5

)1(3

xy

xy

+=+=


a) xy += 3

equationstwothegsubtractinxy

20

5

−=+=

0 = -2 is impossible; therefore, there is no solution. Even from a purely

practical point of view, you can see that there is no way that both of these equations can both be true. How can y be equal to x+3 and x+5 at the same time?

Note: A false statement like 0 2−= indicates a set of equations with no

solution.

b) The two equations are plotted in Figure 2.2. The lines will never meet since they are parallel and thus will never have a point (solution) in common.

34

y xy ++ 5 xy += 3

5

3

-5 -3 0 3 x Figure 2.2 No Solution Infinitely many solutions; a set of equations has infinitely many solutions when there is an infinite number of sets of values that satisfy all equations. Example 13 Given the simultaneous equations

)2(262

)1(3

xy

xy

−=−=


35

a) When equation (2) is divided by 2, the result is exactly the same as

equation (1), since .32

2

2

6

2

2xy

xy −=⇒−=

So, equations (1) and (2) are the same! There is only one equation in two unknowns. If x is given any value, the corresponding y – value can be calculated. For example, when

etcyx

yx

yx

0,3

1,2

2,1

======

There is an infinite number of (x, y) pairs which satisfy equation s (1) and (2).

y

3 xyandxy 2623 −=−= 0 3 x

Figure 2.3 Infinitely many solutions b) Equations (1) and (2) are plotted in Figure 2.3. Note that these equations are

coincident lines, therefore every point on one line is also a point on the other line. Since a line has infinitely many points, there is infinitely many solutions or points in common.

36

Three Simultaneous equations in Three Unknowns The methods used above to solve two equations in two unknowns may be extended to three equations in three unknowns, four equations in four unknowns, etc. The strategy is to eliminate one of the variables first by adding multiples of equations to other equations in two unknowns. See the following Example 14. Example 14 Solve the equations

)3(1435

)2(22

)1(23

=++=−+=−+

zyx

zyx

zyx

The simplest approach is to add equation (3) to equation (1), and hence eliminate z, giving an equation in x and y . Then add equation (3) to equation (2), eliminating z again, giving another equation in x and y .

)1(23 =−+ zyx

)3()1()4(

)3(

16048

1435

+==++=++

yx

zyx

)3()2()5(

)3(

1656

1435

)2(22

+==+=++

=−+

yx

zyx

zyx

Equation (4) and (5) are the usual two equations in two unknowns, so solve for x and y . Then solve for z later.

)5(1656

)4(1648

=+=+

yx

yx

Dividing equation (4) by 4, we have 42 =+ yx as our new equation (4) then subtract (6) and (5). Equation (6) is obtained by multiplying equation (4) by 5.

37

xforsolvingx

x

yx

yx

1

)5(

44

1656

)6(20510

==

=+=+

So, ,1=x substitute 1=x into equation (5), (6) or (4) to solve for y . Substituting 1=x into equation (4) gives .24)1(2 =→=+ yy Finally find z by substituting 1=x , 2=y into any of the equation (1), (2) or (3). For example substituting into (2),

32)2(21 =→=−+ zz , therefore, the values which satisfy all three equations (1), (2) and (3) are .32,1 === zandyx Exercise 3 Solve the following simultaneous Equations. 1. xy = 6. 2=+− zyx xy −= 4 122 −=−+ zyx 922 =++− zyx 2. 13=+ yx 7. 334 −=− qp 3=− yx 5.175.22 =+ pq 3. 1623 =+ yx 8. 143 21 −=+ pp

12

19534

321

32

=++=−=−ppp

ppyx

4. 1853.9725 =+=− yxyx 655.022133 =−=+ yxyx 5. 452.106238 −=−=+ yxqp 8985 =+ qp yx 5.338 −= The method of simultaneous equations and Quadratic equations are now applied to determine equilibrium conditions in various markets; for example the goods, labour and money market. Example 15

38

The demand and supply function for a good are given as Demand function : )1(75.0200 qP −= Supply function : )2(75.020 qP += Calculate the equilibrium price and quantity algebraically and graphically. Market equilibrium occurs when .sdsd PPandQQ == Since the functions are written in

the form )(qfP = with P as the only variable on the LHS of each equation, it is easier to equate prices, reducing the system to an equation inq only, hence, solve for q :

quantitymequilibriuq

q

qq

PP sd

120

1805.1

75.02075.0200

==

+=−=

Now solve for the equilibrium price by substituting q =120 into either equation (1) or (2).

pricemequilibriuP

equationinqngsubstitutiP

110

)1(120)120(75.0200

==−=

Figure 2.4 illustrates Market Equilibrium at point oE with equilibrium quantity, 120, and

equilibrium price K110. The consumer pays K110 for the good which is also the price that the producer receives for the good. There are no taxes! P 200 p = 20 + 0.75q 110 • oE (Goods market equilibrium)

20 p = 200 – 0.75q

-27 0 120 267 q

39

Figure 2.4 Goods Market Equilibrium Example 16 Chipwende made twice Musonda’s salary last month. If their total earnings were K3 172 500, how much did each make? Notice that we asked for two things here, Musonda’s salary and Chipwende’s salary. Let x represent Musonda’s salary. Since Chipwende’s salary is twice Musonda’s we have x2 representing Chipwende’s salary.

Musonda’s plus Chipwende’s salary is K3 172 500 x + 2x = K3 172 500

5000571

50017233

50017232

Kx

x

xx

==

=+

Musonda’s salary is K1 057 500; Chipwende’s salary is twice that or K2 115 000. Example 17 The demand function for a monopolist is given by the equation .5.050 qP −= a) Write down the equation of the total revenue function. b) Graph the total revenue function for 1000 ≤≤ q c) Estimate the value of q at which total revenue is a maximum and estimate the

value of maximum total revenue. a) Since qP 5.050−= and total revenue (TR) = Price × quantity = QP × ,

then 25.050)5.050( qqqqTR −=−=

40

b) Calculate a table of values for 300 ≤≤ q such as those in Table 2.0 the graph is plotted in Figure 2.5.

Table 2.0 points for 25.050 qqTR −=

TR 1250 Maximum Revenue 500 25.050 qqTR −= 450 0 50 100 q

Q TR 0 10 20 30 40 50 60 70 80 90 100 110

0 450 800 1050 1200 1250 1200 1050 800 450 0 -550

41

c) A property of quadratic functions is that the turning point (in this case a

maximum) lies halfway between the roots (solutions) of the quadratic function. The roots of the TR functions are

1000

)100(0

10002

1500

5.050

2

2

2

==−=

−=

−=

−=

qorq

qq

qq

qq

qqTR

The roots are illustrated graphically as the points where the TR function intersect

the x-axis. The turning point occurs halfway between these points, that is, at .50=p Substituting q = 50 into the function and calculate maximum total

revenue as

1250

12502500

)50(5.0)50(50

5.0502

2

=−=

−=−= qqTR

42

EXAMINATION QUESTIONS WITH ANSWERS

Multiple Choice Questions

1) Mulenga made twice Chitalu’s salary last month. If there total earnings were

K1,903,500, how much did Mulenga make? a) K643 000 b) K1 269 000 c) K951 750

c) K634 500

(Natech, 1.2 Mathematics & Statistics, November/December 2000)

2) What is the point of intersection, if any, of the lines represented by the following

pair of equations?

06

023

=−−=−−

xy

xy

a) (2,8) b) (0,0) c) (2,6)

d) none (Natech, 1.2 Mathematics & Statistics, December 1999 (rescheduled)).

3) Evaluate 3 )125/27(

a) 125

27 b)

5

3 c) 0.465

d) 0.216

(Natech, 1.2 Mathematics & Statistics, June 2001)

4) The expression

4

4

3

y equals

a) y b) 3y c) 4

19

y

d) 16

3

y (Natech, 1.2 Mathematics & Statistics, December 2001)

43

5) The price of a product including local sales tax at 19% is K900.68. The price is

reduced by 20%. The new price before sales tax is added, is a) K605.50 b) K720.54 c) K756.87

e) K760.20 (Natech, 1.2 Mathematics & Statistics, June 2005)

6) If ,0)43)(32( =+− xx then the value of x is

a) 4

3

3

2 −or b)

3

4

3

3or

− c)

4

3

3

2or

−

d) 3

4

2

3 −or

(Natech, 1.2 Mathematics & Statistics, June 2005) 7) If 642 =x then x is equal to

a) 6 b) 64 c) 32

d) 6

1

8) If 0852 =++ xx then x is equal to

a) 2

135±− b)

2

135± c)

2

375±−

d) 2

375±

9) 1

2

−−

x

xx is equal to

a) x b) 12 −x c) 21 x− d) 2x 10) One of the following is an irrational number. Which one is it?

a) −3 b) 8− c) 3

d) 16

44

SECTION B

Q1. a) Given the equation .532

52

3 =++− q

q

q Find the value of .q

b) The demand and supply functions for golf lessons at Chinama Golf club

are: Demand function : P = 200 – 5Q Supply function : P = 92 + 4Q Calculate the equilibrium price and quantity algebraically.

(Natech, 1.2 Mathematics & Statistics, Nov/Dec 2000)

c) Solve the 3 x 3 simultaneous equation below algebraically

113

4322

53

=−+=++

=+−

zyx

zyx

zyx

(Natech, 1.2 Mathematics & Statistics, December 2003) Q2. a) The cost equation of a company producing agricultural fertilizers in

Zambia is as follows: 622 ++= xxC where C represent total cost (K’million) and x represents the number of

tones in (hundreds), of fertilizer produced.

i) Construct a table of costs for x values ranging from 1 to 8 inclusive.

ii) Using the result in (i) above, plot the cost on output to produce a

scatter graph and interpret it.

iii) Calculate algebraically the level of output which would result in a total of K25 million rounded to the nearest hundred tones.

(Natech, 1.2 Mathematics & Statistics, December 2004)

45

CHAPTER 3

STATISTICS

3.1 Introduction This Chapter provides the students with a general awareness and understanding of the

collection and presentation of numerical information, including frequency distributions. At the end of the Chapter the student will have a basic ability in the analysis and interpretation of statistical data.

3.2 Sources of Statistics

In many applications of statistics, businesses use internal data – that is data arising from bookkeeping practices, standard operating business procedures, or planned experiments by research divisions with the company. Examples are profit and loss statements, employee salary information, production data and economic forecasts. The data sourced from outside the firm is called external data. Internal data may be of two types. Primary data and Secondary data. By primary data, we obtain data from the organization that originally collected them. An example is the population data collected by and made available from the Central Statistical Office (CSO) Zambia. Secondary data come from a source other than the one that originally collected them. Users of secondary data cannot have a clear understanding of the background as the original investigator, and so may be unaware of the limitations of the data at hand. There are many excellent sources of published (Primary and secondary) data compiled by the state, by business and economic associations, and by commercial sources (periodicals). Some examples are: CSO Journal, Bank of Zambia Journal, A – Z Business Journal etc.

3.3 Descriptive Statistics

When a survey or an experiment has produced a body of data, the original state of data will not generally convey much information about the characteristics of interest. Typically, they will be too many reservations to give on insight into the nature of data. It is necessary to organize and reduce the data into such meaningful forms as graphs and charts or such numerical quantities as averages, totals and percentages. The resulting statistical summaries of the data can be used as a framework for data analysis and interpretation. There are basically two methods of describing data. The graphical method and numerical method. This Chapter focuses on both of these methods.

46

Population

We use the word population to describe possible measurements of the particular characteristic under consideration. A population can be finite (small or large) or infinite (in the sense that it is particularly impossible to count its size). For example, the number of students in a class (small), the yearly output of a certain type of soft drinks (large), the number of particles of sand in the world (infinite).

Sample A sample is a part of a population in which the population characteristic is studied so that inference may be made from the sample to study about the entire population.

Frequency In any population two or more members may have the same value. For example, the height (to the nearest cm) of several members of a school may be the same. The number of members with the same value is known as the frequency and is generally denoted by f.

3.3.1 Frequency Distributions Any data not arranged in a given order is called raw data otherwise it is an array of data.

Example 1

The following data record the number of children under the age working in a certain company 1 1 3 2 0 8 8 6 7 7 8 6 8 8 1 1 0 0 2 9 4 4

Construct an array and also a frequency distribution. The data array is as follows: The data is arranged in increasing order. 0 0 0 1 1 1 1 2 2 3 4 4 6 6 7 7 8 8 8 8 8 9 To construct a frequency distribution start with a tally chart.

47

Tally Chart

Data Value Tally Marks Total 0 1 2 3 4 6 7 8 9

III

IIII

II I

II

II

II

IIII I

3 4 2 1 2 2 2 5 1

Frequency Distribution Table

No. of Children Under Age

No. of Children working (f)

0 1 2 3 4 6 7 8 9

3 4 2 1 2 2 2 5 1

48

When the number of distinct data values in a set of raw data is large (20 or more, say), a simple frequency distribution is not appropriate, since there will be too much information, not easily assimilated. In this type of situation, a grouped frequency distribution is used. An example of a grouped distribution is given below.

Salary Scale ‘K’000,000’ No. of Workers 5 and < 10 10 and < 15 15 and < 20 20 and < 25

5 6 8 3

Table 3.1 Frequency Grouped Distribution for Salary Scale. 3.3.2 Cumulative Frequency Distributions A cumulative frequency distribution describes the number of items that have

values either above or below a particular level. Cumulative frequency distributions come in two different forms:

i) “less than” distributions ii) “more than” distributions.

Example 2

From Table 3.1, construct i) “less than” distribution, ii) “more than” distribution

i)

Salary scale ‘K’000,000’

No. of workers

Salary scale No. of workers

5 and < 10 10 and < 15 15 and < 20 20 and < 25

5 6 8 3

< 10 < 15 < 20 < 25

5 11 19 22

Here, a set of items values is listed (normally the class “upper boundaries”), with each one showing the number of items in the distribution having values less than this item value.

49

ii) Here, a set of item values is listed (normally the class “lower boundaries”) with each one showing the number of items in the distribution having values greater than this item value. See the table below.

Salary scale ‘K’000,000’

No. of workers

Salary scale No. of workers

5 and < 10 10 and < 15 15 and < 20 20 and < 25

5 6 8 3

> 10 > 15 > 20 > 25

22 17 11 3

3.3.3 Results Presentation

One of the most effective ways of presenting information, particularly numerical information, is to construct a chart or a graph. The choice depends on the type of data. A set of data is discrete if we only need to make a count, like the number of customers entering a shop. A set of data is continuous if measurement is made on a continuous scale, such as time, weight etc. For discrete data, we use bar charts, and pie charts while for continuous data, we use a histogram. However, a disadvantage of graphs may be that values may not be read accurately. But graphs are not meant to show up quantitative details as tables do, graphs are meant to show effects. Example 3 The following information shows the total turnover of Mukulumpe plc, analyzed by geographical segment. Sales K’Billion

20 x 3 K’Billion 20 x 4

K’Billion 20 x 5

West Africa East Africa Southern Africa Central Africa

82 41 20.5 61.5 205

78 31.2 18 4.2 131.4

65 22 17 4.5 108.5

Show this as a bar chart.

50

In a simple bar chart, the number observed (counts) whether by ‘geographical segment”, or ‘years” or some other category can be represented as vertical bars. The height of each bar is drawn in proportion to the number (amounts) by a vertical ruler scale. Figure 3.1 shows the sales of each geographical segment. Sales 200 (K’billion) 150 100 50 20x3 20x4 20x5 year Figure 3.1 Sales of Mukulumpe Plc. Component Bar Charts There are used to show the breakdown of a total into components. The bars of the simple bar chart are subdivided to show component parts. They are two kinds of component bar charts. i) Component bar chart (actuals). In these charts the overall heights of the bars and the individual components heights represent actual figures. ii) Percentage component bar chart In these charts the individual component lengths represent the percentage each component forms of the overall bar total. Note that the series of such bars will all be the same total height, i.,e. 100 per cent.

51

Example 4 Construct i) component bar chart, ii) percentage component bar chart, For the data in Example 3 i) Component Bar Chart

Sales 200 (K’billion) 150 100 50 20x3 20x4 20x5 year

West Africa East Africa Southern Africa Central Africa

52

ii) 100 80 Sales (%) 40 20 20x3 20x4 20x5 year West Africa East Africa Southern Africa Central Africa Multiple Bar Charts These are similar to component bar charts but here the components are shown side by side. As this does not give an immediate impression of the change in totals, they should be used where we want to demonstrate the change in size of the components.

53

Example 5 Show the above data as a multiple bar chart. 150 Sales (K’billion) 100 50 20x3 20x4 20x5 years West Africa East Africa Southern Africa Central Africa Histogram This is a bar chart. It is appropriate where there is need to show grouped data which is continuous. There are no gaps between the bars. The total area of each bar represents the frequency of the event.

54

Example 6 The marks obtained by students in a NATech paper were as follows:

Show this as a histogram 20

No of students 16 12 8 4 0 25 - 29 30 – 34 35 – 39 40 - 44 45 - 49 Marks Pie Charts A pie chart is a circle or ‘pie’, divided radically into sectors which represent component parts of the total. The 360o at the center of the circle are divided in proportion to the data thus giving sectors with areas proportional to the values of the components parts. Pie charts can be used to show changes in components where the number of components is too great for a bar chart, though a pie chart with more than seven or eight components would become too clouded for ready interpretation.

Percentage (%) No. of Students 25 – 29 30 – 34 35 – 39 40 – 44 45 - 49

10 15 12 20 3

55

Example 7 For the data in Example 3, for the 20x3, construct a pie chart.

Central West Africa Africa Southern East Africa Africa Figure 3.2. A Pie Chart showing turnover of Mukulumpe Plc. Calculations

West Africa oo 1443605.615.204182

82 =×+++

East Africa oo 723605.615.204182

41 =×+++

Southern Africa oo 363605.615.204182

5.20 =×+++

Central Africa oo 1083605.615.204182

5.61 =×+++

56

Exercise 1

1. Obtain a number of charts and graphs used to describe quantitative data. This is the data produced by ordinal, interval or ratio scales. Sources include, for example, newspaper cuttings, magazines or textbooks. Classify each as being discrete or continuous data and state reasons why you consider them to be informative or misleading. 2. The data below give the scores obtained in an aptitude test by a group of 40 applicants for a particular post in a company

8 9 9 10 11 9 10 8 9 11 12 9 12 6 8 9 8 10 9 8 12 8 9 11 9 12 7 11 9 8 9 8 10 9 8 10 9 8 9 10 construct a frequency distribution from this information.

3. A survey of 55 retail outlets in the Kitwe area gave the following distribution of mango prices.

Price (Kwacha/g)

250 200 150 100 275 185

Number of stores

2 2 7 23 15 6

Construct a bar chart for the given distribution. 4. From sales ledger of a small company, the age of a sample of 100 debts are shown in the distribution below. Construct a histogram of this distribution.

Age of debt(days) 1-10 11-20 21-30 31-40 41-50 51-60 No. of accounts 24 28 22 16 6 4

5. The age distribution of a random sample of 500 people in Ndola is shown below. Construct a histogram from this table.

Age (Years) Under 2 2- 5- 10- 30- Number of people 98 107 170 75 50

57

6. Draw:

i) ‘less than’ distribution ii) ‘greater than’ distribution given the distribution of bonus payments made to 150 employees in a company shown below. Monthly bonus (Kwacha) 0- 10- 20- 30- 40- 50- No. of employees 6 44 36 30 8 6

7. Draw a pie chart to illustrate the expenditure of a large company on a number of advertising methods.

Method of Advertising

Radio

Newspaper

Competitions

Others

Expenditure during 2003 (X ‘K1 000 000)

30

50 20 10

8. Use a bar chart to illustrate the number of workers employed in four factories as tabulated below.

Factory A B C D No. of employees 130 310 260 160

9. Draw a component bar chart of the data given below, for factories, X, Y, Z and W.

No. of Employees

58

X Y Z W Unskilled 30 40 50 40 Semi-skilled 50 110 100 110 Skilled 70 180 130 30

10. Draw a multiple bar chart to illustrate the performances of three companies over a four year period.

Output (X K1, 000,000) 2000 2001 2002 2003 Company X 400 380 365 350 Company Y 285 340 355 340 Company Z 180 200 220 230

3.4 Measure of Central Tendency

This Section describes the most commonly used averages, the arithmetic mean, median and mode. 3.4.1 Measure of central tendency for ungrouped data

The mean is the most used measure of location, with the median and the mode being used for specific (special case) applications. The arithmetic mean is the name given to the ‘simple average’ that most people calculate.

Arithmetic mean = Total value of items Total number of items It is easy to understand and a very effective way of communicating an answer. It

does not apply to categorical data and its interpretation can be difficult when used with ordinal data, but it is often justified for practical reasons. Mathematically it is very useful for further calculations. All the data is included in its calculation. Its disadvantage is that it is easily affected by very high or very low value and cannot be measured or checked graphically. Further more, it may not correspond to any actual value in the distribution itself.

59

Example 8

Consider the following prices of a packet of milk from 12 different retail outlets.

K280 K275 K290 K310 K185 K195 K200 K225 K175 K200 K190 K195 What is the mean price?

The mean of a set of values is their total divided by the number of items. In our example, the mean is 280 + 275 + 290 + 310 + 185 + 195 + 200 + 225 + 175 + 200 + 190 + 185 /12

833.225

12

2710

=

=

We usually employ the symbol x (pronounced, ‘x bar’) to represent the mean of a sample. A general formula for the mean of a sample of n items is therefore

n

xxxxx n...321 +++=

The short form is ∑∑= wheren

xx is the Greek symbol for capital “S” for sum

and ∑ x is simply translated as “add up all the values of x under consideration”.

The mode is the number which appears more times than any number in a given set. It is quoted as a typical value of the variable. The mode can be of great assistance in manufacturing and production. For example production of shoes, clothes, cars, etc. It is not affected by very low or very high values and it is an actual value of the distribution. However, it is not clearly defined when no two items have the same value, or two or more items have the same highest frequency.

60

Example 9

In Example 8, find the mode of the given data set. The mode by definition, is the most ‘common’ number – the value which occurs most often in the data set. There are two numbers which appear more times than any other numbers hence, there are two modes K200 and K195. Note that a distribution can have one mode, two modes (bimodal), three modes etc. The mode is used to describe the size of shoes, clothes or the most popular make of a car, television etc. The median The median is not as widely used as the mean or mode, but has particular applications. For example the use of the IQ scale with the average figure of 100. Also in the real world we must often deal with data, like salary distribution where relatively small numbers of extreme values can distort the arithmetic mean, the median makes it a typical value. It is easily obtained and not affected by high or low values. However, if the number of items is small or the items are not evenly spread, the median loses a lot of its significance. Example 10 Calculate the median for the following data: 310, 290, 280, 275, 225, 195, 200, 200, 190, 185, 175. Recall that there are n items of data in our sample. The position of median is therefore

the thn

2

)1( + from smallest (or largest) when n is odd. Placing out data in increasing

order, we have 310, 290, 280, 275, 225, 200, 200, 195, 190, 185, 175.

The position of the median is 62

)111( =+, hence the median is 200.

61

Example 11

In Example 10, suppose the number 280 is dropped. Find the median of the new data set. Arranging the data, in increasing order, we have 175, 185, 190, 200, 200, 225, 275, 280, 290, 310

10=n is an even number. Hence then position 2

1+n is not a whole number and

so the median is taken as the average of the two middle values. So the median is

the 5.52

)110( =+ from the largest item which is the average of the 5th and 6th

from largest values.

Median is = .2002

200200 =+

3.4.2 Measure Of Central Tendency For Grouped Data

For a grouped frequency distribution, the mean, mode, and median cannot be determined exactly and so must be estimated. This will be illustrated in the following example. Example 12 Given the distribution of ages in a certain firm as shown in the table below: calculate i) mean ii) median iii) mode

Age Number of Employees 15 to 19 20 to 24 25 to 29 30 to 34 35 to 39

3 15 30 45 8

i) In a frequency distribution, the mean xwheref

fxx∑∑= is the middle point of

the class interval.

62

We construct the following table for calculation of the mean

Age No. of Employees (f) Mid-class point )(x

fx

15-19 20-24 25-29 30-34 35-39

3 15 30 45 8

17 22 27 32 37

51 330 810 1440 296

Totals 101 2927

Here ∑ ∑ == 1012927 fandfx

Therefore mean number of ages

98.28

101

2927

=

=

=∑∑

f

fxx

ii) We use the median formula given by

Median = mm

mm C

f

FNL

−+ −15.0

Where =mL lower boundary of the median class interval

=−1mF cumulative frequency of class below the median class interval

=mf Actual frequency in the median class interval

=mC Median class width

63

In our on going example, we need a column of cumulative frequency (F)

Age (years) f F 15-19 20-24 25-29 30-34 35-39

3 15 30 45 8

3 18 48 93 101

Calculate .5N = .5(101) = 50.5. This gives us the position of the median. Therefore the median class interval is 30 to 34. This interval contains the 50.5 th observation. The median can now be estimated using the formula given below.

2222.30

)4(45

485.5030

5.0,

4;45;48;30

1

=

−+=

−+=

====

−m

m

mm

mmmm

Cf

FNLMedianThus

CfFL

i.e median = 30.22 years (two decimal places). iii) An estimate of the mode for a grouped frequency distribution can be obtained using the formula

Mode = Gcab

abL

−−−+

2

Where: L = the lower boundary of modal class interval G = modal call interval width a = frequency of class immediately below modal class interval b = frequency of modal class interval c = frequency of class immediately above modal class.

64

The modal class interval is 30 to 34 it has the highest frequency of 45. Therefore, L = 30, G = 4, a = 30, b = 45, c = 8.

Thus, Mode = Gcab

cbL

−−−+

2

1538.31

)4(830)45(2

304530

=

−−−+=

i.e mode = 31.15 (two decimal places) Example 13 From Example 11, find i) the median ii) the mode, graphically

i) The median

A percentage cumulative frequency curve is drawn and the value of the variable that corresponds to the 50% point (i.e half way along the distribution) is read off and gives the median estimate. The method is shown in the worked example.

Step 1

Age (years) No. of Employees 15-19 20-24 25-29 30-34 35-39

3 15 30 45 8

Table 1. Number of Employees

65

Step 2 Upper boundary F F%

19 24 29 34 39

3 18 48 93 101

3.0 17.8 47.5 92.1 100

Table 2.0 Cumulative frequency of Employees Percentage • Number of 100 Employees • 80 60 50% point 40 • • Median estimate = 30 20 19 24 29 34 39 Age upper boundary Fig 1 Cumulative Frequency Curve ( orgive) of Example 11. The points to remember are: i) Form a cumulative (percentage) frequency distribution. ii) Draw up a cumulative frequency curve by plotting class upper boundary, against cumulative percentage frequency and join the points with a smooth curve. iii) Read off the 50% point to give the median.

66

i) We construct three histogram bars, representing the class with the highest

frequency and the ones on either side of it, we then draw two lines as shown in Figure 1.0. The mode is the value of x corresponding to the intersection of the lines.

x Figure 1.0 The histogram bars in Figure 2.0 represents the following three classes and frequencies. 25 to 29 30

30 to 34 45

35 to 39 8

67

Number of Employees

50 40 30 20 0 25 30 35 39 Age (years) Mode estimate = 31 Figure 2.0 Weighted Averages

Another common problem arises where the means of a number of groups need to be combined to form a grand mean. For example, suppose a company has three outlets and their average sales as as follows, X, K 900 000 per sales from 25 sales, Y, K112 000 per sales from 40 sales and Z, K100 000 per sale from 30 sales. Find the average value per sale overall.

Weighted mean = ∑∑

w

xw where w is the weight assigned to each average,.

For the data given above

95.315578

95

29980000

=

=mean

i.e. average value of all sales = K315 578.95 Relationship Between Measures

68

The relative position of the mean, median and mode will tell us something about the distribution of the data, as shown in the figure below.

Mode median mean Negative skew Mode Median Mean Symmetrical mode median mean Positive skew If the distribution is perfectly symmetric, all three measure will coincide. Skewed Distribution.

69

The three measures will now spread out: Mode - Correspond to the highest point Mean - affected by extreme values, lies down the tail of the distribution

Median - dividing the area under the curve in two, lies between the mean and the mode.

Roughly: mode = mean – 3 (mean – median) Exercise 2

1. Find the arithmetic mean of the following data sets. a) 560, 520, 540, 720, 650, 470, 680, 600 b) 8.8, 9.3, 9.8, 7.9, 10.2, 8.5 c) 12.9, 13.4, 13.8, 14.3, 16.9, 17.1, 13.8, d) 6, 25, -8, 14, -22, 33 e) -3, -4, -10, -18, -9 2. Find the mean of the following frequency distributions. a)

x 20.5 12.5 35.5 f 8 10 14

b)

x 2 3 4 5 6 f 5 6 12 30 32

c)

x 35-40 -45 -50 -55 f 8 20 25 34

70

d) x 0-9 10-19 20-29 30-39 f 2 5 20 25

e)

x 20-30 -40 -50 -60 -70 -80 -90 f 4 60 75 12 15 10 3

3. The mean salaries of 150, 200 and 250 men employed by three different firms are K300 000, K250 000 and K450 000 per month respectively. Calculate the mean salary per month of all the men. 4. The maize yields in a particular region over the past 10 years are (millions of tons): 2.3, 1.5, 1.2, 1.6, 1.7, 2.8, 1.4, 1.2, 1.3, 1.8.

Estimate:

i) The average ii) The median iii) The mode.

5. Using the graphical method, estimate:

i) The median ii) The mode of the distribution given below.,

x 0 1 2 3 4 f 25 28 6 3 3

6. The total price of units ordered from a warehouse of a certain commodity is shown in the distribution below.

Cost of units ordered per day (Kwacha) No. of days 0 and under 50 50 and under 100 100 and under 150 150 and under 200 200 and under 250 250 and under 300

3 8 9 17 10 9

a) Compute the mean b) Using both the graphical and formula method, estimate:

i) the median ii) the mode

71

7. Estimate: a) The arithmetic mean, b) The median, and c) The mode for both of the following frequency distributions.

i) x 0-2 2-4 4-6 6-8 8-10 f 0-2 2-4 4-6 6-8 8-10 ii) x 10-15 15-20 20-30 30-50 50-60 f 5 12 14 4 2

8. A survey of workers in a particular industrial sector produced the following table.

Income (Weekly ‘000’) Number Under K100 K100 but under K150 K150 but under K200 K200 but under K250 K250 and over

180 235 210 150 100

Compute the mean, median and mode.

9. The number of new orders received by a company over the past 30 working days were recoded as follows:

4 0 2 1 2 3 5 3 1 1 4 5 5 6 3 2 6 4 4 0 4 3 3 2 5 3 2 4 5 6 Determine the mean, median and mode.

72

10. Which measure of central tendency would most effectively describe? a) The weight of a person?

b) The most popular make of television set? c) Earnings of part time workers in Zambia?

d) Cost of typical food item at a market? e) Holiday destinations? f) Learning days lost through class boycotts?

3.5 Measure of Dispersion

Having obtained a measure of location or position of a distribution, we need to know how the data is spread about that point. Information about the spread can be given by one or more measures of dispersion. The Range This is the simplest measure of dispersion available in statistical analysis. It uses only two extreme values. The range is defined as the difference between the maximum and minimum values of a given data set. Its advantage lies in its simplicity and its independence of the measure of position. However, it is distorted by the extreme values and tells us nothing between the maximum and minimum values. Example 13 Find the range for the given data set. 1, 3, 4, 10. The range is 10 – 1 = 9 The Quartile Deviation. The median divides the area under the frequency curve in two. The quartiles divide the area in four.

73

Frequency LQ Median UQ

The position of the lower quartile LQ is given by 4

)1( +n. That of the upper quartile UQ

is given by ).1(4

3 +n

The interquartile range is the distance between the quartiles = LU QQ − i.e the range of

the middle 50% of the distribution. The quartile deviation or semi-quartile range is half of the interquartile range.

)(2

1LU QQQD −=

The advantages of the quartiles is that they are easy to understand and are not affected by extreme values. However, they do not cover the whole of the distribution. They give no indication of how many items are dispersed between LQ and UQ .

Example 14 Calculate the first and third quartiles for the following data set: 44, 76, 49, 52, 52, 48, 51. We first arrange the data set in ascending order 44, 48, 49, 51, 52, 52, 76.

74

1Q is the value of the =+th

4

17 2nd item, which is 48.

3Q is the value of the =+ th)17(4

36th item, which is 52.

Notice that if there had been, say, more items in the set, the values of (n+1)/4 and 3(n+1)/4 would not have been whole numbers, which would have necessitated some sort of interpolation formula to obtain (untypical) values. This is beyond this manual. Example 15 Compute the interquartile range and the quartile deviation in Example 14. Interquartile range = 4485213 =−=− QQ

The quartile Deviation = .22

4

213 ==− QQ

Example 16 Compute the median and quartile deviation for the following distribution. x f 3200 – 4000 2 4 000 – 4800 3 4800 – 5600 4 5600 – 6400 8 6400 – 7200 3

x f F(Cumulative frequency) 3200 – 4000 2 2 4 000 – 4800 3 5 4800 – 5600 4 9 5600 – 6400 8 17 6400 – 7200 3 20

75

Using the formula

Median = mm

mm C

f

FNL

)5(. 1−−+

5700

)800(8

)810(5600

8,9,5600

10)20(5.5.

1

=

−+=

===

==

−

Median

fFL

N

mmm

For ;1Q

Position of first quartile = 5)20(4

1

4

1 ==N

In the same formula for the media, we replace

111,

4

15., −− iQmQm FbyFNbyNLbyL and 1Qm fbyf . Therefore, we

have

4600

)800(4

)25(400

)25(.

800,4,2,4000

1

11

1

1

1

1

1

1111

=

−==

−+=

====

−

−

QHence

CQf

FNLQ

CfFL

Q

QQ

QQQQ

76

For 3Q :

Product of third quartile 15)20(4

3

4

3 === N

.6000,

)800(8

)915(5600

)75(.

800,8,9,5600

3

3

13

1

3

3

3

3

3133

=

−+=

−+=

====

−

−

QHence

Q

Cf

FNLQ

CfFL

QQ

Q

Q

QQQQ

Therefore, Quartile deviation

700

)46006000(2

1

)(2

113

=

−=

−= QQ

3.5.1 The Mean Deviation This measure is an average of the deviation of all items from the arithmetic mean. To consider the deviation of an iten from the mean, only the size of the figure is important, the sign is not taken into account i.,e. the modulus is taken. If this is

not done then the sum of the deviation i.e. )(∑ − xx will equal zero.

The following formulas are used depending on the kind of data set given.

Mean deviation = ∑−

n

xx for ungrouped data

∑

∑−

=f

xxf for grouped data.

77

Example 17

A greengrocer owns 10 shops in various parts of a certain town. The distances from the wholesale fruit and vegetables market are 8, 13, 15, 20, 27, 33, 46, 59 , 65 and 72 kilometers.

a) Find the mean deviation of kilometers from the mean b) Find also the mean deviation from the median a)

x xx −

8 13 15 20 27 33 46 59 65 72

27.8 22.8 20.8 15.8 8.8 2.8 10.2 23.2 29.2 36.2

∑ = 358x 8.181=−∑ xx

8.35=x

mean deviation = ∑−

n

xx

18.18

10

8.181

=

=

b) Since n is even, the median is given by the average of the two middle values.

Median = 302

3327 =+

78

x medianx −

8 13 15 20 27 33 46 59 65 72

+22 17 15 10 3 3 16 29 35 42

∑ = 358x 192=−∑ medianx

Median deviation = 2.1910

192 =

Example 18 Given the following data, compute the mean deviation

Weekly Wage K’000 000

Number of Employees

31 and under 36 36 and under 41 41 and under 46 46 and under 51 51 and under 56

7 9 13 19 26

x f xf xx − xxf −

33.5 38.5 43.5 48.5 53.5

7 9 13 19 26

234.5 346.5 565.5 921.5 1391

-13.24 -8.24 -3.24 1.76 6.76

92.68 74.16 42.12 33.44 175.76

74 3459 418.16

79

74.4674

3459===∑∑

f

xfx

Mean deviation = ∑

∑ −

f

xxf

65.5

74

16.418

≅

=

3.5.2 The Standard Deviation The standard deviation is the most widely used measure of dispersion, since it is directly related to the mean. If you chose the mean as the most appropriate measure of central location, then the standard deviation would be the natural choice for a measure of dispersion. The standard deviation measures the differences from the mean; a larger value indicates large variation. The standard deviation is in the same units as the actual observations. For example if the observations are in cm, even the standard deviation will be in cm. To calculate the standard deviation, we follow the following steps.

1) compute the mean x

2) Calculate the differences from the mean )( xx −

3) Square the differences 2)( xx −

4) Sum the squared difference i.e. 2)(∑ − xx

5) Take the average of the sum of the squared differences in (4) to find the

variance i.e.

1

)( 22

−−

= ∑n

xxS for a sample and

N

xx 22 )(∑ −

=σ for a population.

6) Square root of the variance gives the standard deviation

1

)( 2

−−

= ∑n

xxS for a sample and

N

xx∑ −=

2)(σ for a population.

80

Example 19 For the following sample of 7 observations, find the standard deviation. 4, 5, 10, 13, 9, 7 and 8 The calculations are shown in the table below.

x

xx − 2)( xx −

4 5 10 13 9 7 8

-4 -3 2 5 1 -1 0

16 9 4 25 1 1 0

Total 56 0 56

).3(055.3

6

56

1

)(

87

56,7,56

2

placesdecimal

n

xxS

n

xxthereforenx

=

=−−

=

=====

∑

∑∑

Its weakness lies in its calculation and understanding which is more difficult than for other measures. Moreover by squaring, it gives more than proportional weight to extreme values. Other uses of the standard deviation considered in this manual is in the measure of relative standing.

81

1. Coefficient of Variation Coefficient of variation calculates the standard deviation from a set of observation as a percentage of the arithmetic mean.

100×=x

SCv

The higher the coefficient of variation, the more variability there is in the set of observations. 2. Skewness Skewness in a set of data relates to the shape of the histogram which could be drawn from the data.

Pearson coefficient of Skewness = σ

)(3

medianmedian −

Positively skewed if 0>sk Negatively skewed if 0<sk Symmetric distribution 0=sk Example 20

The distribution shown below is the output of the factories of Quality Clothing Plc, for the month of July 2005.

Monthly Output men’s Suits Number of Factories 25 and under 30 30 and under 35 35 and under 40 40 and under 45 45 and under 50 50 and under 55

15 30 30 20 10 15

Calculate the mean and standard deviation

82

Class Interval f x xf fx2 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 45 – 50

15 30 30 30 10 15

27.5 32.5 37.5 42.5 47.5 52.5

412.5 975 1125 850 475 787.5

11343.75 31687.50 42187.50 36125.00 22562.50 41343.75

185250,4625120 2 === ∑ ∑∑ fxxff

This is grouped data and we use the following formulas.

( )

( )

( )

( )

).2(67.7

119120

4625185250

tan

54.38120

4625

2

2

2

2

2

placesdecimal

f

f

xffx

deviationdardS

f

f

xffx

Variance

placesdecimaltwof

xfmean

=

−=

−==

−=

====

∑

∑∑∑

∑

∑∑∑

∑∑

σ

83

Exercise 3 1. Explain the meaning of standard deviation to someone who doesn’t know anything about statistics. 2. The number of new orders received by a company over the past 30 working days were recorded as follows 4 0 2 1 2 3 5 3 1 1 4 5 5 6 3 2 6 4 4 0 4 3 3 2 5 3 2 4 5 6 Determine the range, quartile deviation and standard deviation. 3. For the following results of I.Q test , estimate: a) The mean b) The standard deviation c) The interquartile range d) The coefficient of variation.

Mark 65 85 90 95 99 100 No. of students 5 10 20 45 40 18 Mark 104 108 115 120 125 No. of students 20 19 15 8 3

4. Using the figures given below, calculate:

a) The range b) The arithmetic mean c) The median d) The lower quartile e) The upper quartile f) The quartile deviation g) Pearson’s coefficient of Skewness h) The standard deviation

84

3 16 27 40 48 59 6 18 31 41 52 61 8 19 33 44 54 65 9 23 37 46 56 67 12

5. For the given frequency distribution, find i) mean ii) Mode

iv) Range v) Standard deviation x 3 4 5 6 7 8 f 1 3 4 8 5 6 6. The following data relates to the number of rooms per dwelling in Zambia for two separate years. Number of rooms 1 2 3 4 5 6 7 8 or more Year 1 (%) 2 6 13 28 36 14 4 5 Year 2 (%) 3 5 10 24 31 24 6 5

For each year, calculate the mean, standard deviation and coefficient of variation. Interpret the coefficient of variation based on the data at hand.

7. Explain the term ‘measure of dispersion’ and state briefly the advantages of using the following measures of dispersion.

i) Range ii) Quartile deviation iii) Variance iv) Standard Deviation

85



1.1 What is the arithmetic mean of the following frequency distribution?

Interval 6.1 – 6.5 6.6 – 7.0 7.1 – 7.5 7.6 – 8.0 8.1 – 8.5 Frequency , f 3 16 32 20 9

A) 7.3 B) 7.4 C) 16 D) 32

(NATech, 1.2 Mathematics & Statistics, June 2003) 1.2 What is the quartile deviation of the following set of data?

3 6 8 9 10 12 16 18 19 23 27 20 32 35 40 42 44 A) 27 B) 22 C) 17 D) 11

(NATech, 1.2 Mathematics & Statistics, December 1998)

1.3 What is the variance of the following set of numbers? 4, 6, 8, 9, 13.

A) 2.4 B) 6.78 C) 9.2 D) 40

(NATech, 1.2 Mathematics & Statistics, June 2001) 1.4 A group of people have the following ages, 21, 32, 19, 24, 31, 27, 17, 21, 26 and

42. The median age of the group is A) 31years B) 21years C) 25years D) 26years


1.5 The number of books read by eleven members of the public last year were:

15, 30, 19, 32, 10, 7, 12, 20, 12, 24, 4 What is the quartile deviation of the number of books read? A) 3 B) 8 C) 7 D) 6


86

1.6 The mean wages of 50, 25 and 75 mean employees by three 930 different firms are K40,000, K70,000 and K120, 000 per week. Calculate the mean range per week of all the men. A) K76 000 B) K85 000 C) K31 000 D) K76 667

(NATech, 1.2 Mathematics & Statistics, Nov/Dec 2000)

1.7 What is the approximate mean value per order of the following distribution

Value (K’000) 100 150 200 250 No. of orders 165 190 105 92

A) K110 000 B) K161 000 C) K175 000 D) K180 000

(NATech, 1.2 Mathematics & Statistics, December 1999 (Rescheduled)

1.8 The number of books ready by twelve members of the public last year were: 15,

30, 19, 32, 10, 7, 12, 20, 12, 24, 4 and 28. What is the quartile deviation of the number of books read? A) 3 B) 8 C) 7 D) 6

(NATech, 1.2 Mathematics & Statistics, June 2005)

1.9 A bar chart with three adjacent bars then a gap and three month and a further three

after a final gap is known as: A) A simple bar chart. B) A component part bar chart C) A multiple bar chart D) A percentage bar chart


1.10 The eight accountants in the Standard Chartered Bank have the following years of

experience 5, 8, 5, 19, 7 and 11. Find, for these years of experience the median.

A) 8 B) 19 C) 9.5 d) 12.4


87

SECTION B QUESTION ONE a) Find the first quartile ,1Q the second quartile 2Q and the third quartile 3Q and the

quartile deviation QD of the following data. 18, 2, 5, 13, 4, 8, 11, 7


b) A company trades in five distinct geographical markets. In the last financial year, its turnover was:

(K) Congo DR 59.3 Congo Brazaville 61.6 Tanzania 15.8 Kenya 10.3 Zambia 9.9 Total 156.9 Draw a pie chart using the above figures. QUESTION TWO a) An analyst is considering two categories of companies: X and Y, for possible

investment. One of her assistants has compiled the following information on the price earning ratios of the shares of the companies in the two categories over the past year.

Price – Earning Ratios Number of Category

X Companies Number of Category Y Companies

4.95 to under 8.95 8.95 to under 12.95 12.95 to under 16.95 16.95 to under 20.95 20.95 to under 24.95 24.95 to under 28.95

3 5 7 6 3 1

4 8 8 3 3 4

Required: Compute the standard deviations of these two distributions.

88

b) A College receives the following number of complaints per week.

Complaints per week 0 1 2 3 4 Number of weeks 5 12 7 2 1 What is the median value?

c) After receiving complaints from trade union representatives concerning the disparity between higher and low paid workers in this company, the Personnel manager of the company asks for information on the current salary structure.

He is given the following data:

Basic Wage (K’000) Number of Employees under 100 100 to under 200 200 to under 300 300 to under 400 400 to under 500 500 to under 600 over 800

3 6 11 15 12 7 6

Required: Calculate a statistical measure of mean deviation using the data given above.

(NATech, 1.2 Mathematics & Statistics, June 2005) QUESTION THREE An analysis of access time to a computer disc system was made during the running of a particular computer program, which utilized disc file handling facilities. The results of the 140 access time were as follows:

Access time in Milli seconds Frequency 30 and less than 35 22 35 and less than 40 27 40 and less than 45 21 45 and less than 50 31 50 and less than 55 21 55 and less than 60 18

i) Determine the mean access time for this program

ii) Determine the standard deviation of the access time for this program

iii) Interpret for your superior, who is not familiar with grouped data, what the results in parts (i) and (ii) mean.


89

QUESTION FOUR

a) The times, measured to the nearest second, taken by 30 students to complete an algebraic problem are given below.

47 53 46 68 72 48 41 49 58 45 43 45 48 44 43 61 43 46 48 57 54 63 42 65 44 51 38 46 42 47

i) Group these times into a frequency table using eight equal class intervals, the first of which contains measured times in the range 35 to 39 seconds. ii) Which is the modal class of your distribution?

(NATech, 1.2/B1 Mathematics & Statistics, December, 1999(Rescheduled))

QUESTION FIVE

a) During the 1999/2000 session a college ran 70 different classes of which 44 were ‘English’, with a mean class size of 15. 2 and 26 were ‘History, with a mean class size of 19.2. The frequency distribution of class size were as follows:

Size of Class (No. of students(

No. of English Classes

No. of History classes

1-6 7-2 13-18 19-24 25-30 31-36.95

4 15 11 8 5 1

0 3 10 8 4 1

No student belonged to more than one class.

i) Calculate the mean class size of the college. Suppose now that no class of 12 students or less had been allowed to run. Calculate what the mean class size for the college would have been if the student in such classes: ii) had been transferred to the other classes. iii) Had not be admitted to the college.

(NATech, 1.2 Mathematics & Statistics, Nov/Dec 2000)

90

QUESTION SIX

a) Consider the grouped frequency distribution below.

At Least Value less than Frequency 0 10 20 30

10 20 30 40

0 50 150 100

You are required to estimate the mode graphically.

b) The Director of a large company has decided to analyse the annual salaries that are paid to staff. The frequency distribution of salaries that are currently being paid is as follows: Salary

(million kwacha) Number of Staff

Under 10 10 to under 20 20 to under 30 30 to under 40 40 to under 50 50 to under 70 70 to under 90

16 30 34 22 10 5 3

Records from five years ago include the following statistics about salaries that were paid.

Then: Mean salary = K18.95m Standard deviation K106m Median Salary = K17.0m Quartile deviation K6.2m You are required to help with the analysis by

i) Calculating the mean and standard deviation of current salaries. ii) Interpreting the statistics that you have calculated.


91

QUESTION SEVEN a) The following is a frequency distribution of I.Qs of 100 children at a primary

school.

IQ Number of Children 50 - 59 60 – 69 70 – 79 80 – 89 90 – 99

100 - 109 110 – 119 120 – 129 130 - 139

1 2 8 18 23 21 15 9 3

i) The mean deviation ii) The standard deviation of the IQ scores

b) Compare and comment on the values obtained for the two measure of dispersion in (i) and (ii) above.

(NATech, 1.2 Mathematics & Statistics, December 2002) c) The data in the following Table relates to the number of successful sales made

by the salesmen employed by a large microcomputer firm in a particular quarter.

No. of sales 0-4 5-9 10-14 15 – 19 20 – 24 25 - 29 No. of salesmen 1 14 23 21 15 6

Calculate:

i) The mean, and ii) The standard deviation, of the number of sales.


92

QUESTION EIGHT

a) Given the following data Value Number of Orders

100 000 150 000 200 000 250 000

165 190 105 92

i) Find the mean, and

ii) The modal value per order


b) The number of goals scored per game by a football player during 1997 – 1998 were as follows:

No. of goods, x 0 1 2 3 4 or more No. of games, f 23 14 3 2 0

Calculate i) The mean

ii) Variance, and

iii) Standard deviation of the number of goals per game.


93

c) A sample of estimate of weekly sales for Product A are represented in the weekly sales distribution below.

Weekly Sales (K’000) Number of Weeks

4000 – 5000 5000 – 6000 6000 – 7000 7000 – 8000 8000 – 9000 9000 – 10000 1000 – 11000 11000 – 12000

12000 – 13000 above - 13000

3 7 2 4 6 10 8 4 0 8

Calculate: i) Arithmetic mean ii) Modal sales iii) Standard Deviation iv) Coefficient of Skewness and comment on the distribution


94

CHAPTER 4

PROBABILITY

4.1 Introduction

Statistics is sometimes described as the art of making decisions in the face of uncertainty. Consider the following business problems:

Business Problem Uncertainty Cars How many cars to stock Material prices Economic demand Accounting How many graduates to Employees turnover Train for NATech Exams workloads Exam pass rate

In such situations the uncertain factors can be measured or quantified using probability. In this Chapter, we describe the notion of counting, an experiment and events. Probability is then defined, including the addition and multiplication rules.

Counting The simple process of counting still plays an important role in business and economics. One still has to count 1, 2, 3, 4 . . . , for example, when taking inventory, when determining the number of damaged cases in a shipment of beer from Ndola, or when preparing a report showing how many times certain stock market indexes went up during a month. The process of counting is simplified in this section by means of special mathematical Techniques.

Multiplication of Choices

If a choice consists of two steps, the first of which can be made in m ways and for each of these the second can be made in n ways, the whole choice can be made in nm. ways.

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Example 1 If a firm has 3 warehouses and 5 retail outlets, in how many different ways can it ship an item from one of the warehouses to one of the stores? Since ,53 == nandm there are 3(5) = 15 ways.

Example 2 If a travel agency offers trips to 10 different countries, either by air, rail or bus, in how many different ways can such a trip be arranged? Since ,310 == nandm there are 10(3) = 30 ways. By means of appropriate use of tree diagrams, it is easy to generate the foregoing rule so that it will apply to choices involving more than two steps.

Example 3 Retail Outlets

• • Warehouses • • • 1 • • 2 • • • 3 • • • • • • • Figure 1.0 Tree diagram for Example 1 •

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Permutations and Combinations The rule for multiplication of choices and its generalization is often applied when several choices are made from one set and we are concerned with the order in which they are made.

Example 4 In how many different ways can the judges choose the winner and the first runner-up from among the 5 finalist in a student beauty contest? Since the winner can be chosen in 5=m ways and the first runner up must be one of the other 4=n finalist, there are 5(4) = 20 ways.

Example 5 In how many different ways can the 25 members of a church choose a president, a Vice President, a Secretary, and a Treasurer? Regardless of which position is elected first, second, third and fourth, there are 25(24)(23)(22) = 303 600 ways. In general, if r objects are selected from a set of n objects, any particular arrangement (order) of these objects is called a permutation.

Example 6 Determine the number of possible permutation of two of the three letters A, B and C and list them all. Since m = 3 and n = 2, there are 3(2) = 6 permutations; they are AB AC BC BA CA CB The formula for the total number of permutations of r objects selected from n distinct objectives is given by

)1()1(..).2)(1( →+−−−= rnnnnPrn

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Since products of consecutive integers occur in many problems relating to permutations and other kinds of special arrangements or selections, it is convenient to introduce here the factorial notation. In this notation, the product of all positive integer less than or equal to the positive integer n is called “factorial” and is denoted byn !. Thus,

1!0 = by definition, and

1201.2.3.4.5!5

241.2.3.4!4

61.2.3!3

21.2!2

1!1

====

====

=

and in general .1.2.3...)2)(1(! −−= nnnn In short form,

)2()!(

! →−

=rn

nPr

n

Example 7 Find the number of ways in which 2 of 5 can be selected. For n = 5 and r = 2

20)4(525 ==P from the formula (1)

and )1)(2)(3(

)1)(2)(3)(4(5

!3

!5

)!25(

!52

5 ==−

=P

= 20 Permutations with Similar Items There will be occasions when the items to be arranged will not all be different. If this is the case then the number of permutations will be reduced.

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Example 8 Find the number of distinct permutations that can be formed from all the letters of the word SEE. The arrangements become, where E are numbers as 21 EandE

SEESEE

SEESEEESEESE

1221

21121221

The number of permutations are 3.2.1 = 6. The number of permutations is reduced to 3. If the labels are dropped you can’t tell the difference between the boxed permutations. The version of formula is now

!...!!!

!

321 rnnnn

n

××

Where 1n of the items are of one kind, and 2n of the items are of another type and so on

up to r types. In our example, 32,1 2121 ==+== nnnandnn giving

ways3)1)(2)(1(

)1)(2(3

!2!1

!3 ==×

Combination There are many problems in which we want to know the number of ways in which r objects can be selected from a set of n objects, but we do not want to include in our count all the different orders in which the selection can be made. Note here order is not important.

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Example 9 In how many ways can a committee of three A, B and C be selected? The three persons A, B and C can be assigned to a three-person committee in 3! = 6 orders (ABC, ACB, BAC, BCA, CAB and CBA), but there is only one committee, not six. The formula for a combination of r objects from the given n objects is given by

rn

rn C

r

n

rnr

nC =

−= .

)!(!

! are called binomial coefficients are becoming useful in the

later section on binomial distribution.

Example 10

Find the number of ways in which a person can select 3 stocks from a list of 5 stocks (the number of combinations of 5 things taken 3 at a time). For havewerandn ,3,5 ==

10

)1)(2)(1)(2)(3(

)1)(2)(3)(4(5

!2!3

!5

)!35(!3

!53

5

=

==−

=C

Example 11 In how many ways can a Principal choose 3 of 45 members to review a student grade appeal? For havewerandn ,345 ==

14190

)1)(2(3

)43)(44(45

!42!3

!45

)!345(!3

!453

45

=

==−

=C

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Example 12

Determine the value of

20

30 using the formula

)!(!

!

rnr

nCr

n

−=

435)29(15)1)(2(

)29(30

!2!28

!30

)!2830(!28

!30

28

3028

30

===

=−

=

=

C

Risk

Risk is concerned with the events whose probability of happening and frequency of occurrence can be calculated using statistics and past experience. In a situation of risk, we cannot be sure which of several possible outcomes will occur, but we can at least place values on the different possible outcomes. As an example, an insurance company will be dealing with a situation of risk when issuing a policy on damage to property. The insurer cannot be certain whether or not a certain office building will be destroyed by lightening or fire etc, but at least he knows how much will have to be paid to settle the insurance claim if this happens. A decision maker can be described as one of the following. 1) Risk seeker, one who takes risks to achieve the best outcome no matter how small

the chance of it occurring. 2) Risk neutral, one who only considers the most likely outcome. 3) Risk averse, one who makes a decision based on the worst possible outcomes, for

example an investor who, in spreading his investment over a portfolio of stocks accepts a lower expected return in order to reduce the chances of a larger loss is expressing an aversion to risk.

The main methods of dealing with risk is to estimate the probability and use statistics to look at its overall incidence in order to 1) Ensure that the level of reward (in the long run) is commensurate with the risks

taken.

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2) Reduce the overall incidences to a pre-set accept level (e.g the taking out

of insurance).

Probability Historically, the oldest way of measuring uncertainties is the classical probability concept. It was developed originally in connection with games of chance, and it lends itself most readily to bridging the gap between possibilities and probabilities. This concept applies only when all possible outcomes are equally likely, in which case we can say that. If there are n equally likely possibilities, one of which must occur and s are regarded as

favourable, or as a “success”, then the probability of a “success” is given by the ratio .n

s

Example 13 What is the probability of drawing a 5 from a well shuffled deck of 52 playing cards? Therefore ,4=s “cards numbered five (5)” among the n = 52 cards, so we have

13

1

52

4 ==n

s.

Example 14 What is the probability of rolling a 5 with a balanced die?

Since .6

1,61 ===

n

shavewenands

The major shortcoming of the classical probability concept where possibilities must all be equally likely is that there are many situation in which the possibilities that arise cannot be regarded as equally likely. This might be the case, for example, if we are concerned with the question whether there will be rain, or sunshine, to tell whether a person will receive a promotion or to predict the success of a new business or the behaviour of the stock market or the success of a new marriage. This leads us to the relative frequency probability which is the probability of an event (outcome) is the proportion of the time that events of the same kind will occur in the long run.

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If we say that the probability is 0.53 that a bus from Kitwe to Mufulira will arrive on time, we mean that, buses arrive on time 53% of the time. Also, if the weather man predicts that there is a 20% chance of rain (that is the probability that it will rain is 0.20). It means that under the same whether conditions it will rain 20% of the time. In accordance with frequency concept of probability, we estimate the probability of an event by observing what fraction of the time similar events have occurred in the past.

Example 15 If the records show that (over a period of time) 25 of 125 buses from Lusaka to Kitwe arrive on time, what is the probability that any bus from Lusaka to Kitwe will arrive on time?

Since in the past 20.0125

25 = of the arrivals are on time, we use this fraction as an

estimate of the probability. Example 16 If 85 of 2000 first year student who enter School of Business dropped out before the end of their first year, what is the probability that a freshman entering this college will drop out before the end of his first year?

Since in the past 0425.02000

85 = of the first year dropped out before the end of their first

year, we use this figure as an estimate of the probability. The last method of calculating probability is the personal or subjective evaluation. Hence subjective probability. Such probabilities express the strength of one’s belief with regard to the uncertainties that are involved, and they apply especially when there is little or no direct evidence, so that there really is no choice but to consider information, ‘educated guesses,” and perhaps intuition and other subjective factors. Event An event is a subset A of the sample space S. A is a set of possible outcomes. In our example A = {1} or A = {2, 3} etc. The event S itself is the sure or certain event since an element of S must occur, and the empty set φ , which is called the impossible event because an element of φ cannot occur.

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By using set operations on events in S, we can obtain other events in S. For example, if A and B are events, the 1) BA ∪ is the event “either A or B or both”

BA ∪ is called the union of A and B. 2) BA ∩ is the event “both A and B”.

BA ∩ is called the intersection of A and B. 3) A′ is the event “not A”.

A′ is called the complement of A 4) BA ′∩ is the event “A but not in B.” If the events A and B are disjoint, we write BA ∩ = φ and we say that the two events A and B are mutually exclusive. This means that they cannot both occur.

Random Experiments We are all familiar with the importance of experiments in Science and Engineering. Experimentation is useful to us because we can assume that if we perform certain experiments under very ready same conditions, we will arrive at the results that are essentially the same. Here we discuss random experiments where we don’t have control.

Example 1 Suppose we roll a fair die. The collection of all the possible outcomes ‘S’ is called the sample space and the individual outcome points the sample points. Here S = {1, 2, 3, 4, 5, 6}.

Some Important Theorems On Probability 1) For any event A, 1)(0 ≤≤ AP i.e, a probability is between 0 and 1 inclusive. 2) For φ the empty set 0)( =φP i.e. , the impossible event has probability zero. 3) If A′ is the complement of A, then )(1)( APAP −=′

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4) If A and B are any two events, then ).()()()( BAPBPAPBAP ∩−+=∪ This is called the addition rule for any two events.

5) For any events A and B, )()()( BAPBAPAP ′∩+∩= . 6) If A and B are mutually exclusive events then )()()( BPAPBAP +=∪ . This is

called addition rule for two mutually exclusive events. 7) Conditional Probability Let A and B be two events such that .0)( ≠AP Denote )/( ABP the probability

of B given that A has taken place. Since A is known to have taken place, it becomes the new sample space replacing the original S. From this we have the definition:

)2()/()()(

)1()(

)()/(

→=∩

→∩=

ABPAPBAP

orAP

BAPABP

8) Independent Event If ),()/( BPABP = i.e, the probability of B occurring is not affected by the fact

that A has occurred, then we say that A and B are independent events. This is equivalent to )3()()()( →=∩ BPAPBAP

Note that if this equation holds, then A and B are independent. Equation (3) is

called the multiplication rule. Example 1 Suppose you are going to throw two fair dice. What is the probability of getting a 4 on each die? The first task is to write down the sample space, S. Each die has six equally likely outcomes, and each outcome of the first die can be paired with each of the second die. The sample space is shown in Figure 4.1. The total number of outcomes is 36, and only one is favourables to a 4 on the first die and a 4 on the second. The 36 outcomes are

equally likely, so by ,n

s we have

P(4 on 1st and 4 on 2nd) = 36

1

105

1 2 3 4 5 6 1 (1, 1) (1. 2) (1, 3) (1, 4) (1, 5) (1, 6)

2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 50 (3, 6)

4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

Figure 4.1

Alternatively, using the multiplication rule, since the two events are independent, P( 4 on 1st die and 4 on 2nd die) = P(4 on 1st die) P(4 on 2nd die). There are six faces on a die and on a fair die each is equally likely to come up when you throw the die.

6

14( =dieonP .

Therefore P( 4 on 1st die and 4 on 2nd die) = 36

1

6

1.

6

1 =

The two methods yield the same results.

Example 2 Fairwell Electronics, all 150 employees were asked about their political affiliations. The employees were grouped by type of work, as executive or production workers. The results with row and column total are shown in Table 4.1. Suppose an employee is selected at random from the 150 Fairwell employees. Let us use the following notation to represent different events of choosing: E = executives, PW = Production worker, M = Movement for Multiparty Democracy (MMD) , U = United Party for National Development, (UPND), I = Independent.

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Table 4.1 Employee Type and Political Affiliation Employee Type MMD (M) UPND (U) Independent Row Total Executive (E) Production Worker (PW)

10 65

35 22

10 8

55 95

Column Total 75 57 18 150 a) Compute P(M) and P(E)

367.0150

55)(

5.0150

75)(

===

===

employeesofnumber

executiveofnumberEP

employeesofnumber

MMDofnumberMP

b) Compute P(M/E) i.e. probability of M given E. For the conditional probability, we restrict our attention to the portion of the

sample space satisfying the condition of being an executive.

182.055

10

)(

)()/(

≅=

=∩=executivesofnumber

MMDarewhoexecutivesofnumber

EP

EMPEMP

c) Are the event M and E independent?

One way to determine if the events M and E are independent is to see if P(M) = P(M/E) or equivalently, if P(E) = P(E/M). Since P(M) = 0.05 and P(M/E) = 0.182, we see that event )/()( EMPMP ≠ . This means that the event M and E are not independent. The probability of event M ‘depends on” whether or not event E has occurred.

d) Compute P(M and E) This probability is not conditional, so we must look at the entire sample space. ).()( EMPEandMP ∩=

107

Therefore.

067.0150

10

)(

≈=

=employeesofnumbertotal

MMDareexecutivesofnumberEandMP

Let’s recompute this probability using the rules of probability for dependent

events.

067.0150

10

55

10.

150

55)/()()( ≈=== EMPEPEandMP

The results using the rules are consistent with those using the sample space. e) Compute P(M or E) From part (d) we know that the event MMD and executive are not mutually

exclusive, because .0)( ≠∩ EMP Therefore:

8.0

150

120

150

10

150

55

150

75

)()()()()(

==−+=

∩−+=∪= EMPEPMPEMPEorMP

Example 3 A firm is independently working on two separate jobs. There is a probability of only 0.4 that either of the jobs will be finished on time. Find the probability that: a) both b) neither c) Just one d) at least one of the jobs is finished on time.

108

Let A be the event that the job is finished on time and B the event that the job is not finished on time.

Then P(A) = 0.4 and P(B) = 1 – P(A) = 1 – 0.4 = 0.6. a) The two events are independent hence P(both finished on time)

16.0

)4.0(4.0

)().()(

==

=∩= BPAPBAP

b) P(neither job finished on time)

36.0.

)06(6.0

)()()(

==

=∩= BPAPBAP

c) P(just one job finished on time)

88.024.024.0

)6.0(4.0)6.0(4.0

)()(

=+=+=

∩+∩= ABPBAP

d) P(at least one job finished on time)

64.036.01

)(1

,64.016.024.024.0

)()()(

=−=−=

=++=∩+∩+∩

timeonfinishedjobneitherP

or

BAPABPBAP

Example 4 30% of employees in a company earn over K150 000 per week and 60% earn between K100 000 and K150 000 per week. Find the probability that an employee selected at random earns: a) less that K100 000 per week. b) under K100 000 or over K150 00 per week.

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60.0100

60)000150000100(

30.0100

30)000150(

==

==

KtoKP

KoverP

a) P(under K100 000)

[ ][ ]

%)10(1.0

9.01

30.060.01

)000150()000150000100(1

)000100(1

=−=

+−=+−=

−=KoverPKtoKP

overKP

b) P(under K100 000 or over K150 000)

40.0

30.010.0

)000150()000100(

=+=

+= KoverPKunderP

Exercise 1 1. Students have two independent tests. 20% of students pass test A and 70% pass

test B. Find the probability that a student selected at random passes:

a) Both tests b) Only test A c) Only one test.

2. A computer retailer conducts a survey of 250 computer purchasers and obtains the

information in the table below:

AGE Less than 25 25 – 40 41 and over Male Female

70 45

25 40

50 20

If a customer is selected at random, find the following probabilities:

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a) the customer is female and aged 25 – 40?

b) The customer is male.

c) If the selected customer is aged less than 25, what is the probability that

they are female?

d) Are the events female and aged 25 – 40 independent? 3. I select two cards from a packet of cards. What is the probability that they are

both kings? 4. I toss a fair coin and the throw a dice. What is the probability that I obtain a tail

and a five? 5. A bag contains 6 red counters and 5 blue counters. If one counter is taken at

random, replace, then another is taken, what is the probability that

a) both counters are red?

b) the first is red and the second is blue?

c) the two counters are of different colours?

d) the two counters are of the same colours? 6. If 5 factories in a group of 25 factories in a certain community are violating

environmental regulations and 7 are randomly elected for inspection, what is the probability that:

a) none of the violators will be selected for inspection?

b) all of the violators will be selected for inspection?

7. a) Explain what you understand by

i) Mutually exclusive events;

ii) Conditional probability; iii) Independent events.

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8. Two cards are drawn simultaneously from the same pack of 52 cards.

Find each of the following probabilities.

a) Both cards are picture cards. b) Neither of the cards is a picture card. c) Exactly one of the cards is a picture card.

4.2 Probability Distribution

Many probabilistic situations in business and commercial environments exhibit the same underlying features. These frequently occurring situations can be investigated by a fairly limited number of probability models, a few of which will be discussed in this section. Quantitative variables are either continuous or discrete as defined in Chapter 3. A random variable is a variable which assumes different values depending on the outcomes of an experiment. A discrete random variable has a frequency distribution just like a continuous random variable. A frequency distribution indicates the number of observations of a random variable (i..e the frequency) at each value of the random variable (X). Consider the following example:

Example 5 Consider the rolling of a fair die 150 times as shown in the following table. Face (x) 1 2 3 4 5 6 Number of times (f) 10 15 40 25 45 15

What we have is a frequency distribution. This table can be converted into a probability distribution. Hence, a probability distribution is a table or graph or formula comprising the possible outcome with its associated probability. Thus Face (x) 1 2 3 4 5 6 P(x)

150

10

150

15

150

40

150

25

150

45

150

15

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The probabilities in a probability distribution has the following properties: i) ∑ = 1)(XP

ii) 1)(0 ≤=≤ XP iii) ∑= )()( xxPXE

iv) [ ]22 )()()( ∑∑ −= xxPxPxXVar

Properties (iii) and (iv) give us the expectation and variance of a random variable respectively. Example 6 What is our mathematical expectation if we win K75 000 if a balanced coin falls head and lose K50 000 if it falls tails?

The amounts are 00050000750 21 KxandKx −== , the probabilities are 2

1)( 1 =xP and

2

1)( 2 =xP and the mathematical expectation is

.50012

500120002550037

2

1)00050(

2

100075

)()(

2211

K

PxPx

XPxXE

==−=

−+

=

+=

=∑

In the long run, he is expected to win K12 500.

Example 7 The probabilities are 0.25, 0.15, 0.22 and 0.38 that a speculator will be able to sell a house within a year at a profit of K8 800 000, at a profit of K500 000, at a profit of K400 000, or at a loss of K600 000, respectively. What is the expected profit?

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Substituting ,000600,000400,000500,000800 4321 −==== xKxxx

000135

0002280008800075000200

)38.0(000600)22.0(000400)15.0(000500)25.0(000800)(

),(

)(int38.0)(22.0)(,15.0)(,25.0)( 4321

K

XE

havewexxP

XEoxPandxPxPxP

=−++=

−++=

=

====

∑

Example 8 If the probabilities are 0.25, 0.35, 0.06 and 0.30 that a certain office will receive 0, 1, 2, 3, or 4 complaints about theft on any one day. How many such complaints can be expected per day? The expected number is )(XE = 0(0.04) + 1(0.25) + 2(0.35) + 3(0.06) + 4(0.30) = 3.33 Example 9 On a particular day, a trader expects the sales of cabbages to follow the pattern.

Sales 0 50 100 150 Probability 0.01 0.28 0.37 .34

Calculate: i) the expected sales, ii) the standard deviation.

i) The expected sales = 0(0.01) + 50(0.28) + 100 (0.37) + 150(0.34) = 102

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ii) The standard deviation = ( )22 )()(var ∑ ∑−= xxPxPxiance

Since ∑ = 102)(xxP , from (1), we compute

12050

650770037000

)34.0()150()37.0()100()28.0()50()01.0(0)( 22222

=++++=

+++=∑ xpx

Therefore standard deviation = 57.401646)102(12050 2 ==−

The probabilities associated with values of the random variables can be computed

from a well-established equation. The Standard Deviation: Measuring risk Risk can be defined as the chance that an outcome other than expected will occur or it’s the variability in the returns or outcomes from the investment. Standard deviation is used to measure risk. This is because the amount of scatter or variability in the probability distribution is measured by the standard deviation.

iancedeviationdards vartan ==σ Another useful measure to risk is the coefficient of variation (CV), which is the standard deviation divided by the expected return (mean value) that is

100

100var

×=

×==

µσreturn

RiskCViationoftCoefficien

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Example 10 Stocks A and B have the following probability distributions of future returns

Probability A(K’000) B(K’000) 0.03 0.20 0.50 0.27 0.10

10 20 12 35 5

15 25 18 30 4

a) Calculate the expected return for each stock. b) Calculate the standard deviation of the returns for stock A and Stock B. Now

calculate the coefficient of variation for Stock A and Stock B. Which stock is risky? Explain.

a) E(A) = Expected return from Stock A = ∑ )(xxP

= 10(0.03) + 20(0.20) + 15(0.50) + 35(0.17) + 5 (0.10) = 0.3 + 7.500 + 5..950 + 0.5 = 18.25 = K18 250

E(B) = Expected return from Stock B = ∑ )(xxP

= 15(0.03) + 25(0.20) + 18(0.50) + 30 (0.17) + 4 (0.10) = 0.45 + 5 + 9 + 5.100 + 0.4 = 19. 95 = K19 950

b) [ ]22 ))(()(∑ ∑−= xxpxpxσ

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For A, we have

x )(xP 2x )(2 xPx 10 20 15 35 5

0.03 0.20 0.50 0.17 0.10

100 400 225 1225 25

3 80

112.5 208.25

2.5 ∑ = 25.406)(2 xPx

555.8

1875.73)25.18(25.406 2

=

=−=

σ

σ

For B, we have

x )(xP 2x )(2 xPx 15 25 18 30 4

0.03 0.20 0.50 0.17 0.10

225 625 324 900 16

6.75 125 162 153 1.6

∑ = 35.448)(2 xPx

0956.7

347531875.50)95.19(35.448 2

=

=−=

σ

σ

%5.36;%9.46

10019950

6.7095;100

18250

8555100

==

×=×=×=

BA

BA

CVCV

CVCVµσ

The coefficient of variation of stock B is less compared to that of Stock A. Hence one would conclude that Stock B is less risky than Stock A.

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Binomial Distribution If P is the probability of a “success’ in an individual trial, then in n independent trials the probability of x successes is given by the Binomial formula:

xnxx

n PPCxP −−= )1()(

where x = 0, 1, 2, . . . , n . The number of successes has a mean np and standard

deviation = )1( pnp − .

The features of a binomial experiment are as follows: 1) The number of trials are fixed. We denote this by the number n as in the formula

above. 2) The n trials are independent and repeated under identical conditions. 3) Each trial has only two outcomes: success, denoted by S and failure denoted by F. 4) For each individual trial, the probability of success is the same. We denote the

probability of success by p and that of failure by q . Since each trial results in either success or failure .11 pqandqp −==+

5) The main problem of a binomial experiment is to find the probability of x

success out of the n trials. Alternatively, when we multiply out the power of a binomial bracket such as

nba )( + we say that we are expanding the term and the result is called a binomial expansion.

543223455

4322344

32233

222

1

510105)(

464)(

33)(

2)(

)(

babbababaaba

babbabaaba

babbaaba

bababa

baba

+++++=+

++++=+

+++=+++=+

+=+

Let us now make some observations regarding the above expressions of nba )( +

looking at the results, we see four things:

118

1. the “sum” of the powers is constant on each line for example 4)( ba + total power on each term is 4.

2. the power of a is decreasing from left to right for example 1234 aaaa . 3. the power of b is increasing from left to right for example 4321 bbbb 4. The numbers form a triangle. The triangle formed is know as Pascal’s Triangle, which can be used to obtain the binomial coefficients. The numbers are arranged as follows: n = 0 l n =1 1 1 n =2 1 2 l n =3 l 3 3 l n =4 1 4 6 4 1 n =5 1 5 10 10 5 1 n =6 1 6 15 20 15 6 1 n =7 1 7 21 35 35 21 7 1 n =8 1 8 28 56 70 56 28 8 1 etc. Starting with this triangle and then using the rules given in (1), (2) and (3) above, we can write down any expansion of nba )( + .

119

Example 11 Expand 3)43( ba+ We start with the line of Pascal’s triangle for 3=n which gives us 1 3 3 1 we get

3223

323

6414410827

)4(1)4)(3(3)4()3(3)3(1

babbaa

bbabaa b

+++=+++

Example 12 Expand 5)32( ba − We start with the line of Pascal’s triangle for 5=n which gives us 1 5 10 10 5 1 We get:

54322345

54322345

243810108072024032

)3()3)(2(5)3()2(10)3()2(10)3()2(5)2(1

babbababaa

bbabababaa

−+−+−=−+−+−+−+−+

The Binomial Expansion. The Binomial Theorem A power of a binomial can also be expanded by means of the Binomial Theorem.

...!3

)2)(1(

!2

)1(

!1)(

33221

+−−+−++=+−−− bannnbannbna

abannn

nn

120

Example 13 Use the Binomial Theorem to expand 3)43( ba +

3223

32233

6414410827

!3

)4)(1)(2(3

!2

)4)(3)(2(3

!1

)()3(3)3()43(

babbaa

bbabaaba

+++=

+++=+

which agrees with the answer in Example 11. You will notice that in the last term

!3

)1)(2(3 cancels out.

Example 14 Use the Binomial Theorem to expand 5)32( ba −

.243810108072024032

!5

)3)(1)(2)(3)(4(5

!4

)3)(2)(2)(3)(4(5

!3

)3()2)(2)(4(5

!2

)3()2)(4(5

!1

)3()2(5)2()32(

54322345

54

3223455

babbababaa

bba

bababaaba

−+−+−=

−+−+

−+−+−+=−

You will notice that this time !5

)1)(2)(3)(4(5 cancels out.

The Binomial Theorem in Terms of Combinations The Binomial Theorem can also be expressed in terms of combinations.

...)( 333

222

11 ++++=+ −−− baCbaCbaCaCba nnnnnnn

onn

121

Example 15 Use the Binomial Theorem to expand 3)43( ba+ .

3223

3223

33

322

331

3333

6414410827

)4()4)(3(3)4()3(3)3(

)4()4)(3()4()3()3()43(

babbaa

bbabaa

bCbaCbaCaCba o

+++=

+++=

+++=+

Which agrees with the answers obtained in example 11 and 13. Example 16 Use the Binomial Theorem to expand 5)32( ba −

54322345

544

5323

5232

541

53555

243810108072024032

)3()3)(2()3()2()3()2()3()2()2()32(

babbababaa

bbaCbaCbaCbaCaCba o

−+−+−=

−+−+−++−+−+=−

Example 17 If ,4.03 == pandn use Binomial formula to find: a) P(2), b) P(0), c) P(1) a) 232

23 )4.011()4.0()2( −−= CP

12 )6.0()4.0(!1!2

!3=

Therefore P(2) = 0.288

122

b) 30

03 )4.01()4.0()0( −= CP

216.0)0(

)6.01()4.0(!3!0

!3 30

=

−=

P

c) 20

13 )4.01()4.0()1( −= CP

432.0)1(

)6.01)(4.0(!2!0

!3 1

=

−=

P

Alternatively If 6.0,4.0,3 === qpn Using the binomial expansion

)0()1()2()3(

216.0432.0288.064.0

)6.0()6.0()4.0(3)6.0()4.0(3)4.0()6.04.0( 322233

PPPP

+++=+++=+

Note: At this stage it is very advisable to check that the sum of the individual

probabilities (the whole probability) is 1. Example 18 Construct a Binomial distribution when n = 4 and p = 0.2

4096.0)8.0()2.0()0( 400

4 == CP

4096.0)8.0()2.0()1( 31

14 == CP

1536.0)8.0()2.0()2( 22

24 == CP

4096.0)8.0()2.0()3( 13

34 == CP

0016.0)8.0()2.0()4( 04

44 == CP

123

Example 19 From past experience it is known that approximately 70% of applicants pass an initial assessment test. In a group of six applications find the probability that: a) all applicants pass the test, b) only four applicants pass, c) more than four applicants pass. We have n = 6, p = 0.70. a) 06

66 )30.0()70.0()6( CP =

%)8.11(118.0

)1176490.0(!0!6

!6

=

=

b) 24

46 )30.0()70.0()4( CP =

%)4.32(324.0

)09.0)2401.0(15

)09.0)(2401.0(!2!04

!6

==

=

c) P(more than four) = P(5 or 6) = P(5) + P(6)

118.0)30.0()70.0()6(

303.0)30.0()70.0(6

)30.0()70.0()5(

066

6

5

55

6

==

==

=

CPand

CnowP

Therefore P(more than four) = 0.303 + 0.118 = 0.421 (42.1%)

124

Example 20 One fifth of all accounts are found to contain errors. In a batch of 6 accounts, find the probability that the number of accounts containing errors is: a) less than three b) more than three Find the mean and standard deviation of the accounts containing errors.

We have 20.05

1,6 === Pn

a) P(less than three) = P(0) + P(1) + P(2) Now 2621.0)80.0()20.0()0( 606 == oCP

andCP ,3932.0)80.0()20.0()1( 51

16 ==

2458.0)80.0()20.0()2( 42

26 == CP

Therefore P(less than three) = 0.2621 + 0.3932 + 0.2458 = 0.9011 b) P(more than three) = P(4) + P(5) + P(6) = 1 – P(less or equal to three) = )]3()2()1()0([1 PPPP +++− Now 0819.0)80.0()20.0()3( 33

36 == CP

Therefore P(more than three ) = 1 – [0.2621 + 0.3932 + 0.2458 + 0.0819] = 1 – 0.983 = 0.017

125

Alternatively, 0154.0)80.0()20.0()4( 24

46 == CP

0015.0)80.0()20.0()5( 15

56 == CP

0001.0)80.0()20.0()6( 06

66 == CP

Therefore (more than three) = 0.0154 + 0.0015 + 0.0001 = 0.017.

Mean = 2.15

16 =

=np

Standard deviation =

=−5

4

5

16)1( pnp

9798.096.0 ==

Poisson Distribution To be able to use the binomial distribution, one must be able to count the number of successes and the number of failures. In some situations however, it is not really sensible to speak of successes and failures. For instance the number of telephone calls received at a switch board in a unit interval, say may be of interest, but the number of time they are ‘received’ is almost meaningless. In such situations the binomial probability distribution is no longer appropriate. Fortunately another probability distribution the Poisson is available. This models a situation where there is an interest in the number of times and rate at which events occur. If events occur at random at an average rate of λ per unit time then the probability of x event is given by Poisson formula

.tan...,2,1,0!

)( λλλλ

====−

deviationdardsandmeanxwherex

exP

x

126

Example 21

,3=λ find the Poisson probabilities P(0), P(1), P(2), P(3) and P(4).

.4,3,2,1,0,!

)( ==−

xx

exP

xλλ

0498.0!0

3)0( 3

03

=== −−

ee

P

1494.03!1

)3()1( 3

13

=== −−

ee

P

2241.02

9

!2

)3()2( 3

23

=== −−

ee

P

2241.06

27

!3

)3()3( 3

33

=== −−

ee

P

1681.024

81

!4

)3()4(

343

===−− ee

P

Example 22 Show that the Poisson formula give a good approximation of the Binomial probabilities when n = 200 and p = 0.02. We have 4)2.0(200,02.0,200 ===== nPsoPn λ Using the Binomial formula we have:

0176.0)98.0()02.0()0( 20000

200 == CP

0718.0)98.0()02.0()1( 1991

1200 == CP

1458.0)98.0()02.0()2( 1982

2200 == CP

etcCP 1964.0)98.0()02.0()3( 1973

3200 ==

127

Now using the Poisson formula, we have

0183.0!0

4)0(

40

==−e

P

0732.0!1

4)1(

41

==−e

P

1465.0!2

4)2(

42

==−e

P

Since n is large and p small, the Binomial and poisson probabilities are in close agreements. They agree to two or three decimal places.

Example 23 An assembly line produces approximately 3% defective items. In a batch of 150 items, find the probability of obtaining: a) only two defective b) less than two defectives c) at least two defective We have n = 150 and p = 0.03. Since n is large and p is small we can use Poisson distribution to approximate these Binomial probabilities. Now 5.4)03.0(150 === npλ .

a) 1125.0!2

)5.4()2(

25.4

==−e

P

b) ).1()0()2( ppthanlessp +=

1952.0!3

4)3(

43

==−e

P

128

Now ( ) ( )0111.0

!0

5.40

5.4

==− oe

P

And ( ) ( )04999.0

!1

5.41

5.4

==−e

P

Therefore ( ) 0611.004999.00111.02 =+=thanlessP

( ) ( )( )

9889.00111.01

01

1)

=−=−=

−=P

nonePdefectivetwoleastatPc

Example 24 On average, 5 boys are absent from school each day. Find: i) The probability that, on a day selected at random, exactly 5 boys are absent.

ii) The expected number of days on which there will exactly be 5 boys absent if the

school is open for 400 days each day.

i) We have 5=λ

( ) 1755.0!5

55

55

==−e

boysP

ii) np=λ from (i) ( ) .1755.05 =boysP Then expected number of boys is

( ) 187.701755.0400 = 70≅

129

Exercise 2 1) Use the Pascal’s triangle to expand the following expressions: i) 75 )7())( −+ aiiba iii) 4)53( ba − iv) 5)25( t− v) 623 )2( yx − 2) Use the Binomial Theorem to expand i) 45 )54())32( sriidc ++ iii) 3)1( x−

iv) 53)25( ba + v) 42

2

3

+ c

c

3) If ,4.0,6 == pandn use the Binomial formula to obtain the following

probabilities:

a) ( )0P b) ( )2P c) ( )4thanlessP d) ( )4leastatP e) ( )4P f) ( )4mostatP

4) If ,5.3=λ use the poisson formula to obtain the following probabilities:

a) ( )0P b) ( )2mostatP c) ( )2thanlessP d) ( )2thanmoreP 5) An assembly line produces approximately 5% defective items. In a sample of

seven items, find the probability of obtaining:

a) No defectives b) only one defective c) More than one defective d) at least one defective

130

6) Based on information from the Post, it is estimated that 85% of cars on the dual carriage highway are going faster than the speed limit. A random sample of five cars is observed. What is the probability that:

a) none of the cars is speeding? b) at least one is speeding?

7) If 25 per cent of the packages produced by an automatic machine are defective,

find the probability that out of four packages chosen at random:

a) one is defective, b) three are defective c) at most three are defective

8) A supermarket uses several cash registers to check out the orders of its customers

but has assigned one cash register to an express lane, which serves customers who have purchased only less than or equal to ten articles. The probability that a customer in this store will use the express lane is 0.25. Find the probability that among six randomly selected customers there are zero, one, two, three, four, five or six who will use the express lane.

9) If the probability is 0.45 that any one person will dislike the taste of a new

toothpaste, what is the probability that at least 3 of 20 randomly selected persons will dislike it?

10) If a bank received on the average 5 bad cheques per day, what is the probability

that it will receive 3 bad cheques a given day? 11) The number of patients who are received per hour in the emergency room of a

hospital is a random variable having the poisson distribution with 2.5. Use the poisson distribution to compute the following probabilities that in any given hour the emergency room will receive:

a) no patients b) at least one patient c) two patients d) four patients

12) Items produced from a machine are known to be 2% defective. If the items are

boxed into lots of 500, what is the probability of finding that a single box has 3 or more defective?

131

Normal Distribution Many business and accounting applications involve continuous or near enough continuous variables. It is a distribution of “natural phenomena”, such as:

• Weight measurements • Time measurements • Interest rates • Financial ratios • Income levels • Exchange rates.

The main characteristics of the distribution are: a) It is symmetrical, with mean, median and mode equal. Symmetry implies that

50% of the area under the curve is below this point and 50% is above this point. b) It is unimodel, i.e., the normal distribution peaks at a single value. c) It approaches the horizontal axis on either side of the mean. In other words, the

normal distribution is asymptotic to the x – axis. Consider a Normal distribution with mean =µ and standard deviation = σ as shown below.

x σz

The area shaded can be obtained from tables of the Normal distribution by finding the standardized unit.

σ

µ−= xZ

Note: the total area under the Normal Curve = 1.

µ

132

Example 1 Given a Normal distribution with mean = 20 and standard deviation = 3.5, find the areas under this Normal Curve. a) above 30 b) below 23 c) above 12 d) between 15 and 28.

a) 5.3,20 == σµ

2320 == xµ From the Normal tables, we are given the area between 0 and 2.86. That is

0.4979. The shaded area = 0.5000 – 0.4979 = 0.0021 b)

2320 == xµ

86.05.3

3

5.3

2023 ==−=Z

The Normal tables give the area from 0 to 0.86 = 0.3061. The shaded area = 0.5000 + 0.3061 = 0.8061.

86.25.3

10

5.3

2030 ==−=Z

133

c) 12 20=µ

29.25.3

8

5.3

2012 −=−=−=Z

The Normal tables give the area from 0 to 2.29. Because of the symmetric nature

of the distribution about o, this is the same as the area from 0 to –2.29. That is 0.4890.

The shaded area = 0.5000 + 0.4890 = 0.9890 d) 15 20=µ 28

29.25.3

8

5.3

202843.1

5.3

5

5.3

2015 ==−=−=−=−= ZZ

The Normal tables give the area from 0 to –1.43 and 0 to 2.29. The shaded area is

0.4296 + 0.4890 = 0.9186.

134

Example 2 The wages of blue-collar workers in a large company are Normally distributed with a mean of K104, 500 per week and a standard deviation of K14, 250 per week. Find the probability of a worker, selected at random, earning: a) over K123, 500 per week, b) between K95, 000 and K133, 000. a) 500104=µ 123 500

33.125014

500104500123 =−=Z

The required probability 4062.05000.0 −= 0938.0=

135

b) 95 000 5000104=µ 133 000

67.0

0.225014

500104000133

25014

50010400095

=

=−=−= ZZ

The required probability = area shaded

7258.0

4772.02486.0

=+=

Example 3 The Bank has analyzed the number of transactions processed by each of its branches. A typical branch will process 750 transactions a week. Experience has shown that there is a standard deviation of 2.1. Transaction processing is approximately normally distributed. a) Compute the probability of the number of branches who can be expected to

process more than 775 transactions per week. b) What is the probability that the number of transactions expected to be processed

are between 705 and 795?

136

a) 750=µ 775

19.121

750775 =−=Z

The required probability is 0.5000 – 0.383 = 0.117. Hence, 11.7% of the

branches process more than 775 transactions per week. b) 705 750=µ 795

14.221

75079514.2

21

750705 =−=−=−= ZZ

The required probability is 0.9676 or 96.8%. Example 4 In an assessment of job performance the marks awarded are Normally distributed with a mean of 65 and a standard deviation of 13. a) In a group of 500 employees, how many would you expect to obtain over 85

marks? b) From past performances it can be seen that approximately 25% of employees

obtain unsatisfactory gradings. What is the minimum ‘satisfactory grade?

137

a) 65=µ 85

54.113

6585 =−=Z

The required area = 1.5000 – 0.4382 = 0.0618. Therefore in a group of 500

employees: 500 (0.0618) = 30.9. Hence approximately 31 employees obtain over 85 marks.

b) 25% 65=µ X is the minimum satisfactory grade. From tables

42.56

58.865

66.013

65

=

−=−

−=−

X

X

X

Therefore the minimum grade = 56.

138

Exercise 3 1. Given a Normal distribution with mean = 70 and standard deviation = 20.

a) over 80 b) under 72 c) over 64 d) between 72 and 82 e) between 65 and 90

2. A population is Normal with 2855 == σµ and . a) If one item is taken at random from this population, find the probability

that it is: (i) greater than 70 (ii) less than 47 (iii) between 40 and 65.

b) Find the 90% confidence limits for this item.

3. The lengths of the fish received by a certain cannery have a mean of 5.62cm and a standard deviation of 0.28cm.

(i) What percentage of all these fish is longer than 6.00cm? (ii) What percentage of the fish is between 5.36 and 5.86cm long? 4. A baker knows that the daily demand for brown bread is a random variable with a

distribution, which can be approximated closely by a normal distribution with the mean 5.54=µ and the standard deviation 8.5=σ . What is the probability that the demand for brown bread will exceed 60 on any given day?

5. A bank manager has determined from experience that the time required for a

security guard to make his rounds in a bank building is a random variable have an approximately normal distribution with 0.20=µ minutes and 6.3=σ minutes. What are the probabilities that a security guard will complete his rounds of the bank building in:

a) less than 13 minutes b) 17 to 22 minutes c) more than 22 minutes

139

6. A department store sales clerk knows that the number of sales she will make on a business day is a random variable having approximately a normal distribution with 9.30=µ and 1.5=σ . Find the probability that during a business day the sales clerk will make:

a) more than 31 sales b) fewer than 26 sales. 7. On the college soccer team, the mean weight of men is 65.6kg and a standard

deviation of 0.95kg. a) What is the probability that a certain man weighs over 63kg?

b) Below what weight will 35% of the men weigh? c) What is the probability that he weighs between 63kg and 66kg? d) Above what weight will the heaviest 5% of the men be?

Normal Approximation To The Binomial The Normal distribution can also be used as an approximation to the Binomial when n is large and when p is not too small or large. Using:

( )pnp

normalofdeviationdards

npnormalofmean

−=

===

1

tanσµ

Example 5 A fair coin is spun 15 times. Calculate the probability of the result giving exactly 8 heads. The Binomial Distribution is a discrete distribution. It is the distribution of integers or whole numbers. The Normal Distribution is a continuous distribution. It is the distribution of an infinite number of values.

140

When we use the Normal Distribution as an approximation to the Binomial Distribution, we have to use a correction factor of 0.5. For example:

5.45.34

5.35.23

<<=<<=

xbyreplacedisx

xbyreplacedisx

Hence in our example, 5.72

115 =

== npµ

( )

( ) ( )

−<<−=

<<==

=

=−=

936.1

5.75.8

936.1

5.75.7

5.85.78

936.12

7

2

1151

ZP

xPxP

pnpσ

7.5 8.5

( )1985.0

52.00

=<<= ZP

141

Check: The binomial distribution would be:

( )

( )( )( )

1964.0

108125.71090625.36435

2

1

2

18

33

78

815

=

××=

=

−−

CheadsP

142

EXAMINATION QUESTIONS

Multiple Choice Questions No. 1

1.1 In the normal curve of a distribution, what is the approximate area enclosed by

one standard deviation either side of the mean?

A. 68.27% B. 75.84% C. 95.45% D. 99.73%


1.2 Robot machines used in welding car bodies, in an automated car factory, carry out

1 000 operations per vehicle. It is expected that 1 operation in 40 will fail inspection. If more than 35 corrective operations are required on a single car body, it is removed from the production lines. What is the proportion of car bodies likely to be removed?

A. 5% B. 4.9% C. 2.5% D. 0.025%

(NATech, 1.2 Mathematics & Statistics, December 1998) 1.3 A coin is spun eight times. What is the expression, which would give the

probability that the coin gives exactly two heads?

A. ( ) ( )26 5.05.021

78

××

B. ( ) ( )35 5.05.0321

678

××××

C. ( ) ( )44 5.05.04321

5678

××××××

D. ( ) ( )53 5.05.054321

45678

××××××××

(NATech, 1.2 Mathematics and Statistics, December 1998)

1.4 A student is asked to do five questions from an exercise containing eight

questions. How many different selections are possible?

A. 56 B. 5 C. 336 D. 31

81


143

1.5 Each of four bags contains three coloured marbles, two red and one green. A

marble is drawn at random from each bag. What is the probability that four red marbles are chosen?

A. 5

1 B.

9

4 C.

3

1 D.

3

2


1.6 A box contains 10 electric bulbs, two of which are defective and the remainder

sound. What is the probability of selecting a sound bulb first and a defective bulb second?

A. 5

1 B.

5

4 C.

45

8 D.

25

4


1.7 A study of Excelsior Furniture Limited regarding the payment of invoices revealed that, on average, an invoice was paid 20 days after it was received. The standard deviation equated 5 days. What percentage of the invoice were paid within 15 days of receipt?

A. 20% B. 34.13% C. 15.87% D. 84.13%

(NATech, 1.2 Mathematics and Statistics, December 2002) 1.8 A pair of dice is tossed together. What is the probability that the sum of points on

the two dice is more than 8?

A. 36

5 B.

18

5 C.

9

2 D.

9

5

1.9 A local city council puts 10, 000 light bulbs on the streets in a city. If lives of

bulbs follow a normal distribution with a mean of 60 days and a standard deviation of 20 days, how many bulbs will have to be replaced after 20 days?

A. 228 B. 167 C. 477 D. 100

(NATech, 1.2 Mathematics and Statistic, June 2003)

1.10 Students take two independent tests. 30% of students pass test A and 60% pass

test B. What is the probability that a student selected at random passes only test A?

A. 0.12 B. 0.3 C. 0.6 D. 0.18

144

(NATech, 1.2 Mathematics and Statistics, December 1999 Rescheduled)

Multiple Questions No. 2

1.1 What is the third term of the binomial expansion ( )55.05.0 + ?

A. 32

5 B.

16

5 C.

8

5 D.

4

5

(NATech, 1.2 Mathematics, December 1999 (Rescheduled))

1.2 Statistics show that, on average, 5 boys are absent from Lusaka Boys’ High

School each day. Which of the following is the expression, which gives the probability that, on a day selected at random, exactly 7 boys are absent?

A. 757× B. 7

57

C. !7

575 ×−e D.

!5

755 ×−e

(NATech, 1.2/B1 Mathematics, December 1999 (Rescheduled))

1.3 HIV/AIDS vaccine is administered to 35 subjects and found to be effective in 7 cases. What is the approximate probability that the vaccine will be effective?

A. 35

7 B.

42

7 C.

35

28 D.

42

35

(NATech, 1.2 Mathematics and Statistics, Nov/Dec 2000)

1.4 Find the probability that an item drawn at random from the normal distribution

with mean 5 and standard deviation 3 will be between –1.24 and 1.37.

A. 0.2039 B. 0.9793 C. 0.385 D. 0.0204


1.5 Expand ( )32 qp +

A. 3223 662 qpqqpp +++ B. 3223 33 qpqqpp +++

C. 3223 6128 qpqqpp +++ D. ( ) ( ) 3223 3232 qpqqpp +++

(NATech, 1.2 Mathematics and Statistics, June 2002)

145

1.6 A normal distribution has a mean of 65 and a variance of 144. The probability of

a score of 80 or less is approximately:

A. 0.8944 B. 0.3944 C. 0.1056 D. 0.5398

(NATech, 1.2 Mathematics and Statistics, December 2001) 1.7 The number of accidents occurring on average each year in a factory is 36. They

occur completely randomly. Using the normal approximation to the poisson distribution, what is the probability that in 2002, there were more than 40 accidents?

A. 0.2266 B. 0.6622 C. 0.2626 D. 0.6226

(NATech, 1.2 Mathematics and Statistics, December 2004) 1.8 In a group of 100 NATech students 30 are male, 55 are studying for the

Foundation Stage and 6 of the male students are not studying for the Foundation Stage. A student chosen at random is female. What is the probability that she is not studying for the Foundation Stage?

A. 0.70 B. 0.56 C. 0.20 D. 0.45


1.9 Four coins are tossed. What is the probability of getting precisely three heads?

A. 4

2 B.

4

3 C.

4

1 D.

16

1


1.10 If IQ scores are normally distributed with a mean of 100 and a standard deviation

of 15, what proportions of people have IQs above 125?

A. 40% B. 0.4525 C. 0.048 D. 1.667


146

1.11 A box contains 400 bulbs of which 50 are red and 80 are blue. What is the

probability of drawing a ball that is neither red nor blue?

A. 8

1 B.

5

1 C.

40

13 D.

40

27


1.22 A manufacturer claims that his mass produced goods are no more than 2%

substandard. A potential buyer agrees to place an order if a random sample of 100 of the units gives no more than 2 defective items upon thorough testing. What is the probability of the order being place if the manufacturer’s claim is valid?

A. 0.677 B. 0.5 C. 0.02 D. 0.133


SECTION B (1) QUESTION ONE a) The distribution of the sales is normal with a mean of 150 items per week, and a

variance of 100 items.

Required: Find the probability that sales are less than 170 items in any week.

b) The probability that an item is defective is 0.02. How many defectives would you

expect to find in a batch of 4, 000 items? c) i) What is the major difference between a combination and a permutation?

ii) In how many ways can number five question be selected out of a total number of seven questions?


147

QUESTION TWO a) The time required to completely move operations from Mongu to Lusaka will

vary with a mean of 300 days and a standard deviation of 9 days. Assuming that the estimated durations are approximately normally distributed. Find the probability that the project will take to relocate:

i) Less than 280 days ii) More than 310 days iii) Between 280 and 310 days.

b) In order to manage product quality, all material receipts from suppliers are

inspected by inspection and quality control department and returned to the supplier once found to be defective. The number of orders randomly made in a month varies.

i) Assume a monthly average rejection rate of four (4) orders, what is the

probability that exactly five (5) orders will be rejected in one particular month? (To two decimal places)

ii) If the probability of rejecting an order is 30%, determine the probability

that at least three (3) out of eight (8) orders in a specific month made will be rejected. (To two decimal places).


c) Over a long period of time a drug has been effective on 55% of cases in which it has been prescribed. If 6 patients are treated by this drug, find the probability that it will be effective for:

i) At least 5 patients; ii) None of the patients; iv) 1 or 3 patients.

(NATech, 1.2 Mathematics and Statistics, December 2001) QUESTION THREE a) The weights of a certain item produced in large quantities over a long period of

time, is normally distributed with a mean of 8kg and a standard deviation of 0.02.

i) Any item whose weight lies outside the range 7.985 – 8.035 kg is taken to be faulty. What is the probability of a faulty item?

ii) If it is required to reduce the weight 8.035kg by 2%, find the new mean

weight.

148

b) A firm produces 55 percent of items on production line A and 45 percent of the

items on production line B. In general, 3 percent of the products of line A and 5 percent of the products of line B are found to be defective. If an item is subsequently returned as faulty, what is the probability that it was from line A?

c) In the manufacturing of clay pots, it was found that 8% were rejected. Calculate

the probability that a box of ten clay pots contained at most 2 rejects.


QUESTION FOUR a) The probability that a NATech student will graduate is 0.4. Determine the

probability that out of 5 students chosen at random:

i) None will graduate. ii) At least 1 will graduate. iii) At most 2 will graduate.

b) A company coach carries passengers by road regularly between two cities. The

average journey time between the two cities is 190 minutes; the standard deviation for journey taken is 20 minutes. Assume that the journey taken follows a normal distribution.

i) Find the probability that a coach journey between the two cities will take

less than 160 minutes. ii) If 150 journeys were made, how many journeys would take longer than

210 minutes? v) The coach company claims that 95% of all its journey take less than 3

hours 35 minutes. This claim is incorrect. Calculate the true percentage figure.


vi) The average number of errors on a page of a book is 1.2. Calculate the probability that a page will have 0, 1, 2, 3, 4, 5, 6 or 7 errors. You may assume that the probability of r successes in a possion distribution is given by the formula:

( )!r

erP

r λλ −

=

λ stands for the average. Check the answers from a table of poisson probabilities 3012.02.1 =−e .

(NATech, 1.2 Mathematics and Statistics, November/December 2001)

149

QUESTION FIVE a) Two (2) machines produce the same type of product. The older machine produces

45% of the total output but five (5) in every hundred are normally defective. The newer machine produces 55% of the total output and three (3) in every hundred are defective.

Determine the probability that a defective product picked at random was produced by the older machine.

(NATech, 1.2 Mathematics and Statistics, Nov/Dec 2000)

b) A banker claims that the life of a regular saving account opened with his bank averages 18 months with a standard deviation of 645 months.

i) The probability that there will still be money in 22 months in a savings

account opened with the said bank by a depositor? ii) The probability that the account will have been closed before two years?

(NATech, 1.2 Mathematics and Statistics, November/December 2000) c) One quarter of all accounts are found to contain errors. In a batch of 8 accounts

find the probability that the number of accounts containing errors is

i) more than 2; ii) less than 2.

(NATech, 1.2 Mathematics and Statistics, December 1999 (Rescheduled))

QUESTION SIX a) An urn contains 10 red and 5 white balls. What is the probability of drawing 2

balls, one of each colour?

b) An assembly line produces approximately 2% defective items. In a batch of 140 items, find the probability of obtaining:

i) Only two defective; ii) Less than two defective.

150

c) The number of errors made by a computer operator during 240 ten minute

intervals are given below.

No. of Errors 0 1 2 3 4 5 6 7 8 9 10 11 No.of Intervals

4

18

35

47

47

38

25

14

7

3

1

1

i) Calculate to 3 significant figures, the arithmetic mean number of errors per

interval. ii) Given that this distribution is possonian, calculate the poisson distribution

with the mean found in (i), above.

(NATech, 1.2 Mathematics and Statistics, December 1999 (Rescheduled)) QUESTION SEVEN a) Four normal (that is, fair) 6-sided dice are thrown simultaneously. Calculate the

probabilities of the result being 4 sixes, 3 sixes, 2 sixes, 1 six, or 0 six. b) The wages of workers in a large Lusaka Company are normally distributed with a

mean of K110, 000 per week and a standard deviation of K15, 000 per week. Find the probability of a worker selected at random, earning

i) Over K130, 000 per week; ii) Between K100, 000 and K140, 000 per week.

(NATech, 1.2 Mathematics and Statistics, December 1999 (Rescheduled)) c) 45% of voters are known to be African Congress Party (ACP). What is the

probability that out of a sample of 8 voters exactly 5 vote ACP?

(NATech, 1.2 Mathematics and Statistics, June 2003) QUESTION EIGHT a) During a recent African Cup final match, the number of footballers being attended

to by first-aiders during a 10-minute interval is known to have a mean of 3. What is the probability that 2 or more footballers arrive for attention during a 5-minute interval?

151

b) The number of patients admitted to a local hospital each day with TB has a mean

of 2. On a particular day, what is the probability that:

i) no patients are admitted to the hospital? ii) More than two patients are admitted to the hospital?

c) A man’s chance of winning a game is 5

3. If he plays five games, what are the

probabilities that he will win 0, 1, 2, 3, 4, 5 games respectively?


SECTION B (2)

QUESTION ONE a) A company does not estimate gross sales for the coming year as a single value,

but forecasts, instead, the probabilities of various amounts of gross sales. Suppose that this forecast shows the frequency distribution with a mean of K50, 000, 000 and a standard deviation of K3, 500, 000.

Calculate: i) The probability that gross sales will be less than K40, 000, 000 according

to this forecast. ii) The probability that gross sales will exceed K55 000 000. iii) The expected value of gross sales for the coming year.

b) In a traffic study, observations were made on the number of mini buses passing a

particular point each minute over a period of five hours. It was found that, on the average, one mini bus passed the point every minute. Calculate:

i) The probability that in any minute, the number of minibuses passing is 0,

1, 2, 3 and 4.

ii) The probability that at least one minibus passes the point in anyone minute.

152

c) A student, asked to answer three compulsory questions from an examination containing eight questions, ignores these instructions and answers four questions selected at random. Find the probability that he has answered:

(i) All the three compulsory questions set; (ii) Exactly two of the three compulsory questions set assuming that all

combinations were equally likely to be selected.

(NATech, 1.2 Mathematics and Statistics, December 2002) QUESTION TWO

a) A worker, on the average, spends 4

1of his/her time not working, and

4

3of his/her

time working. Five observations are made at random during a day at which time it is noted whether the worker is working or not working. What is the probability of making 0, 1, 2, 3, 4 and 5 observations when the worker is not working?


b) A quality product undergoes a strength test (S) and a reliability test (R). 100

products examined yield the following results.

45 pass S and pass R 16 pass S and fail R 19 fail S and pass R 20 fail S and fail R Find: i) The probability of a component chosen at random passing the strength

test. ii) The probability of a component, chosen at random, passing both the

strength test and the reliability test.

c) If X is normally distributed variable with a mean of 100 and a standard deviation

of 15, translate the following interval into an interval of Z scores.

13070 << X


153

QUESTION THREE a) Dr. Mulenga works on eight days each calendar month at Easy Go Batteries (z)

Limited. The probability that he turns up for his appointments at Easy Go on time is 0.75.

Use the binomial distribution to find the probability that Dr. Mulenga will turn up for his appointments on time on 0, 1, 2, 3, 4, 5, 6, 7, 8 days in the same month.

b) An average of 3 cars arrive at Chirundu Border Post every minute. If this rate is

approximated by a poisson process, what is the probability that exactly 5 cars will arrive in a 1-minute period?


c) On average, a bank cashier in the People’s Bank serves 1.5 customers per minute.

Find the probability that during a minute chosen at random, the bank cashier would serve:

i) No customers; ii) Exactly 3 customers.


QUESTION FOUR a) Three fair cubical dice are thrown. Find the probability that:

i) The sum of the scores is 5; and ii) None of the three dice shows a 6.

b) From past experience it is known that approximately 60% of applicants pass in

initial assessment test. In a group of five applicants, find the probability that:

i) All applicants pass the test; ii) Only three applicants pass; and iii) More than three applicants pass.

154

c) The heights of 250 boys at Kabwata High School are normally distributed with mean 150cm and standard deviation 16cm.

i) Estimate the number of boys whose heights are between 162.8cm and

170cm. ii) Estimate the number of boys whose heights are between 140cm and

145cm. (NATech, 1.2 Mathematics and Statistics, June 2001)

QUESTION FIVE a) Two fair dice are thrown. What is the probability that:

i) The total score is 2; and ii) The total score is 4?

b) A small machine shop has three sets of welding equipment. On average, during

an hourly period, one set is in use. Find the probability that:

i) No sets are in use; and ii) All sets are in use and there is a demand for a fourth set.

c) If a large grass lawn contains on average 1 weed per 600cm2, what will be the

poisson distribution of the number of weeds in an area of 400cm2? (NATech, 1.2 Mathematics and Statistics, June 2001)

QUESTION SIX a) According to The Journal of Nutrition , the mean mid-arm muscle circumference

(MAMC) for males is a normally distributed random variable with a mean of 273mm and a standard deviation of 29.18mm. The MAMC for John Banda has been measured at 341mm.

What is the probability that a male in the population will have a MAMC measure greater than Banda’s 341 mm?

b) The mean number of babies born per day in Chambeshi Teaching Hospital is 5.5.

i) Find the probability that there will be 0, 1, 2, 3, 4, 5, … babies born in the hospital on a day selected at random.

155

ii) Find the number of day in a period of six months (183 days) on which

there will be 0, 1, 2, 3, 4, 5, … babies born.

(NATech, 1.2 Mathematics and Statistics, December 1998) c) A Non-Governmental Organization (NGO), that provides counseling services to a

population of people who recently took an AIDS test, wishes to determine the “power” of the test. To this end, it hires a Research Analyst to investigate the frequency of “false positive” and “false negatives”. The analyst has observed that there are four possible categories. These categories, and the corresponding proportions are summarized in the table below.

Person’s HIV Status Results of AIDS Test Has HIV Has No HIV

HIV Positive 0.89 0.02 HIV Negative 0.05 0.04

One person from this population will be contacted today. Compute the probability that the person contacted:

i) Actually has HIV if it is known that he tested HIV positive; and

ii) Tested HIV positive if it is known that he actually has HIV.


156

4.3 Methods of Sampling and Survey Design

Sampling is a process of examining a representative number of items (people or things) out of the whole population or universe. The reason for sampling is to gain an understanding about some feature or attribute of the whole population, based on the measures of the sample. To find a good sample is often not easy. For example, we sample the fruit at the top of a basket we may have no idea if there is a bad fruit in the middle or at the bottom of the basket. If we are studying college students’ attitudes and we interview students on the steps of the School of technology, we may never encounter a business student. A sample only provides an estimate of a population measure and accuracy of the estimate will depend on: a) The right size for a sample, the larger the sample size the greater is the

probability that the sample is representative of the population. b) Selecting the right sampling method so that the sample represents the

population.

c) The extent of variability in the population.

Why do we sample? 1) Time and cost are probably the two most important reasons for sampling.

If one wants to determine if customers in a given market will buy a product, one doesn’t usually have the time or funds to interview all potential customers. For example, try interviewing everyone who uses protex in Zambia.

2) Testing may prove destructive. If you want to test the durability of your

product, stress tests on the entire output may leave you with no product to sell.

3) Accuracy is another reason for sampling. One would think that a study of

the entire population would be more accurate than a study of a sample. If one has a very large population and wishes to take a complete count, then one must hire a large number of inventory takers who must work at a rapid rate. As more personnel are hired, it is likely that they may be less efficient than the original employees. Thus, a limited number of skilled workers, studying a well defined sample, may provide more accurate results than a survey of the entire population.

157

Sampling Designs

Probability Samples

Taking a sample to use for the study of a population means that one must make sure that the sample really represents the population. The first problem is to decide what is the population that should be sampled. We need a sampling frame. A sampling frame is a list of every item or member of the population to be sampled. Sampling of individual items should be done at random i.e random sampling procedure yielding a representative sample requires that everyone in the population has an equal chance of being chosen.

Simple Random Sample Choosing people from a list may be done using a table of random numbers. If we were interested in sampling a group of 10 people, for example, we could number 10 pieces of paper from 0 through 9 and put them in a container. Then, we could mix the pieces of paper and draw out a number. This is essentially what has been done when a random number table is constructed. Such a table uses numbers of one, two, three, four or more digits, and a computer picks a number at random. These numbers are then recorded in a table such as Table 1 in the Appendix..

Table 1 shows a list of randomly selected numbers. Let us say that we wanted to select a sample of 20 people from a list of 500. First we number everyone on the list from 1 to 500. Now we might decide, which number to choose as the start of our sample. Assume we close our eyes and put our pencil down upon a number, say 62463, the sixth number in third column. Now, assume that we continue reading down this column from this point on and then start at the top of the next column. We discard numbers that are too large for our list (that is, numbers with three digits and less than or equal to 500 are retained). We could just as easily read across the rows. Any procedure is acceptable as long as we follow it consistently. Now let us pick 20 numbers we have, 022, 077, 065, 484, 140, 135, 239, 074, 037, 275, 474, 075, 145, 401, 264, 076, 449, 374, 230 and 041. Note that if the number was selected previously, we discard it. We now look up these numbers on our numbered list and interview the appropriate persons. Nowadays, many hand calculators have random number programs that can be used to generate a set of random numbers. The method just described above is called Simple random sampling.

158

Systematic Random Sampling

This method consists of starting with a random number from the random number table counting down to this number on our list. Then selecting additional units at evenly spaced intervals (every kth population unit )1>K until the sample has been formed. If the list follows some numerical pattern, this method can’t be used. The regular pattern might cause bias.

Stratified Random Sampling Another random sampling technique is called stratified random sampling , in which we divide our population into groups and take a selected sample size in each of the groups. This technique is used for two reasons: i) it can lead to reduced sampling error; and ii) it ensures a large enough sample in each stratum (class) for study of the particular

strata. As an example, we may be planning a survey to determine the profitability of offering a refuse collection service in a town. Recognizing that wealthy home owners might stratify the population of homes on the basis of value, forming three strata: homes valued at K30,000,000 and less, more than K30,000,00 but less than K100,000, and homes valued at K100,000 and more. From each stratums, we then take a random sample of selected homes.

Cluster Sampling Clusters are identified in the populations, a set of clusters is randomly sampled, and a complete census is taken of each to form the sample. Clustering tends to decrease costs and increase sampling error for the same size sample, because people who live close together are more likely to be similar than others. For example, a few geographical areas (perhaps a township or a road in a town) are selected at random and every single household of the population. In general, population units may be stratified on any number of characteristics common variables used for stratifying in ‘people” surveys are age, income, and location of residence.

159

Non Probability Samples

These samples do not rely on the laws of probability for selection, but depend on the judgment of interviewers or their supervisors. Convenience samples consist of studies of people who happen to be available or who call in their results. Suppose we are interested in conducting a study of the attitudes of shoppers at a new shopping center on the kinds of stores in the center, the attractiveness of the center, parking difficulties, and so on. To collect sample information, we ask persons to participate in the survey who happen to walk past the central area of the center. The sample in this instance is a convenient sample – the people are not being selected according to a probability plan and, presumably judgment is not being used in selecting those to participate in the survey. Convenience samples are prone to bias by their very nature – selecting population elements that are convenient to choose almost always makes them special ro different from the rest of the elements in the population in some way.

Judgment Sampling Judgment sampling is done by an expert who is familiar with the population measures. He selects the units from the population. The quality of judgment sample depends on the competence of the expert who selects the population units to be sampled.

Quota Sampling Quota sampling attempt to ensure that the sample represents the characteristics of the population. The interviewer is free to select any one who meet the given specifications. He or she may choose in a non random fashion. We cannot make good estimate of sampling error because we haven’t used a random sampling procedure. The method is cheap and reasonably effective and in consequence is widely used.

160

4.4 Statistical Inferences Introduction Statistical inference can be divided into two parts namely estimation and

hypothesis testing. Firstly we deal with estimation which is the procedure or rules or formulae used to estimate a population characteristics (parameter). Sample measures (or statistics) are used to estimate population measures (such as population means, µ the Greek symbol ‘mu’, population variance 2σ the Greek

symbol ‘Sigma’). The corresponding sample measures are sample mean,‘x ’ pronounced x-bar, and sample variance ‘2S ’ respectively.

Hypothesis testing is the process of establishing theory or hypothesis about some characteristics of the population and then draw information from a sample to see if the hypothesis is supported or not

Estimation

Properties of good Estimators

There are four properties of a good estimator

a) Unbiasness. An estimator is said to be unbiased if the mean of the sample

mean x of all possible random samples of size n , drawn from a population of size N, equals the population mean )(µ . Therefore the mean of the distribution of the sample means will be the same as the population mean.

b) Consistency. An estimator is said to be consistent if, as the sample size

increases, the accuracy of the estimate of the population parameter also increases.

c) Efficiency. An estimator is said to be efficient than any other if, it has

the smallest variance among all the estimators. d) Sufficiency. An estimator is said to be sufficient if it utilizes all the

information about it to estimate the required population parameter.

In practical situation, it is not possible to have all the four qualities on one estimator. The researcher choose which qualities he/she wants the estimator to have.

161

Distribution of Sample Means Consider a population with mean = µ and standard deviation = σ . Samples of size n

are taken from this population and the sample means x are found. The distribution of

sample means has mean µµ =x

and standard deviation (standard error) nx

σσ = .

If the population is normally distributed, or if the sample size is ‘large’ (i.e. )30>n , then the sample means is approximately normal.

Example 1 A normal population has a mean = 500 and standard deviation = 125. Find the probability that a sample of 65 values has a mean greater than 538. We have .125500 == σµ and The distribution of sample means has mean

50.1565

125tan500 ====

nerrordardsand

xx

σσµ

50.15=

xσ

500=

xµ 538Sample means

45.250.15

500538 =−=Z

Therefore the probability that a sample mean greater than 538 is equal to the area shaded = 0.5000 – 0.4929 = 0.0071 from the tables.

162

Example 2 The daily output from a production line has a mean 7500 unts with a standard deviation of 500 units. What is the probability that during the next 125 days the average output will be under 7400 units per day? We have .5007500 == σµ and The population of sample means is approximately

Normal (since n is large) with 72.44125

5007500 =====

nand

xx

σσµµ

72.44=

xσ

7400 7500=

xµ Sample means

24.272.44

75007400 −=−=Z

0125.0

4875.05000.0

)7400(

=−=

=< shadedareaSamplemeanP

Confidence Intervals

If we have chosen a good sample and have calculated the mean from the sample for the effect we wish to study, we may offer this estimate to the public or a company as an estimate for the population mean. This is called a point estimate. The only problem is that we offer evidence from one sample about the nature of the population, and we have no idea how reliable this estimate is. Reliability depends on sample size n and the amount of variability in the original population σ . Even more helpful would be a combination of the estimate, the standard

163

error of the estimate and some notion of the probability of being correct. This information is contained in what is called a confidence interval. Since sample means are normally distributed, we can use normal curve probabilities to describe our estimates. The following confidence intervals hold for the normal curve

IntervalconfidenceX

IntervalconfidenceX

x

x

%9065.1

%9596.1

=±

=±

σ

σ

In general the )%1( α− confidence interval is given by x

ZX σα2

± where

errordardstheis

isrightthetoareawhosepocriticaltheisZ

meansampletheisX

xtan

.2

int2

σ

αα

Example 3 Find the 95% confidence limits for the average daily output over 125 days given in Example 2. We have .5007500 == σµ and Also .72.44,7500 ==

xxσµ The 95%

confidence limits for a sample mean are:

65.758735.7412

65.877500

)72.44(96.1750096.1

to

xx

=±=

±=± σ

We are 95% sure that the average output over 125 days will be between 7412.35 and 7587.65 units per day. The principles involved in setting confidence limits can be used to determine what sample size should be taken, if we wish to achieve a given level of precision.

164

Example 4 In Example 2, if the daily output from a production line has a mean of 8000 units with a standard deviation of 534 units. If the probability is 0.99 that the error will be at most 116 points on the test scale for such data, how large should the sample size be? d = the error term which is half the width of the confidence interval. Hence d = 2.5

Therefore; .85.2,005.2

,01., 005.0

2

tablesthefromZn

Zd ==== αασα

Hence, n

)534(85.2116=

173

13.172116

)534)(85.2(

≅=

=

n

n

n

Round up to the next whole number Finite Population Correction Factor (FPC) For random samples of size n taken from a finite population having the mean µ and

standard deviation σ , the sampling distribution x has the mean µµ =x

and the standard

deviation 1−

−=N

nN

nx

σσ

Where N is the population size and the sample size is n > 5% of the population size.

165

Example 5 A sample of 120 is drawn from a population of 1200 with a sample standard deviation of 9 centimeters. What is the finite population correction factor? What is the standard error of the mean?

cmN

nN

n

fpc

ThereforeNof

NifnN

nNfpc

x7797.0)949.0(

120

9

1

949.011200

1201200

.60)1200(05.%5

%51

==−−=

=−

−=

==

>−−=

σσ

Estimation of Population Proportions So far the process of statistical inference has been applied to the arithmetic mean. The standard error of a sample proportion is given by

Pandn

pqpp == ˆˆ µσ

where p is the sample proportion, P is the population proportion and .1 pq −= Using this value, we are able to set confidence interval for the estimate of the population proportion based on the sample proportion in exactly the same way as outlined previously for the mean as:

n

pqZp

2

ˆ αµ ±

166

Example 6 A manufacturing process produced approximately 5% defective items. a) Find the mean and standard deviation of the proportion of defectives obtained in

sample 500 items. b) Find the 95% confidence limits for the proportion of defectives in a sample of 500

items.

a) We have 500,05.0 == nandp

00975.0500

)95.0(05.0

)1(tan

05.0

ˆ

ˆ

==

−==

===∴

n

PPdeviationdardS

Pmean

p

p

σ

µ

b) Assuming Normal distribution for the proportion of defectives in a sample, the

95% confidence limits are given by:

06911.003089.0

01911.005.0

)00975.0(96.105.096.1ˆ

to

p

=±=

±=±µ

Thus the 95% confidence limits for the proportion of defectives in a

sample are 3.1% to 6.9%.

Example 7 In a random sample of 375 employees, 68% were found to be in favour of strike action. Find the 99% confidence limits for the proportion of all employees in the company who are in favour of such action.

167

We are given 37569.0ˆ == nandp

0241.0375

)32.0)(68.0()ˆ1(ˆ

68.0ˆ

ˆ

ˆ

=

=−=

==∴

n

pp

p

p

p

σ

µ

Assuming a normal distribution the 99% confidence limits are:

7422.06178.0

0622.068.0

)0241.0(58.268.058.2 ˆˆ

to

pp

+=

±=± σµ

The 99% confidence limits are 61.78% to 74.22%. Estimation From Small Samples. When sampling is done where sample size is less than 30 and the population variance is unknown (i.e. S, the sample standard deviation is used as an estimate of σ , the population standard deviation). The confidence limits for the time population mean are given by

n

Stx

n 1,2

−± α

Where 1,

2−n

tα is student’s t distribution value and 1−n is the degree of freedom.

1

tan

22

−−

= ∑n

xnxS

meansampletheofdeviationdardstheS

meansampletheisx

the t variable is defined by the following formula

168

nS

x

t

µ−

=

where x is the mean of a random sample of n measurements, µ is the population mean of the x distribution, and S is the sample standard distribution. The characteristics of the t-distribution are: i) It is an exact distribution which is symmetrical and mould shaped. ii) It is flatter than the Normal distribution i.e. the area near the tails are greater than

the Normal Distribution. iii) As the sample size becomes larger the t distribution approaches the normal

Distribution. To use the t distribution the tables in Appendix 1.

Example 8 A random sample of 12 men is taken and is found to have a mean height of 1.67cm and a standard deviation of 0.48cm. Find: i) 99% ii) 95%

confidence limits for the population mean height.

x = 1.67, n = 12 and S = 48. .005.02

01.,991 ===− ααα and Hence

106.311,005.01,

2

==−

ttn

α from Table 2. The 99% confidence limits are given by:

169

cmtocm

n

Stx

n

1.224.1

430.067.1

12

48.0106.367.1

1,2

=

±=

±=

±−

α

Thus the 99% confidence limits for the population mean are between 1.24 and 2.1cm.

ii) 201.2.025.02

,05.0,95.1 11,025.01,

2

=====−−

ttThereforen

αααα , from

Table 2 the 95% confidence limits are given by

975.1365.1

305.067.1

12

48.0201.267.1

1,2

to

n

Stx

n

=

±=

±=

±−

α

Thus the 95% confidence limits for the population mean are between 1.365 to

1.975cm.

Exercise 4 1) A study of a sample of 500 bank accounts is made to estimate the average size of

a bank account. The sample mean is calculated to be K500 000. From previous studies of bank accounts, it is known that the standard deviation is K67 000. Construct a 99% confidence for the mean size of bank accounts.

170

2) An ice cream factory wishes to know the average number of women per block of houses in a given compound. A sample of 120 blocks of houses within the compound indicates that the average number of women is 94. When the standard deviation is estimated for those 120 block of houses, it is found to be S = 15.04. Calculate a 95% confidence interval for the number of women.

3) Assume that some college students want to find out what percent of the

population will vote for the MMD candidate. A sample of 125 voters reveals that 65 will vote MMD. Should we predict that the MMD candidate will win (assuming that there are just two parties)? Construct a 98% confidence interval that the MMD will win.

4) A study of a sample of 125 customers at the college bookstore indicated

20% preferred new books while 80% wanted used ones.

a) Estimate the standard error of a proportion for those favouring new books. What is the standard error of a proportion for those favouring old books?

b) Construct a 99% confidence interval for the proportion of the

population favouring new books. c) Construct a 95.5% confidence interval for the proportion of the

population favouring new books.

5) A group of college student is trying to determine the appropriate sample size to use. They wish to be within 2% of the true proportion with 95% confidence. Past records indicate that the proportion of defective is 9 in 300. What sample size should they use?

6) Past experience has indicated that the salaries of factory workers in a

certain industry are approximately normally distributed with a standard deviation of K225 000. How large a sample of factory workers would be required if we wish to estimate the population mean salary µ to within K27 000 with a confidence of 95%?

7) The length of time required for persons taking a Mathematics test is

assumed to be normally distributed. A random sample of 25 persons taking the test is conducted and their test times are recorded, yielding an average test time 120 minutes with a standard deviation of 24 minutes. Find a 99% confidence interval for the population mean test time µ .

8) The Local Authority have tested the durability of a new paint for white

center lines, a highway department has painted test strips across heavily travelled roads in nine different locations, and automatic counters showed that they disappear after having been crossed by 200, 245, 235, 225, 220,

171

230, 235, 248 and 250 cars. Construct a 99% confidence interval for the average number of crossings this paint can withstand before it disappears.

4.5 Hypothesis Testing

Hypothesis testing or significance testing is in many ways similar to the process of estimation dealt with in the previous section. Random sampling is involved and the properties of the distribution of sample means and proportion are still used. A hypothesis is some testable belief or opinion and hypothesis testing is the process by which the belief is tested by statistical means.

This belief about the population parameter is called the null hypothesis denoted oH . The

value specified in the null hypothesis is often a historical value, a claim or a production specification. The opposite of the null hypothesis is the alternative hypothesis denoted by .1HorH a

For example, if the average score of Mathematics students is 85, 10 years ago, we might use a null hypothesis 85: =µoH for a study involving the average score of this year’s

mathematics class. If television networks claim that the average length of time devoted to commercials in a 45-minutes program is 10 minutes, we would use 10: =µoH

minutes as our null hypothesis in a study regarding the length of time devoted to commercials. The alternative hypothesis is accepted if the null hypothesis is rejected. In the two examples above, if we believe the average score of mathematics students is greater than it was 10 years ago, we could use an alternative hypothesis 85:1 >µH while in the commercial case if the length of time devoted to commercials is not10 minutes, we could use an alternative hypothesis 10:1 ≠µH . If we reject the null hypothesis when it is in fact true, we have an error that is called a type 1 error. On the other hand, if we accept the null hypothesis when it is in fact false, we have made an error that is called a type II error . Table 4.1 summarizes these results. Table 1.1 Type I and II Errors

Our Decision

Truth of Ho Accept Ho Reject Ho

Ho is true Correct decision Type I error

172

Ho is false Type II error Correct decision

To investigate the significance of sampling mean we evaluate

n

xZei

xZ

x/

.σ

µσ

µ −=−=

A selection of ‘significant’ values of Z together with the significance level α are given below.

α .05 .025 .01 .005

Z 1.64 1.96 2.33 2.58

Example 9 Consider a Normal population with a standard deviation = 30. A random sample of 24 items is found to have a mean of 168. Test the assumption at the 5% significance level that the population has a mean of 150. Assumption: population mean = .150=µ Alternative mean .150≠µ We write

150:

150:

≠=

µµ

a

o

H

H

This type of a test, we call it two tailed test. If 150≠µ , it could either be less than 150 or greater than 150, hen the term two tailed test.

5.02

=α 025.

2=α

95.1 =−α -1.96 1.96

173

We are given α = .05. Since it is a two tailed test, we share this area equally in the two

tails i.e .025.2

05.

2==α

The shaded area is called the rejection region and the unshaded

area the acceptance region. The point separating the rejection region from the acceptance region is called the critical point.

We are given that nowxandn ,16824,30 ===α

94.212.6

18

24/30

150168

=

=

−=−=

n

xZ σ

µ

A ‘significant” value of Z at the 5% level is 1.96, i.e the 95% confidence limits for Z are –1.96 and +1.96 (see diagram above). Therefore, our value of Z is significant (i.e it is outside the confidence limits). We reject oH . We thus accept that the population mean

is not equal to 150.

Example 10 A random sample of 12 family toy cars is found to have an average retail price of K300 000. Assuming that toy car prices are Normally distributed with a standard deviation of K50 000.00, test the assumption (at the 5% level) that the average price of a family toy car is: a) K35 000, and b) more than K35 000

We are given 00030012,00050 === xandnα a) 00035: =µoH

00035:1 ≠µH this is a two-tailed test. We have

174

46.3

4641.3

0005000050

12

50000350000300

−=

−=

−=−=

n

xZ σ

µ

025. 025. 95.1 =−α −1.96 1.96

Now a significant value of Z at the 5% level is 1.96. i.e. we reject oH if Z value is

greater than +1.96 or less than –1.96. Therefore, our value of Z(-3.46) is significant. We reject the assumption oH . Thus the sample shows that the

average price of a family car is significantly different from K350 000. b) 00035: =µoH

00035:1 >µH This is a one-tailed test. We have

46.3

12

50000350000300

−=

−=−=

n

xZ σ

µ

175

5% 0 1.65 Z

Therefore, our value of Z(= -3.46) is not significant. We accept oH . We thus

accept the assumption that the average family car is equal to K350 000.

Example 11

A random sample of 12 items is obtained from a Normal population and is found to have a mean = 50 with a standard deviation = 7. Test the assumption, at the 5% significance level, that the population mean is 40.

We are given 5012,7 === xandnS

40: =µoH

40:1 ≠µH (two tailed test). Now the sample size is small and hence the test statistic is no longer Z but tgiven by

95.4

021.2

10

12

74050

=

=

−=−=

n

Sx

tµ

Now, a significant value of t at the 5% level (two tailed ) with n –1 = 12 = 11 is 2.201. Therefore, our value t = 4.95 is significant. We reject oH . We thus conclude that the

population mean is significantly different from 40.

176

Example 12 An assessment test is given to all prospective employees in a company. Test scores are known to be Normally distributed. A random sample of 7 participants obtained the following results: 69, 58, 68, 66, 75, 85, 80. Test the assumption that the mean test score is 65 using the 5% significance level.

65: =µoH

65:1 ≠µH This is a two-tailed test. We have x : 49, 58, 66,75, 85, 82. Now

823.0

86.4

4

7

858.126569

858.12

6

)69(734319

1

,697

483

222

=

=−=−=∴

=

−=−−

=

===

∑

∑

t

n

Sx

t

n

xnxS

andn

xx

µ

From the tables, a significant value of t at 5% level with n – 1 = 7 – 1 = 6 is 2447. Therefore, our value of t = 0.823 is not significant. We can accept oH . We thus

conclude that the average test score is not significantly different from 65.

Example 13

The amount of monthly income tax paid by employees is approximately Normally distributed. A random sample of 25 employees paid an average of K350 000 per month in income tax, with a standard deviation of K160 000. At the 5% significance level test the assumption that the average amount of income tax paid is greater than K250 000 per month per employee.

177

We are given 35025,160 === xandnS We write: 000250: =µoH

000250: >µaH

This is a one tailed test. Now

125.3

25

000160000250000350

=

−=−=

t

n

Sx

tµ

From the tables, a significant value of t at the 5% level with n – 1 = 24 is 1.111. Therefore, our value of t = 3.125 is significant. We can reject oH and accept the

assumption that the average income tax paid by employees in the company is significantly greater than K250 000 per month.

Hypothesis Testing of Proportion

A population contains proportion P of ‘successes”. Random sample of size n are taken from this population. The proportion of ‘successes” in the samples are distributed with a

meann

PPdeviationdardsandP pp

)1(tan ˆˆ

−== σµ

If n is ‘large’ the sample proportion are approximately Normally distributed. The significance of a sample proportion p can be examined using the formula

p

pPZ

ˆ

ˆˆ

σµ−

=

Example 14

It is assumed that over half of the employees in a large company are in favour of a proposed new salary structure. A sample of 250 employees found that 42% were in favour. Does this sample verify the assumption? (use the 1% significance level) We have 250,42.0ˆ == np

50.0: =PHo

178

50.0: <PH i

Now,

53.20316.0

50.042.0ˆ

0316.0250

)5.0(5.0)1(

ˆ

ˆ

ˆ

−=−=−

=∴

==−=

p

p

p

PZ

n

PP

σµ

σ

A significant value of Z at the 1% level (one-tail) is –2.33. Therefore, our value of Z = -2.53 is significant. We reject oH . We thus accept the assumption that the population

does have a proportion less than 50%, i.e. less than half of the employees are in favour of the proposed new salary structure.

Example 15 It is required to test the hypothesis that 56% of households have a television set. A random sample of 500 households found that 75% of the sample had television sets. The significance level is 1%.

We have %.1,75.0ˆ,500,06.0 ==== αPnP This is a two tailed test because we wish to test the hypothesis as it is and not against a specific alternative hypothesis that the real proportion is either larger or smaller. i.e 56.0: =PHo

56.0: ≠PH i

Now,

56.80222.0

56.75.0

0222.0500

)44(.56.0ˆ

=−=∴

==

Z

pσ

At the 1% level of significance for a two-tailed test the appropriate Z value is 2.58.

179

Therefore, our value of Z = 8.56 is significant. We reject oH . We thus accept the

assumptions that the proportion of household who have a television set is not 56%. Exercise 5

1) A Normal population has a standard deviation of 50. A random sample of 30

items is found to have a mean of 270. Using the 1% significance level examine the assumption that the population has a mean of 280.

2) A machine makes twist-off caps for bottles. The machine is adjusted to make

caps of diameter 1.87cm. Production records show that when the machine iis so adjusted, it will make caps with mean diameter 1.87 cm and with standard deviation .045.0 cm=σ During an inspector checks the diameters of caps to see if the machine is not functioning properly in which case the diameter is no longer

1.87cm. A sample of 65 caps is taken and the mean diameter for this sample x is found to be 1.98cm. Is the machine working properly i.e 87.1≠µ . (Use a 5% level of significance).

3) Monthly salaries in a company are normally distributed with a standard deviation

of K94500. A sample of 20 employees is found to have a mean salary of K756 000 per month. Using the 1% level of significance would you conclude that the average salary in this company is significantly higher than K720 000 per month?

4) A random sample of seven bank accounts show balances equal to: K270 000,

K120 000, K1600 000, K620 000, K1980 000, K3200 000, K2600 000. Test the assumption that the mean bank balance is K1250 000. (use the 5% significance level).

5) A sample of 28 items from a normal population is found to have a mean of 550

with a standard deviation of 69. At the 5% level test the assumption that the population has a mean of 520.

6) From a random sample of 95 Zambian companies, it is found that 36 companies

had annual turnovers in excess of 30 million kwacha. Using a 1% significance level, test the assumption that 45% of all Zambian companies have over 30 million kwacha annual turnover.

7) A team of eye surgeons has developed a new technique for a risky eye operation

to restore the sight of people blinded from a certain disease. Under the old method, it is know that only 45% of the patient who undergo this operation recover their eyesight. Suppose that surgeons in various hospitals have performed a total of 230 operations using the new method and that 98 have been successful (the patients fully recovered their sight). Can we justify the claim that the new method is better than the old one? (use a 5% level of significance).

180

Hypothesis Testing of the Difference Between Two means The distribution of sample mean differences is normally distributed and remains normally distributed whatever the distribution of the population from which the samples are drawn. When 30>n i.e. large samples, the Normal area tables are used. When 30<n the t distribution are used.

The standard errors of the 2

22

1

21

)( 21ˆ

n

S

n

Sxx

+=−

σ difference of means where:

1S = Standard deviation of sample 1, size 1n .

2S = Standard deviation of sample 2, size 2n and the Z score is calculated thus:

21

ˆ21

xx

xxZ

−

−=σ

Example 1

A psychological study was conducted to compare the reaction times of men and women to a certain stimulus. Independent random samples of 50 men and 50 women were employed in the experiment. The results were shown in the table below. Do the data present sufficient evidence to suggest a difference between time and mean reaction times for men and women? Use 05.0=α

Men Women

20

43

50

21

1

1

==

=

S

x

n

12

37

50

22

2

2

==

=

S

x

n

181

We have

)(:

:

21

21

tailedtwoH

H

a

o

−≠=

µµµµ

( )

8.050

12

50

20

55

2

22

1

21

21

=

+=

+=−

∧

nnxxσ

Now

5.7

8.0

3743

21

21

=

−=−=

− xx

xxZ

σ

A significant value of Z at the 5% level with 96.1=Z . Therefore our value of 5.7=Z is significant. We reject 0H . We thus conclude that there is a significant difference

between the average earnings in the two companies.

Example 2 A consumer group is testing camp stoves. To test the heating capacity of a stove, the group measures the time required to bring 2 litres of water from 10ºc to boiling (at sea level). Two competing models are under consideration. Thirty-seven of each model are tested and the following results are obtained.

Model 1: mean time 5.121 =x min; standard deviation 6.21 =s min

Model 2: mean time 1.102 =x min; standard deviation 2.32 =s min Is there any difference between the performances of these two models (use a 1% level of significance)?

182

We have

21

21

:

:

µµµµ

≠=

a

o

H

H

54.3678.0

4.2

37

)2.3(

37

)6.2(

1.105.1222

2

22

1

21

21

==

+

−=+

−=

n

S

n

S

xxZ

Z = 3.54 is greater than the critical value of Z = 2.58. Therefore , we reject oH and

accept the assumption that there is a significant difference between the performances of these two models. Hypothesis Testing of the Difference Between Proportions In a similar manner it may be required to test the difference between the proportions of a given attribute found in two random samples. The following symbols will be used. Sample 1 Sample 2

Sample proportion of successes 1P 2P

Population proportion of successes 21 PP

Sample size 21 nn The assumption is that the two sample are from the same population. Hence the pooled sample proportion.

183

)1(ˆ

)()ˆˆ(

,ˆtan

)(

2121

21)(

21

2211

21

21

→−−−=

+=++=

−

−

PP

Pp

PPPPZ

andn

pq

n

pqiserorrdardstheand

nn

nPnPP

σ

σ

But under the null hypothesis ,: 21 PPH = hence (1) is reduced to:

)(

21

21ˆ

)ˆˆ(

PP

PPZ

−

−=σ

Example 3 The following results have been recorded from random samples of candidates taking two Institute examinations.

Examinations No. of candidates sampled No. of passes

Communication Mathematics

50 85

35 42

Use the 1% level of significance to examine whether there is a significant difference in the proportions of candidates passing the two examinations. Let Communication be sample 1 and mathematics be sample 2

184

38.20883.0

49.07.0

568.0135

65.76

8550

)85(49.0)50(7.0

0882.0

85

)432.0)(568.0(

50

)432.0)(0568(.ˆ

:

:

49.085

42ˆ7.050

35ˆ

8550

)(

21

21

21

21

21

=−=∴

==++=

=

+=

≠=

====

==

−

Z

PNow

PPH

PPH

PP

nn

PP

a

o

σ

A significant value of Z at 1% level (two tailed is 2.58. Therefore, our value of Z = 2.38 is not significant. We accept oH . There is no significant difference between the

proportion of candidates passing the two examinations. Example 4 A college committee wishes to know if the proportion of students who received A grades was decreasing as a result of the committee recent report to the Principal that showed that grades had risen since 2002 and that Mampi College grades had risen faster than grades in other college in the country. A sample of grades in 2002 and 2003, after the committee’s report was given, were studied to see if the proportion of A grades had gone down significantly. Use .05.0=α

Year Proportion of A grades Number of students 2002 2003

0.70 0.50

120 110

Let 2002 results be sample 1, and 2003 results be sample 2 We have:

185

604.0110120

)110(50.)120(70.0

:

:

12

12

=++=

<≥

PNow

PPH

PPH

a

o

065.0110

1

120

1)396.0)(604.0(ˆ )( 21

=

+=− PPσ

This is a one tailed test. The critical Z value is –1.65.

08.3065.0

70.050.0 −=−=Z

Since –3.08 is less than –1.65, we reject oH . We therefore conclude that the grades have

gone down since the grade committee reports was issued. Exercise 6 1) A quality inspection of two production lines gave the following results

Production line Sampling size No. of defectives X

Y

45

35

4

9

Use a 5% significant level to test the claim that line A is more reliable than line B. 2) In a survey of voting intentions prior to a local government elections, 45% from a

random sample of 400 voters said that they intend to vote for the MMD candidate. In a second area in the same constituency there were 32% intending to vote MMD in a sample of 370. Use a 1% level of significance to determine whether there is a significant difference between voting intentions in the two areas.

3) The graduating class to two prestigious business schooks were surveyed about

their average starting salary with the following results.

186

School Average starting salary(K’ million)

Standard deviation

(K’ million)

Sample size

X

Y

80

85

1.8

1.44

150

115

At a 0.05 confidence level, do we have adequate reason to believe that graduate of

School Y have equal starting salaries? 4. A calculator Company was trying to decide between two brands of batteries to

recommend in its calculators. If the batteries were of equal life, the company preferred brand 1 because of its better distribution network. Based on the following data and using 5% confidence level, which battery should the quality control engineer recommend?

Battery Mean life (hr) Standard Deviation(hrs)

Sample size

Brand 1

Brand 2

110

115

15

20

150

150

5) Consider the following null and alternative hypotheses.

0:

0:

21

21

>−≤−

µµµµ

a

o

H

H

Sample of size 3030 21 == nandn are planned to test this null hypothesis.

Suppose it is known that .24,12 21 == σσ Further, suppose that the two population are taken from each population independently with mean

2030 21 == xandx respectively. Should the null hypothesis be rejected or not rejected. Explain. Use .05.=α

187

EXAMINATION QUESTION WITH ANSWERS


1.1 A simple random sample of size 25 is drawn from a finite population consisting of 145 units. If the population standard deviation is 10.5, what is the standard error of the sample mean when the sample is drawn with replacement?

A. 5.0 B. 2.1 C. 1.45 D. 1.92

(Natech, 1.2 Mathematics and Statistics, June 2003) 1.2 A Corkhill machine is set of fill a small bottle with 9.00 grams of medicine. It is

claimed that the mean weight is less than 9.0 grams. The hypothesis is to be tested at the 0.01 level. A sample revealed these weight (in grams): 9.2, 8.7, 8.9, 8.6, 8.8, 8.5, 8.7 and 9.0. What are the null and alternative hypotheses?

A. 0.9:,90:90:,90: 11 >≠<= µµµµ HHBHH oo

C. 0.9:,0.9:0.9:,0.9: 11 =≠<> µµµµ HHDHH oo

(Natech, 1.2 Mathematics and Statistics, June 2001)

1.3 A study of Excelsior Furniture limited regarding the payment of invoices revealed

that, on average, an invoice was paid 20 days after it was received. The standard deviation equaled 5 days. What percentage f the invoices were paid within 15 days of receipt?

A. 20% B. 34.13% C. 15.87% D. 84.13%

(Natech, 1.2 Mathematics and Statistics, December 2002)

1.4 A procedure based on sample evidence and probability theory used to determine

whether a statement about the value of a population parameter is reasonable and should not be rejected, or unreasonable and should be rejected is called:

A. Forecasting B. Hypothesis C. inferential statistics D. Hypothesis testing.


188

1.5 If, in a sample of 150, 60 respondents say they prefer product P to productQ ,

then the standard error of the sample proportion is: A. 0.0016 B. 0.04 C. 0.4 D. 0.6

(Natech, 1.2 Mathematics and Statistics, December 2001) 1.6 Which of the following does not require a sampling frame?

A. Cluster sampling B. Quota sampling C. Stratified sampling D. Random sampling

(Natech, 1.2 Mathematics and Statistics, December 2003) 1.7 The finite correction factor is used in computing the standard error of the mean

when: A. X is infinite. B. N is finite. C. n is finite D. σ is finite. 1.8 In a “5 percent two-tail test” Concerned with the value of the population mean,

the area in each tail (region of rejection) of the standard-normal distribution model is:

A. 0.025 B. 0.05 C. 0.10 D. 0.95

1.9 Type I error refer to the error of:

A. Accepting a true null hypothesis B. Rejecting a true null hypothesis C. Accepting a false null hypothesis D. Rejecting a false null hypothesis

1.10 In the general procedure of hypothesis testing the “benefit of doubt is give to the: A. Sample statistics B. Test statistic C. Null hypothesis C. Alternative hypothesis

189

SECTION B

QUESTION ONE a) State the null and alternative hypotheses given the following information. During the last year, the average quarterly charge on a current account held at a bank was

K25,000. The bank wishes to investigate whether the amount paid in charges has increased or not. So they sample 50 accounts and obtain a mean of K25,500.

b) The mean breaking strength of the cables supplied by a manufacturer is 1,800

with a standard deviation of 100. By a new technique in the manufacturing process it is claimed that tbe breaking strength of the cables has increased. In order to test this claim a sample of 50 cables is tested. It is found that the mean breaking strength is 1,850. Can we support the claim at 0.01 level of significance?


c) A company manufactures support struts which have a mean breaking strength of 125 KN with a standard deviation from the mean of 185KN. As a result of trials with more expensive raw material, a batch of 25 struts with a mean breaking strength of 1310KN is produced. Is this evident that the new material is producing stronger struts?

(Natech, 1.2 Mathematics and Statistics, June 2003) QUESTION TWO a) Distinguish between a point estimate and an interval estimate. b) ZACB, a training Institution runs a number of promotional activities for their

programmes, one of which is called “ZACB hour”. In order to ascertain the effectiveness of this promotional activity, a sample of 500

new students was randomly selected and asked how they came to know about the institution’s programme. 185 of the 500 students indicated that it was through “ZACB hour”.

Required:

i) Compute a 95% confidence interval for the population proportion of

students to be interviewed if the sample promotion is to lie within 1% of the true population proportion at 95% confidence interval level.

190

ii) Calculate the number of students to be interviewed if the sample promotions is to lie within 1% of the true population proportion at 95% confidence level.

(Natech, 1.2 Mathematics and Statistics, December 2004) QUESTION THREE a) i) The mean and the standard deviation of the height of a random sample of

100 students are 168.75cm and 7.5cm respectively.

Required: Calculate the 99% confidence interval for the mean height of all students

at the college.

ii) In measuring the reaction of individuals, a psychologist estimates that the standard deviation of all times is 0.05 second.

Required: Calculate the smallest size necessary in order to be 95% confidence that the error in the estimate will not exceed 0.01 seconds.


b) A sample of 10 measurement of the diameter of a marble gave a mean of 4.38 mm

and a standard deviation of 0.06mm. Construct a 99% confidence interval of the population diameter.


c) A weekly turnover of a small retail company is Normally distributed with an

average of K42 000 per week.

Following an advertising campaign a seven-week period produced an average turnover of K49 000 per week. With a standard deviation of K4 500. At the 1% level test whether there has been a significant increase in the turnover.


191

d) A random sample of six (6) bank accounts showed balances equal to

K232,200; K2,752,000; K843,200; K1, 376,000; K421,600; K1,346, 400. Test the assumption that the mean bank balance is K1,032,000. (use the 1% significance level).

(Natech, 1.2 Mathematics and Statistics, Nov/Dec 2000) QUESTION FOUR a) The Manager of a coach company wishes to estimate the average daily number of

kilometers covered by his coaches. He requires a confidence limit of 80%, but the error must be within plus or minus 20 kilometers of the true mean.

Assuming that the previous investigation have indicated that a very good estimate

for the population standard deviation is 130 kilometers.

i) Find the required sample size. ii) Suppose the sample size is greater than his fleet of coaches, what steps

should be taken?

(Natech, 1.2 Mathematics and Statistics, December 2001) b) XYZ Ltd has developed a cleaning detergent for use by housewives in their

homes. A decision must now be made whether or not to market the detergent. Marketing the product will be profitable if the mean number of units ordered per household is greater than 3.0 and unprofitable if less than 3.0. The decision will be based on the sales potential shown in the home demonstrations of the detergent with a random sample of the housewives in the target market.

i) Specify the null and alternatives hypotheses if the more costly error is to market

the detergent when it is not profitable. ii) Specify the null and alternative hypotheses if the more costly error is to

fail to market the detergent when the mean number of units ordered per household exceeds 3.0.

192

c) An instructor wishes to determine whether or not student performance has changed over the duration of the course. Scores in two equivalent test in “mathematics proficiency”, one before and the other after the course, are summarized in the Table below.

Student 1 2 3 4 5 6 7 8 9 10 11 Before 29 61 73 51 49 71 33 48 44 75 55 After 55 67 73 35 48 93 59 47 42 60 46

Is there any statistical evidence that the course has produced some learning? Use

05.0=α .


QUESTION FIVE a) Formulate the appropriate null and alternative hypotheses in each of the following

situations.

i) A large mining company has a mean level of absenteeism of 98 workers per 1,000 at anyone time. To help reduce this rather high rate of absenteeism, the company has introduced a new attendance bonus. The company now wishes to determine whether absenteeism ahs declined.

ii) Suppose you own a book shop that sells a variable number of books per day, and that if the mean number of books sold is less than 20 per day, you will eventually be bankrupt. If the mean number of books sold exceeds 20 per day, you care financially safe. You wish to determine whether your sales (from the bookshop) are leading to a financial disaster.

iii) When it is operating correctly, a metal lathe produces machine bearing

with a mean diameter of 1.27cm. Otherwise, the process if out of control. A quality control inspector want to check whether the process is out of control.

b) A relief organization knows from the previous studies that the average distance

that each family has to walk to fetch water is 5.6km. A small capital investment programme is initiated to sink boreholes to address this problem.

Two months after the completion of the programme, a sample of thirty families

revealed that the mean distance is now 5km with a standard deviation of 1.4km. Is this a significant improvement? Conduct your test at 5% level of significance.


193

QUESTION SIX a) From a recent survey conducted on environmental issues, out of 200 persons

sampled. 1600 favoured more strict environmental protection measures. What is the estimated population proportion?

b) It is known that the distribution of efficiency ratings for production employees at

Bearing Centre Ltd is normally distributed with a population mean of 200 and a population standard deviation of 16. The research department if challenging this mean, stating it is different from 200. As a result the efficiency ratings of 100 production employees were analyzed and then mean of the sample was computed to be 203.5.

With this information and using the 0.01 level of significance, test the hypothesis

that the population mina is 200.


194

4.6 Regression and Correlation Analysis This section introduces regression analysis which is a method used to describe a

relationship between two variables and goes on to explain about correlation analysis which measures the strength of the relationship between two variables. This manual uses Spearman’s rank correlation coefficient and Pearson’s product moment coefficient of correlation as a measure of strength between two variables. Regression analysis is concerned with the estimating of one variable (dependent variable) on the basis of one or more other variables (independent variable). If an analyst for instance is trying to predict the share price of a particular sector there will be a whole range of independent variables to be considered. In this manual, we will restrict our attention to the particular case where a dependent variable y is related to a single independent variable x .

The Regression Equation

When only one independent variable is used in making forecast, the technique used is called Simple Linear regression. The forecasts are made by means of a straight line using the equation bxay +=

xinchangetuniawithchangesythatamounttheslopeb

xwhenerceptythea

===−= 0int

The linear function is useful because it is mathematically simple and it can be shown to be reasonably close to the approximation of many situations. The first step to establish whether there is a relationship between variables is by means of a scatter diagram. This is a plot of the two variables on an yx − graph. Given that we believe there is a relationship between the two variables, the second step is to determine the form of this relationship.

195

Example 1

Consider the following data of a major appliance store. The daily high temperature and of air conditioning units sold for 8 randomly selected business days during the hot dry season.

Daily High Temperature

(x) oc Number of Units

(y) 27 35 18 20 46 36 26 23

5 6 2 1 6 4 3 3

Draw a Scatter diagram for the data.

y

6 • Number of units 5 • used

4 • 3 • • 2 • 1 • • 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 x Daily High temperature (oc) Figure 1.0

196

The distribution of points in the Scatter diagram suggests that a straight line roughly fits these points. The most straight forward method of fitting a straight line to the set of data points is ‘by eye’. The values of ‘a’ and ‘b’ can then be determined from the graph, ‘a’ is the intercept on the vertical axis and ‘b’ is the slope. The other method is that of semi averages. This technique consists of splitting the data into two equal groups, plotting the mean point for each group and joining these points with a straight line.

Example 2 Using data of Example 1, fit a straight line using the method of semi-averages. The procedure for obtaining the y on x regression line is as follows:

Step 1 Sort the data into size order by x - value.

x y 18 20 23 26 27 35 36 46

2 1 3 3 5 6 4 1

Step 2 Split the data up into two equal groups, a lower half and upper half (if there is an odd of items, drop the central one).

Lower half of Data Upper half of data x

18 20 23 26

y 2 1 3 3

x 27 35 36 46

y 5 6 4 1

Totals 87 Averages 21.75

9 2.5

144 36

16 4

197

Table 1.0

y 6 •

5 • 4 3 • • • 2 • 1 • •

18 26 34 42 x Method of Semi-average for Example 1.

Figure 2.0 Step 3 Calculate the mean point for each group Step 4 Plot the mean points in Step 3 on a graph within suitably scaled axes and joining them with a straight line. This is the required y on x regression line. Least Square Line Let us consider a typical data point with coordinates ),( ii yx (See Figure 3.0). The error

in the forecast (y coordinate of data point-forecasted coordinate as given by the straight

line ) is denoted by ie . The line which minimizes the value of ie is called the “least

198

square line” or the regression line. This can be shown by using calculus. Here we just give the ‘best estimates of ‘a ’ and ‘b ’ by the following formula.

( )

spodataofnumbertheisnwherexbya

n

xx

n

yxxy

b

int

2

2

−=

−

−=

∑ ∑

∑ ∑ ∑

The values of ‘a ’ and ‘b ’ are then substituted into equation bxay +=ˆ y least squares line y

• ie

• iy

x Figure 3.0 The least squares line with the error term ie .

Example 3 Fit the least squares line to the data in Example 1.

199

Table 2 shows the calculations for the estimates of a and b.

x y 2x 2y xy

18 20 23 26 27 35 36 46

2 1 3 3 5 6 4 1

324 400 529 676 729 1225 1296 2116

4 1 9 9 25 36 16 1

36 20 69 78 135 210 144 46

∑ = 231x ∑ = 25y 72952 =∑ x 1012 =∑ y ∑ = 738xy

125.3,8 == yn , 1.23=x

200

( )

( )

38.2

)875.28(0258.0125.3

0258.0

875.624

125.16

8

2317295

8

)25)(231(738

2

2

2

=

−=

−=

=

=

−

−=

−

−=

∑ ∑

∑ ∑ ∑

xbya

b

b

n

xx

n

yxxy

b

giving the equation for the regression line of xy 0258.038.2 += Forecasting Using the Regression Line Having obtained the regression line, It can be used to forecast the value of y for a given value of x . Suppose that we wish to determine the number of units sold if we have a daily temperature of co42 . From the regression line the forecasted value y is 46.3)42(0258.038.2ˆ =+=y i.e. the expected number of units sold is 3. Now suppose that we wish to determine the number of units sold if the temperature is

.49 co The forecasted value of y is the given by 64.3)49(0258.038.2ˆ =+=y i.e. the expected number of units sold is 4. The two examples differ due to the fact that the first y value was forecasted from an x value within the range of x values, while the second value outside the range of x values in the original data set.. The first example is a case of interpolation and the second is that of extrapolation. With extrapolation, the assumption is that the relationship between

201

the two variables continue to behave in the same way outside the given range of x values from which the least square line was computed. Exercise 7 1. For the following data x 2 5 6 8 10 11 13 16 y 2 3 4 5 6 8 9 10

a) Draw a scatter diagram b) By eye, fit a straight line to the data (ensuring it passes through the mean

value)

c) Fit the equation of the line by the method of semi-averages. 2. The following data have been collected regarding sales and advertising

expenditure. Sales Advertising expenditure

(K’ms) (K’ 000s) 10.5 230 11.2 280 9.9 310 10.6 350 11.4 400 12.1 430

a) Plot the above data on a scatter diagram. b) Fit the regression line using the method of least squares.

c) Estimate the sales if K530 000 is spent on advertising expenditure.

Note that advertising expenditure is the ‘x ’ variable and sales is the y variable.

202

3. Fit a least square line to the data in the table below. x 5 7 8 10 11 13 y 4 5 6 8 7 10 4. The table below shows the final grades in Mathematics and Communication

obtained by students selected at random from a large group of students.

a) Graph the data

b) Fit a least-squares line

c) If a student receives a grade of 85 in Mathematics, what is her expected grade in Communication?

d) If a student receives a grade of 65 in Communication, what is her expected

grade in Mathematics?

Mathematics (x) 80 86 97 70 89 75 99 69 87 78 Communication (y) 75 65 80 65 80 70 79 45 70 80 5. The table below shows the birth rate per 100 population during 1999 – 2005

year 1999 2000 2001 2002 2003 2004 2005 Birth rate per 1000 14.6 14.5 13.8 13.4 13.6 12.8 12.6

a) Graph the data.

b) Find the least squares line fitting the data. Code the years 1999 to 2005 as the whole number 1 through 7.

c) Predict the birth rate in 2009, assuming the present trend continues.

4.7 Correlation Analysis

Correlation analysis is used to determine the degree of association between two variables. Having obtained the equation of the regression line, correlation analysis can be used to measure how well one variable is linearly related to another. The coefficient of correlation r can assume any value inclusive in the

203

range 11 +− to . A value of r is close to or equal to – 1 , we have a negative correlation. The sign of the correlation coefficient is the same as the sign of the slope of the regression line. The following scatter diagrams illustrate certain values of the correlation coefficient. x x x x x x x x x x x x 1=r 0=r

x x x

x

1−=r The method of investigating whether a linear relationship exists between two variables x and y is by calculating Pearson’s product moment correlation coefficient (PPMCC) denoted by r given by the formula

( ) ( )

−

−

−=

∑ ∑∑ ∑

∑ ∑ ∑

n

yy

n

xx

n

yxxy

r2

2

2

2

Example 4 By calculating the PPMCC find the degree of association between weekly earnings and the amount of tax paid for each member of a group of 10 manual workers. Weekly Wage (K’ 000) 79 81 87 88 91 92 98 98 103 113 Income Tax (K’ 000) 10 8 14 14 17 12 18 22 21 24

204

The PPMCC is calculated in the Table below

x y 2x 2y xy

79 81 87 88 91 92 98 98 103 113

10 8 14 14 17 12 18 22 21 24

6241 6561 7569 7744 8281 8464 9604 9604 10609 12769

100 64 196 196 289 144 324 484 441 576

790 648 1218 1232 1547 1104 1764 2156 2163 2712

=∑ x 930 ∑ = 160y ∑ = 874462x ∑ = 28142y ∑ = 15334xy

( ) ( )

( )( )

( ) ( )

921.0

)254)(956(

454

10

1602814

10

93087446

10

16043015334

22

2

2

2

2

=

=

−

−

−=

−

−

−=

∑ ∑∑ ∑

∑ ∑ ∑

r

n

yy

n

xx

n

yxxy

r

r is ‘near’ 1 and indicates a strong positive linear correlation between the two

variables. Example 5

205

Evaluate the PPMCC for the following data.

x 15 20 25 30 35 y 143 141 144 149 148 The PPMCC is calculated in the Table below.

x y 2x 2y xy

15 20 25 30 35

143 141 144 149 148

225 400 625 900 1225

20449 19881 20736 22201 21904

2145 2820 3600 4470 5180

∑ x =125 ∑ = 725y ∑ = 33752x ∑ = 1051712y ∑ = 18215xy

( ) ( )

( )( )

( ) ( )

839.0

)46)(250(

90

5

725105171

5

1253375

5

72512518215

22

2

2

2

2

=

=

−

−

−=

−

−

−=

∑ ∑∑ ∑

∑ ∑ ∑

r

r

n

yy

n

xx

n

yxxy

r

The Coefficient of Determination The coefficient of determination is the square of the coefficient of correlation r. In words, it gives the proportion of the variation (in the y - values) that is explained (by the variation in the x - values).

206

In Example 10, the correlation coefficient is r = 0.839. Therefore coefficient of determination:

704.0

)839.0( 22

=

=r

( 3 decimals) This means that only 70.4% of the variation in the variable y is due to the variation in the variable x . Note that the coefficient of determination 2r is between 0 and +1 inclusive. Spearman’s Rank Correlation Coefficient. An alternative method of measuring correlation is by means of the Spearman’s rank correlation coefficient obtained by the formula.

)1(

61

2

2

−−= ∑

nn

dr

where d = difference between rankings.

Example 6 Two members of an interview panel have ranked seven applicants in order of preference for a specified post. Calculate the degree of agreement between the two members. Applicant A B C D E F G Interviewer X 1 2 3 4 5 6 7 Interviewer Y 4 3 1 2 5 7 6 The differences in rankings are shown below. D -3 -1 2 2 0 -1 1

2d 9 1 4 4 0 1 1

207

6429.0

3571.01

336

1201

)149(7

)20(61

)1(

61

20,0

2

2

2

=

−=

−=

−−=

−−=

==

∑

∑∑

r

nn

dr

dd

Example 7 The results of two tests taken by 10 employees are shown below (figures in %) Employee A B C D E F G H I J

Test X 50 52 58 66 70 74 77 86 92 94

Test Y 56 51 53 65 64 81 76 78 80 92

Rank each employee in order of performance in the two tests and calculate the rank coefficient . Ranking the employees in each test we have Employee A B C D E F G H I J

Test X 10 9 8 7 6 5 4 3 2 1

Test Y 8 10 9 6 7 2 5 4 3 1

d 2 -1 -1 1 -1 3 -1 -1 -1 0

2d 4 1 1 1 1 9 1 1 1 0

208

8788.0

1212.01

990

1201

)1100(10

)20(61

)1(

61

20,0

2

2

2

=

−=

−=−

−=

−−=

==

∑

∑ ∑

r

nn

dr

dd

Exercise 8 1. Draw a scatter diagram of each of the sets of values given below, and calculate

the PPMCC in each case.

x 6 7 8 9 10 a) y 3 6 9 12 15

b) x 1 3 5 7 9 11 y 8 7 6 5 4 3

c) x 2 4 6 8 10 12 14 y 12 8 8 14 9 7 13 2. The following table gives the percentage unemployment figures for males and

females in 9 regions. Draw a scatter diagram of these data and calculate PPMCC.

Region Unemployment Luapula Northern Eastern Central Lusaka Copperbelt N.Western Western Southern

209

% Male 3.4 3.5 4.5 4.4 12.5 12.8 3.2 4.2 4.8 Female 3.2 3.8 4.6 3.8 11.8 11.5 4.0 3.8 3.5

3. In a job evaluation exercise an assessor ranks eight jobs in order of increasing health risk. The same jobs have also been ranked in decreasing order on the basis of the number of applicants attracted per advertised post.

Job A B C D E F G H Health 1 2 3 4 5 6 7 8 Applicant 4 3 2 1 6 5 8 7

Calculate the rank correlation coefficient for this information. 4. The table below gives the Shorthand and Typing speeds of a sample of seven

secretaries

Secretary 1 2 3 4 5 6 7

Typing 42 44 47 47 50 54 57 Speed

(words /min) Shorthand 97 84 95 96 107 98 117

Calculate the degree of correlation between the two skills by:

a) the PPMCC, and

b) the rank correlation coefficient. 5. On the different days (picked at random) the following values were obtained for

the price of a share for a particular company together with the index on that day

210

Share price

(K)

260 250 350 200 150 100 115 120 135 145

Index 115 135 140 120 105 110 106 165 175 115

Calculate Spearman’s rank correlation coefficient and say whether the index and indicate whether the index is a reasonable indicator for the price of the company’s share.

EXAMINATION QUESTIONS WITH ANSWERS Multiple Choice Questions

1.1 If ,8102 ==∑ nandd the Spearman’s rank correlation coefficient to 3 decimal

places will be? A. 0.188 B. 0.841 C. 0.821 D. 0.881

(Natech , 1.2. Mathematics & Statistics, December 2004)

1.2 The prices of the following items are to be ranked prior to the calculation of

Spearman’s rank correlation coefficient. What is the rank of item G?

Item E F G H I J K L Price 18 24 23 19 25

A. 5 B. 4 C. 3 D. 2.5


211

1.3 8 x x 6 x x

4 x x x

2 x x x 0 4 8 12 16 On the basis of the Scatter diagram above, which of the following equations

would best represent the regression line of Y on X? A. y = −x−8 B. y = x + 8 C. y = −x + 8 D. y = x − 8

(Natech , 1.2. Mathematics & Statistics, December 2003) 1.4 An investigation is being carried out regarding the hypothesis that factor X is a

cause of ailment Y. Which coefficient of correlation between X and Y gives most support to the ailment?

A. -0.9 B. -0.2 C. +0.8 D. 0

(Natech , 1.2. /B1Mathematics & Statistics, December 1999 (Rescheduled)) 1.5 If ∑ ∑ ∑ ∑∑ ===== 19635,46075,10436,555,216 22 xyyxyx and n

=8, then the value of r, the coefficient of correlation to two decimal paces, is

A. 0.79 B. 0.62 C. 1.01 D. 1.02 (Natech , 1.2. Mathematics & Statistics, December 2001)

212

1.6 The Scatter diagram below shows • • • • • • • A. High positive correlation B. Very high correlation C. Very high negative correlation D. Perfect correlation.

(Natech , 1.2. Mathematics & Statistics, June 2005)

1.7 Find the value of ‘a’ in a regression equation if ∑ ∑ === 400,150,7 yxb and

n = 10. A. 145 B. -65 C. 25 D. -650


1.8 In regression analysis, the variable whose value is estimated is referred to as the:

A. Simple variable B. Independent variable C. Linear variable D. Dependent variable

1.9 The value of the coefficient of determination is interpreted as indicating

A. The proportion of unexplained variance B. The proportion of explained variance C. The extent of causation D. The extent of relationship

1.10 Of the following coefficient of correlation, the one that is indicative of the

greatest extent of relationship between the independent and dependent variables is

213

A. 0 B. +.20 C. 95.− D. +.70

SECTION B QUESTION ONE a) Derive the product moment correlation coefficient from the following data and

comment on your results.

Pupil A B C D E F G H I J K Mathematics marks, x

41 37 38 39 49 47 42 34 36 48 29

Physics mark, y

36 20 31 24 37 35 42 26 27 29 23

b) Find the estimated line, by method of least squares, fitting the following results

from a Physics experiment.

Load, x (Newtons)

0.1 0.1 0.2 0.2 0.3 0.3 0.4 0.4 0.5 0.5

Extensions, y (mm)

18 11 25 22 35 50 54 45 52 68


c) A company has the following data on its profit (y) and advertising expenditure 9x) over the last six years.

Profits

(Million (K) Advertising Million (K)

11.3 12.1 14.1 14.6 15.1 15.2

0.52 0.61 0.63 0.70 0.70 0.75

i) Use two (2) methods to justify your assumption that there is a relationship

between the two variables.

ii) Forecast the profits for next year if an advertising budget of K800 000 is allocated.


214

QUESTION TWO a) In the context of regression analysis explain what is meant by the following terms.

i) Regression coefficient ii) Explanatory variable.

b) The following data shows the monthly imports (I) of apples and average prices (P) over a twelve-month period.

Monthly Imports (I)

(‘000 tonnes) Average Monthly Prices (P)

(K/tones) 100

120

125

130

128

126

120

100

90

90

95

98

232

220

218

210

210

212

217

240

242

238

230

230

i) Determine the regression equation if imports (I) of apples on the price (P) and use it to forecast monthly imports when the average monthly price is K250 per tonne.

ii) If the correlation coefficient of the data is –0.95, interpret the results.


QUESTION THREE Hungry Lion is a major food retailing company, which has recently decided to open several new restaurants. In order to assist with the choice of sitting these restaurants the management of fast foods limited whished to investigate the effect of income on eating habits. As part of their report a marketing agency produced the following table showing the percentage of annual income spend on food y, for a given annual family income ((K) ‘x’)

215

x (‘K’000,000)

y

18

27

36

45

54

72

90

62

48

37

31

27

22

18

a) Plot, on separate Scatter diagrams.

i) y against x ii) ,loglog 1010 xagainsty and comment on the relationship between income

and percentage of family spent on food.

b) Use the method of least squares to fit the relationship baxy = to the data. Estimate a and b.

c) Estimate the percentage of annual income spent on food by a family with an

annual income of K64,800,000.

(Natech , 1.2. Mathematics & Statistics, December 2001) QUESTION FOUR a) Sales of product A between 0 and 4 years were as follows: Year Units sold (‘000s)

2000 20 2001 18 2002 15 2003 14 2004 11

Required: i) Calculate the correlation coefficient r. ii) Comment on the result in (i) above. iii) Calculate the coefficient of determination and comment.

216

iv) Use a regression equation to estimate the sales in the year 2005. b) The table below shows the respective masses X and Y of a sample of 12 fathers

and their oldest ones.

Mass X of father (Kg)

65 63 67 643 68 62 70 66 68 67 69 71

Mass Y of son (Kg)

68 66 68 65 69 66 68 65 71 67 68 70

From the data given above:

i) Construct a scatter diagram ii) Calculate the rank correlation coefficient using Spearman’s method.

(Natech , 1.2. Mathematics & Statistics, June 2005) c) Find the degree of correlation between the Bank of Zambia base lending rate and

the dollar exchange rate taken over the past six months using:

i) The product moment coefficient of correlation. ii) The coefficient of rank correlation. Month Jan Feb Mar Apr May Jun Base % as on 1st of each month

14 14 13.5 12.5 12 12

Average rate ($) 1.90 1.91 1.86 1.84 1.84 1.83 (Natech , 1.2. Mathematics & Statistics, Nov/Dec 2000) QUESTION FIVE a) The following table shows the number of units of a good product and the total

costs incurred.

Units Produced 100 200 300 400 500 600 700

Total Costs (K) 40 000 45 000 50 000 65 000 70 000 70 000 80 000

Draw a scatter diagram b) Find the appropriate least squares regression line so that the costs can be predicted

from production levels and estimate the total costs when production is 250 units. c) State the fixed costs of production.

217

d) Calculate r and explain how much of the variation in the dependent variable is

explained by the variation of the independent variable.


QUESTION SIX a) A sample of eight employees is taken from the Production Department of an

electronics factory. The data below relates to the number of week’s experience in the soldering of components, and the number of components, which were rejected as unsatisfactory last week.

Employee A B C D E F G H Weeks of experience (x)

4 5 7 9 10 11 12 14

No. of rejections (y)

21 22 15 18 14 14 11 13

i) Draw a Scatter diagram of the data.

ii) Calculate a coefficient of correlation for these data and interpret its value.

iii) Find the least squares regression equation of rejects on experience.

Predict the number of rejects you would expect from an employee with one week experience.

(Natech , 1.2. Mathematics & Statistics, December, 1999 Rescheduled))

b) i) Distinguish between ‘regression’ and ‘correlation’.

ii) A experiment was conducted on 8 children to determine how a child’s reading, ability varied with his/her ability to write. The points awarded were as follows:

Child A B C D E F G H Writing 7 8 4 0 2 6 9 5 Reading 8 9 4 2 3 7 6 5

Calculate the coefficient of rank correlation and interpret the results.

(Natech , 1.2. Mathematics & Statistics, December, 2002)

c) The mass of a growing animal is measured, in g, on the same day each week for with weeks. The results are given below.

Week x 1 2 3 4 5 6 7 8

218

Mass (g) y 480 504 560 616 666 702 759 801

i) Using 2cm to represent week 1 on the x-axis and 2cm to represent 100g on the y-axis, plot a scatter diagram of mass y against week x.

ii) Find the equation of the regression line of y on x.

(Natech , 1.2. Mathematics & Statistics, December, 1998) QUESTION SEVEN a) The following Table gives the cost price and number of faults per annum

experienced with seven brands of video recorders.

Video Recorders Brand Price (K’000’) No. of Faults per Annum

A B C D E F G

492 458 435 460 505 439 477

2 6 7 4 3 5 1

i) Determine Spearman’s rank Correlation coefficient.

iii) Interpret your answer in (i) above.

(Natech , 1.2. Mathematics & Statistics, December,1998)

b) The following Table gives a set of ten pairs of observation of inspection costs per thousand articles produced recorded on a number of occasions at several factories controlled by a single group and producing comparable products.

Observation Inspection costs per thousand articles

Number of defective articles per thousand

1

2

3

4

5

0.25

0.30

0.15

0.75

0.40

50

35

60

15

46

219

6

7

8

9

10

0.65

0.45

0.24

0.35

0.70

20

28

45

42

22

Putting inspection costs = x and number of defectives = y You are required to:

i) Represent these pairs of observations on a scatter diagram.

ii) Find the regression line of y on x.

(Natech , 1.2. Mathematics & Statistics, December,1997)

QUESTION EIGHT A Quality Control Manager has been hiring temporary workers to check all the surgical needles before they are dispatched (in boxes) to the customers. He believes that there is a relationship between the number of defective needles (per 1000) dispatched to customers and the experience (in weeks) of the workers. To test this theory, he randomly selects a sample of eight workers and gives then a box each of surgical needles to check. Unknown to the workers, the inspected boxes are returned to the Manager for some more checking. He checks all the surgical needles in each of these boxes and records the number of defective needles (per 1,000) contained in it. This information is summarized in the table below.

220

Worker A B C D E F G H

Experience (in weeks) x 4 5 7 9 10 11 12 14

Defective (Per 1,000) y 21 22 15 18 14 14 11 13

You are required to: a) Draw a scatter plot of y against x b) Calculate the coefficient of correlation ad interpret its value. c) Find the least squares regression equation of the number of defectives on

experience. d) Estimate the number of defectives in a box inspected by a worker with 6 weeks of

experience.


221

CHAPTER 5

SERIES

5.0 SERIES A Sequence is a list of numbers which follow a definite pattern or rule. If the rule is that

each term, after the first, is obtained by adding a constant, d , to the previous term, then the sequence is called an arithmetic sequence, such as 3, 7, 11, 15, 19, . . . where d = 4. d is known as the common difference. If the rule is that each term, after the first is obtained by multiplying the previous term by a constant, r , then the sequence is called a geometric sequence such as 2, 6, 18, 54, 162 . . . ., where r = 3, r is known as the common ratio.

A series is the sum of the terms of a sequence. A series is finite if it is the sum of a finite number of terms of a sequence. That is 2 + 6+18+ 54+162 and the number of terms is a finite series. A series is infinite if it is the sum of an infinite number of terms of a sequence.

5.1 Arithmetic Series

An arithmetic series is the sum of the terms of an arithmetic sequence, such as 3, 7, 11, 15, 19, . . . with d = 4. Denote the first term of an arithmetic sequence with the value ‘a ’ and progressing by adding the value ‘d ’ to each previous terms, the arithmetic sequence can be outlined as in Table 5.1.

Table 5.1 Arithmetic Sequence

Sequence dnadadaa )1(...2 −+++ Term number nTTTT ...321

Each element of a sequence an be identified by reference to its term number.

Example 1

In the sequence 3, 7, 11, 15, 19, . . . ...,112,7,3 321 =+==+=== daTdaTaT etc.

222

The value of any term can be calculated knowing that the nth term of the arithmetic sequence is )1.5()1( →−+= dnaTn .

Example 2 In the above example, the value of 15th term is

15123

)4)(14(3)1(

=+=−+=−+ dna

The sum of the n terms, ,nS of an arithmetic series is given by the formula

[ ] )2.5()1(22

→−+= dnan

Sn

Formula 5.2 can be rewritten as:

−++=

n

n

T

dnaan

S

)1(

2

Note that .)1( dnaTn −+= Hence [ ] )3.5(2

→== nn Tan

S where ‘a’ is the first term

and nT is the nth term of an arithmetic series.

Example 3 Find the sum of the first 20 terms of the series 40 + 36 + 32 + 28 + . . . This is an arithmetic series since a = 40, n = 20, and d = 36 – 40 = – 4. Using formula (5.2)

[ ][ ]

40

768010

)4)(120()40(22

2020

=−=

−−+=S

223

Example 4 Find the sum of the series 6, 11, 16, 21, . . . if series consists of 15 terms. We have a = 6, and d = 5 with n = 15

78)5(1461415 =+=+= daT

Here the sum is [ ] .6157662

1515 =+=S

5.2 Geometric Series A geometric series is the sum of the terms of a geometric sequence, such as 2, 6,

18, 54, 162, . . ., with r = 3. Denoting the first term of a geometric sequence with the value ‘a ’ and progressing by multiplying the previous term by a common ratio r , the geometric sequence can be outlined as in Table 5.2.

Table 5.2 Geometric Sequence Sequence 12 ... −nararara Term nTTTT ...321 In this pattern it can be seen that when any term is divided by the previous term,

the result is a common ratio, r . The nth term of a geometric sequence is )4.5(1 →= −n

n arT The sum of the first n terms of a geometric series is given by the formula,

)6.5(1

1

)5.5(1

1

... 12

→

−−=

→

−−=

+++== −

r

raS

orr

raS

arararaS

n

n

n

n

nn

224

It is more convenient to use formula (5.5) when 1<r and formula (5.6) when

.1>r When the geometric series has an infinite number of terms and r is less than 1, if n becomes very large nr approaches 0. Therefore the sum of an infinite series as long as

1<r is given by the formula:

)7.5(1

→−

=∞ r

aS

Example 5 Find the 8th term of the series 3, 9, 27, . . ., The 1st term is 3 and the common ratio is 3, that is a = 3, and r = 3. The 7th term is .2187)3(3)3(3 617

7 === −T

Example 6 Find the sum of the series 1, 4, 16, 64, . . . which has 8 terms. Since the common

ration, r, is 4 (obtained y from )16

64

1

4or we shall use formula 5.6,

.1

)1(

−−=

r

raS

n

n In this problem a = 1, and n = 8. Hence

2184514

)14.(1 8

8

=−

−=S

Example 7

Find the sum of the first 15 terms of the series 5 + 1 + ++25

1

5

1 . . .

This is a geometric series. Since the ratio 5

1=r . Therefore ,15,15,5

1 === nar and

hence, using formula (5.5)

225

−=

−

−=

15

15

15 5

11

4

25

5

11

5

11

5S .

Example 8

Find the sum of the given series ...25

1

5

115 ++++

4

24

5

45

5

11

55

1,5

==−

=

==

∞S

ra

Exercise 1 1. Find the 15th terms of the series 2, 7, 12, . . . 2. Find the 7th term of the series 6, 13, 20, . . . 3. Find the number of the term which is 43 in the series 3, 8, 13, 18 . . . 4. The 4th term of a series A.P is 13 and the 8th term is 25. Find the 13th term. 5. The 9th term of a series in A.P is 17 and the 16th term is 31. Find the 28th term. 6. In a series in G.P the first term is 4 and the common ratio is 3. find the tenth and

eleventh terms.

7. Find the sum of the series 7, 12, 17, . . . if there are 18 terms in the series.

8. Find the sum of the series in arithmetical progression, which has the 1st term of 75 and an 8th term and last term of 110.

9. Find the 8th term of the series 4, 12, 36, . . .

10. Determine the sum of the infinite G.P series ...64

1

16

1

4

11 ++++

11. Determine the sum of the infinite G.P series

...16

5

4

5520 ++++

226

12. Determine the sum of the infinite G.P series ...72

1

24

1

8

1

8

3 +−+−

13. The first term of a series in G.P. is 3.5 and the 6th term is 0.00112. if the

series has nine terms, find its sum. 14. Find the sum of the series 6, 9, 13.5, 20.25, . . . which has 16

terms. 5.3 Time Series

Introduction

This section defines a time series and describes the structure within which time series’ movements can be explained and understood.

Definition of a time series: Any variable that is measured overtime in sequential order is called a time series. Business people, economists, and analysts of various kinds all look back at the sequence of events that occurred over the past year or years in order to understand what happened and thereby (they hope to) be in a better position to anticipate what may happen in the future. Examples of time series are total monthly sales, yearly unemployment figures, daily average temperatures, etc. The classical time series model focuses on the decomposition of the time-dependent variable into four components: trend (T), cycle (C), seasonal variation (S), and residual or irregular (I). The model may be additive in its components CISTY +++= or multiplicative in its components ISCTY ×××= . The movements of a time series may be classified as follows: 1. A trend (also known as a secular trend) is a long-term relatively smooth

pattern or direction that the series exhibits. By definition, it has a duration of more than one year. For example, data for beer sales show to have an upward trend to the right, increase or decrease in population and technological changes etc.

2. A cycle is a wavelike or oscillatory pattern about a long-term trend that is

generally apparent over a number of years. By definition, it has a duration of more than one year. Examples of cycles are well known business cycles that

227

record periods of economic recession, booms and inflation. Long-term product demand cycles and cycles in the monetary and financial sectors.

3. Seasonal Variations. These are the oscillations, which depend on the season of

the year. The interval of time can be any length (minutes, hours, days, etc). the changes are of a periodic type. Examples, employment is usually higher at harvest time at Nakambala Sugar Estate in Mazabuka, rainfall will be higher at some times of the year than at others, goods sold during the week and at Christmas are higher than at other sale times, etc.

4. Residual or Irregular Variation is the random movement that a series exhibits after

the trend, cycle and seasonal variations are removed. Unpredictable events such as strikes, fires, breakdowns, illness, etc., are some of the examples of random variations.

The motivation behind decomposing a time series is twofold. On the one hand, we wish

to see whether a particular component is present in a given time series and to understand the extent to which it explains some of the movements in the variable of interest. On the other hand, if we wish to forecast a particular variable, we can usually improve our forecasting accuracy by first breaking it into component parts, then forecasting each of these parts separately, and finally combining the individual effects to produce the composite overall forecast.

Business Forecasting is concerned with estimating the future value of some variable of interest. This maybe done for the short-term or for the long-term, and different forecasting models are more appropriate for one case than for the other.

Decomposition Analysis

Decomposition analysis seeks to breakdown the time series into its components, which are then used as a basis for the forecast.

Trend Calculation The basic purpose behind the trend calculation is to find the line (or its equation) which best fits the given data.

There are two methods of finding a line already discussed in the earlier section on regression analysis. These were:

• Method of semi averages • Least squares method.

Here we show an example of least squares method.

228

Example 1 The following table gives the company sales of a certain product over a 7 – year period. Forecast the sales in 2004. Year 1984 1985 1986 1987 1988 1989 2000 Sales (K’000 000’) 32 26 27 22 21 21 30

The data is plotted on graph

• • • 30 • • Sales • • 20 10 1984 1985 1986 1987 1988 1989 2002 year How we measure time along the horizontal axis (it turns out) is irrelevant in time-

series analysis. We can suit ourselves, picking whatever numbers serve to reduce the computational burden. A common practice is to measure the time periods consecutively (1, 2, 3, …), and we shall do it here.

Calculations for Example 1

Year Sales

y Time

x 2x

xy

1984 32 1 1 32 1985 26 2 4 52 1986 27 3 9 81 1987 22 4 16 88 1988 21 5 25 105 1989 21 6 36 126 2000 30 7 49 210

∑ = 179y ∑ = 28x 1402 =∑ x ∑ = 694xy

229

These values can be used to fit a ‘least squares’ line of best fit and produce a trend line equation of

xy 786.071.28ˆ −= This equation can then be used to give the trend figures shown in the table below:

(e.g., where ( ) 93.271786.071.28ˆ,1 =−== yx )

Time Trend Value

^

y

1 27.93 2 27.14 3 26.36 4 25.57 5 24.79 6 24.99 7 23.21

Moving Averages Moving averages smooth out a time series in order to isolate the trend. Example 2 Year 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 Profit (x K1 000 000

25

28

37

50

39

25

27

37

60

70

55

Isolate the trend in the time series of annual profit given above by finding: a) three-point moving averages, b) five-point moving averages. First it is advisable to draw a graph of the time series so that the overall picture can be seen clearly.

230

Profits (K’000’000’) 80 • • • 40 • • • • • •

• 94 95 96 97 98 99 20 01 02 03 04 years The moving averages are calculated as follows: A moving average is a simple arithmetic average computed over any number of time periods. For a three period moving average, we would take the first three (1, 2 and 3) and average them. Then we would move to the next grouping (2, 3 and 4) and averaging them; and so on. In a similar fashion, we can compute 5 moving averages, as shown in the table below, or any other number of month’s averages. Notice that, the longer the time period, over which we average, the smother the series becomes. Eventually it becomes a straight line moving average. Reducing the number of observation points for the 3 moving average, we lose the first and last observation; for the 5 moving average, we lose both the first 2 observation and the last 2 observations. Calculations for 3 and 5 Moving Averages for Example 2

Year Profit Three-Point Moving Average (3MA)

Five-Point Moving Average (5MA)

1994 25 1995 28 30.0 1996 37 38.3 35.8 1997 50 42.0 35.8 1998 39 38.0 35.6 1999 25 30.3 35.6 2000 27 29.7 37.6 2001 37 41.3 43.8

60

20

•

231

2002 60 55.7 49.8 2003 70 61.7 2004 55

When n is odd, there will be a middle item opposite which to associate the original observation. When n is even, for example for the data involving quarterly sales in

Example 2 the four moving total for year 1, falls between quarters 2 and 3. The second total of 127 is recorded between the 3 and 4 quarters. The process continues. None of

this totals correspond to the actual quarter. To bring a total number of sales opposite the actual quarter we take the average of the two totals. Each of these new totals is the sum of two sums of four numbers, i.e. a sum of eight numbers, therefore we need to divide by

8 to get the average. This set of figures are called centered moving average (CMA). Here, they are the four centered moving average (4CMA).

Year Quarter Sales 4MT 4CMA 1 1

2 3 4

124

127

133

31.4

32.5

2 1 2 3 4

136

138

139

33.6

34.3

34.6

Seasonal Variation Once the trend has been calculated the amount of seasonal variation about the trend can be determined. By the addition model this is equal to:

ICSTY ++=−

232

and by the multiplication model as:

ICST

Y ××=

Whilst season fluctuations are readily identified, they may be combined with irregular or random disturbances. In order to remove such random influences from the data, it is corrected for seasonal variations by calculating the average departure of the actual data from the trend over several years. Since random influences may operate in opposite directions in succeeding years, this method of averaging variations from the trend should do something to eliminate then. These average variations are calculated from the data of column(6) in Example 2. It will, of course, be a mere accident if the sum of the variations in column (6) is zero, and so the average quarterly variations will need adjusting so that their sum is zero. This is effected by adding one-quarter of the sum of the actual averages (if this sum is negative) to each of them. If the average is positive, then one quarter of this sum taken as negative and must be added to each average. In order to obtain a result to one decimal place, this may entail correcting the differences also to one decimal place, the larger differences being subtracted from (or added to) the averages with larger absolute magnitudes. Example 2 The following table gives the sales from a grocery store in each quarter for the last four years. (The figures are given in thousands of copies).

Quarters 1 2 3 4 1 20 20 60 24 2 23 26 63 26 3 24 27 63 27

Year

4 25 29 67 30 Calculate the trend figure and seasonal variations, using the moving averages. Assuming an additive model.

Year

Quarters

Sales

4-quarterly moving Totals

Trend figures (T) 4CMA

Deviations (S)

1 1 2 3 4

20 20 60 24

124 127

31.4 32.5

+28.6 -8.5

2 1 2 3 4

23 26 63 26

133 136 138 139

33.6 34.3 34.6 34.9

-10.6 -8.3

+28.4 -8.9

3 1 2

24 27

140 140

35.0 35.1

-11.0 -8.1

233

3 4

63 27

141 142

35.4 35.8

+27.6 -8.8

4 1 2 3 4

25 29 67 30

144 148 151

36.5 37.4

-11.5 -8.4

The deviations are used to estimate the seasonal variations in the following table.

Quarters Year 1 2 3 4

1 - - +28.6 -8.5 2 -10.6 -8.3 +28.4 -8.9 3 -11.0 -8.1 +27.6 -8.8 4 -11.5 -8.4

1 2

3

4

Total Average

Total Deviations -33.1

-24.8

84.6

-26.2

Average Deviations -11.03

-8.27

28.2

-8.73

+0.17

Adjusted Deviations 11.0725

-8.3125

28.1575

-8.7725

0425.04

17.0 −=

−

Therefore the forecasted seasonal variations (s) are: 1: -11 2: -8 3: 28 4: -9 Note that if you were required to find the seasonally adjusted figures or deseasonalised figures. For the additive model, we have: Seasonally adjusted values = Original values – Seasonal variations The results are shown in the table below:

Quarter

Sales (Y) Seasonal

Variations Adjusted Sales (Y-S)

1 1 2 3 4

20 20 60 24

-11 -8

+28 -9

31 28 32 33

2 1 2 3 4

23 26 63 26

-11 -8

+28 -9

44 34 35 35

3 1 2 3

24 27 63

-11 -8

+28

35 35 35

234

4 27 -9 36 4 1

2 3 4

25 29 67 30

-11 -8

+28 -9

36 37 39 39

Suppose we now want to forecast the sales in the first, second, third and fourth quarters of the 5th year. The trend is shown in the graph below. 80 • • 60 • • • 40 • • 20 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 Time From the graph the trend estimates for the 5th year are: 1st quarter 39.4 2nd quarter 39.7 3rd quarter 40.1 4th quarter 40.4 Combining the trend and seasonal variations we have: 1st quarter: 39.4 – 11 = 28.4 2nd quarter: 39.7 – 8 = 31.7 3rd quarter: 40.1 + 28 = 68.1 4th quarter: 40.4 – 9 = 31.4 Therefore, the forecasted sales in the 5th year are:

Trend Values

235

Quarter 1st 2nd 3rd 4th Sales K28 000 K32 000 K68 000 K31 000

In example 2, assuming the model was multiplicative. The trend values are the same. The difference starts with the seasonal variations these are now:

ICST

Y ××= and we call them seasonal indices. For year 1, quarter 3, we have

91.14.31

60 = , quarter 4, is 74.05.32

24 = . The rest of the results are shown in the

table below:

Year

Quarters

Sales (Y)

Trend figures (T)

4CMA

Seasonal

Indices (T

Y)

1 1 2 3 4

20 20 60 24

31.4 32.5

1.91 0.74

2 1 2 3 4

23 26 63 26

33.6 34.3 34.6 34.9

0.68 0.76 1.82 0.74

3 1 2 3 4

24 27 63 27

35.0 35.1 35.4 35.8

0.69 0.77 1.78 0.75

4 1 2 3 4

25 29 67 30

36.5 37.4

0.68 0.78

The ratios are used to estimate the seasonal variations in the following table.

Quarters

Year 1 2 3 4

Total

Averages

1 - - 1.91 0.74

2 0.68 0.76 1.82 0.74

236

3 0.69 0.77 1.78 0.75

4 0.68 0.78 - -

Total Seasonal Indices 2.05 2.31 5.51 2.23

Average Seasonal Indices 0.69 0.77 1.84 0.73 4.03

For the multiplicative model, seasonally adjusted values iationsseasonal

valuesoriginal

var= .

Quarter

Sales (Y)

Seasonal Indices

Adjusted Sales

S

Y

1 1 2 3 4

20 20 60 24

0.68 0.76 1.83 0.73

29.4 26.3 32.8 32.9

2 1 2 3 4

23 26 63 26

0.68 0.76 1.83 0.73

33.8 34.2 34.4 35.6

3 1 2 3 4

24 27 63 27

0.68 0.76 1.83 0.73

35.3 35.5 34.4 37.0

4 1 2 3 4

25 29 67 30

0.68 0.76 1.83 0.73

36.8 38.2 36.6 41.1

Suppose we want to forecast the sales in the first, second, third and fourth quarters of the 5th year. This can be done in two ways: 1. By fitting a line of best fit through the trend found by moving averages as in the

previous example.

Adjusted 1 2 3 4 Seasonal Indices

68.069.03.4

4 =

× 0.76 1.83 0.73

237

2. By using linear regression.

The forecast value = Trend Value Extrapolated × Seasonal Indices

Using (1) again we plot as graph a in the Example 2. Combining the trend and seasonal variations, we get:

1st quarter 39.4 (0.69) = 27.2

2nd quarter 39.7 (0.77) = 30.6

3rd quarter 40.1 (1.84) = 73.8

4th quarter 40.4 (0.73) = 29.5

Therefore the forecasted sales in the 5th year would be:

Quarter 1st 2nd 3rd 4th

Sales K27, 000 K31, 000 K74, 000 K30, 000

Exercise 2 1. The sale of Local Authority houses to existing tenants over a four-year period is

tabulated below.

No. of Houses Sold Year Jan - April May - Aug Sept - Dec

1991 45 87 65 1992 65 102 71 1993 80 126 89 1994 92 144 104

Plot these values on to a suitable graph. Plot on the same graph the three-point

moving averages in order to highlight the trend in this time series.

238

2. The quarterly production figures for a large manufacturing company are given

below. Use four-point moving averages to isolate the trend. Plot the centred moving averages on to a graph.

Total Production (Thousand Units) 2000 2001 2002 2003 2004

1st quarter - 154 150 110 136 2nd quarter - 156 152 142 - 3rd quarter 140 132 132 124 - 4th quarter 150 147 142 140 -

3. The table below shows the staff turnover (the number of workers leaving

employment as a percentage of the total work force) experienced by a company over a four-year period.

Quarter

Year 1st 2nd 3rd 4th 1 5.1 3.8 5.8 5.3 2 5.5 3.7 6.3 5.8 3 4.2 4.2 6.9 5.9 4 4.4 5.0 7.0 6.4

By finding average deviations from the trend, estimate the seasonal variation for

each of the four quarters. (Use an additive model). 4. The number of package tours booked through a travel agency over a three-year

period is given in the table below.

2000 2001 2002 1st quarter 150 148 152 2nd quarter 100 95 85 3rd quarter 50 60 6985 4th quarter 100 102

Estimate the number of tours booked during each quarter in 2003. (use a

multiplicative model).

239

5. Quarterly production at a copper mining plant was reported as follows:

Quarterly Production (000 tonnes)

Quarter Year 1 2 3 4 2002 2003 2004 2005

40.5 46.4 48.0 51.4

42.3 46.4 48.0 51.4

39.5 42.5 45

46.5 50.8 52.2

Using the method of moving averages, find the average seasonal deviations (use

the additive model). 6. Calculate the five-month moving average for the following sales information.

Show both the original series and the calculated moving averages on the same graph. What purpose is served by such calculations and such a graph?

Sales (X K 000 000) 2003 2004 January February March April May June July August September October November December

25 40 42 32 21 21 43 48 37 26 33 35

47 39 31 38 50 37 35 33 41 51 43 38

240

7. Sales from Company GBM’s motor division have been monitored over the past

four years and are presented below Year Quarter 1 2 3 4 1 10 20 29 40 2 20 31 40 50 3 30 35 42 55 4 40 48 50 60

a) Construct a graph of this data.

b) Find a centered four-point moving average trend and place it on your graph.

c) Calculate the seasonal variations using additive model.

.

241



1.1 What is a seasonal variation?

A. The square of the standard deviation of a series. B. Dispersion C. The regular pattern of change expected over any time period. D. The comparison of the variability of two or more sets of figures.


1.2 In a Time Series analysis, the multiplicative model is used to forecast sales and

the following seasonal variations apply.

Quarter 1 2 3 4 Seasonal Variations 1.3 1.4 0.5 ?

The seasonal variation for the fourth quarter to one decimal place is: A. -1.4 B. +1.4 C. -0.8 D. +0.8

(Natech, 1.2 Mathematics and Statistics, December 2003) 1.3 The components associated with conventional time series are all:

A. Long-run factors B. Causative factors B. Short-run factors D. Time-related factors

1.4 The time series component most useful in long-term forecasting is the

A. Trend B. Seasonal C. Cyclical D. Irregular 1.5 The time series component most useful in short-term forecasting is the:

A. Trend B. Seasonal C. Cyclical D. Irregular 1.6 A seasonal index number would never be computed by:

A. Year B. Week C. Quarter D. Month

242

1.7 A seasonal index of “65” for a particular month indicates that the level of values

for that month generally is:

A. 35 percent higher than the monthly average for the year; B. 65 percent lower than the monthly average for the year; C. 65 percent higher than the monthly average for the year; D. 35 percent lower than the monthly average for the year.

1.8 When annual data are used for the purpose of identifying the seasonal component

of a time series, then the procedure that is followed can be represented by the algebraic expression:

A. ICST

Y ××= B. ICTS

Y ×=

C. SICT

Y =××

D. ICY

TS ×=

1.9 In the linear trend equation ,^

bxaY += the x is most often represented by:

A. Week B. Quarter C. Month D. Year 1.10 The model generally used in conjunction with the time series analysis is:

A. ICSTY ×××= B. ICSTY −−−=

C. CI

TSY = D. ICSTY +++=

243

SECTION B QUESTION ONE a) Index of output of coal mining in Zambia 1998 – 2000 (1995 = 100).

Quarters Year 1 2 3 4 1998 99 92 84 73 1999 95 82 71 87 2000 91 82 67 87

By means of a moving average, estimate the trend and seasonal movements.

(Natech, 1.2 Mathematics and Statistics, December 2001) b) (i) Assume a four-year cycle and calculate the trend by the method of moving

averages from the following data relating to production of sugar at a sugar estate.

Year

Production (kg)

Year

Production (kg)

1991 464 1996 540 1992 515 1997 557 1993 518 1998 571 1994 467 1999 586 1995 502 2000 612

(ii) From the trend values obtained in part (i) above, what would you say is

the main disadvantage of obtaining a time series trend using the method of moving averages?


244

QUESTION TWO a) The sales of M & G Limited are given below:

Year

Quarter

Sales

Centered Moving Average (T)

Seasonal Variation (Y)

1996 1 2 3 4

26.8 36.3 38.1 47.5

- -

37.4750 38.98.75

- -

1.017 1.218

1997 1 2 3 4

31.2 42.0 43.4 55.9

40.3625 42.0750 44.2250 46.1750

0.773 0.998 0.981 1.211

1998 1 2 3 4

40.0 48.8 54.0 69.1

48.3500 51.3250 54.8125 57.7750

0.827 0.951 0.985 1.196

1999 1 2 3 4

54.7 57.8 60.3 68.9

59.6875 60.4500

- -

0.916 0.956

- -

Required: (i) Evaluate the seasonal component for each quarter (the average of the

seasonal variations). (Note: Do not adjust, round to 2 decimal places). (ii) Forecast sales for the four quarters of the year 2000 using trend forecasts

of 66.7, 68.8, 70.9 and 73 (assume an additive model).

(Natech, 1.2 Mathematics and Statistics, December 2003) b) The characteristics movements of a time series include: (i) Long-term movements

(ii) Cyclical movements

(iii) Seasonal movements

245

Explain what is meant by these movements.

Natech, 1.2 Mathematics and Statistics, November/December 2000) QUESTION THREE a) The following are the sales figures for Moonga Brothers Engineering Limited for

the fourteen years from 1991 to 2004:

Year Sales (K’m) 1991 491 1992 519 1993 407 1994 452 1995 607 1996 681 1997 764 1998 696 1999 751 2000 802 2001 970 2002 1026 2003 903 2004 998

Using the five-year moving averages method, establish the general trend of the

sales, and comment on the results.

(Natech. 1.2 Mathematics and Statistics, June 2005). b) With which movement of a time series would you mainly associate each of the

following? (i) A fire in a factory delaying production for 3 weeks. (ii) An increase in employment during the rainy season. (iii) A recession. (iv) A need to increase maize production due to a constant increase in

population. c) Find the moving average of order 3 for the following set of data. 2, 6, 1, 5, 3, 7, 2.

246


d) The daily revenue of a department store is tabulated below over a three-week

period.

Daily Revenue (K’000) Mon Tue Wed Thurs Fri Sat

Week 1 28 25 29 29 36 54 Week 2 30 25 30 34 37 56 Week 3 31 28 33 34 40 59

You are required to; i) Find the six-point moving average and plot the centered values on to a

graph. ii) Use your graph to forecast the trend values during the fourth week.

(Natech, 1.2 Mathematics and Statistics, June 2001) 5.4 Simple Index Numbers

Introduction

This section introduces index numbers and describes the most simple form. Laspeyres and Paasche are discussed. Finally, we discuss the time series analysis.

Index Number An index number is a simple device, which attempts to explain the changes in an economic activity overtime. The use of index numbers is important in the calculations of the inflation rate. ZESCO and ERB have incorporated the inflation in their computation of electricity tariffs and gasoline prices respectively.

Simple Indices Price and Quantity Indices A price index measures the change in the money value of a group of items overtime. While a quantity index measures the changes in the quantity of an item. The comparison is made between the given year relative to the base year. The formulas are as follows:

247

Simple Price Index 1000

×=P

Pi

Simple Quantity Index 1000

×q

qi

Where: iP is the price of the item at time i (current year).

0P is the price of the item at time 0 (base year).

iq is the quantity of the item at time i (current year).

0q is the quantity of the item at time 0 (base year).

Example 1 If a commodity costs K3 000 in 2005 and K2 500 in 2004, calculate the simple

price index for 2005 ( )iP using 2004 ( )0P as the base year.

Simple Price Index 1201002500

3000100

0

=×=×=P

Pi

The price index is 120. This means that 2005 (current year) price of the

commodity is 120 percent of 2004 (base year) commodity. In other words the commodity’s price has gone up by 20%.

Example 2 12 000 of a certain commodity was bought in January 2004 and 14 000 of the

same commodity was bought in December 2005. Calculate the simple quantity index for December 2005 using January 2004 as base year.

Simple quantity index

20010000012

00014

1000

=×=

×=q

qi

This means that the quantity sold has increased by 100% of its 2004 figure.

Usually, an index number is required to show the changes in a number of items at once rather than just one item. In this case we use a weighted average of price relatives or quantity relatives. The formulas are:

Weighted average of price relatives

248

1000 ×

=∑

∑

w

P

Pw i

or weighted average of quantity relative

1000 ×

=∑

∑q

qw i

Example 3 Calculate an index of crop prices in 2004 based on 2003 given the table of information below.

Price per tonne (k) Crop Weighting 2004 2003

Wheat 70 250 155 Maize 35 200 115 Rice 15 150 100

The price indices (relatives) for each crop are calculated below.

Wheat: 29.161100155

250 =×

Maize: 91.173100115

200 =×

Rice: 150100100

150 =×

The weight average of price relatives is:

249

( ) ( ) ( )( )

56.163120

15.19627120

225085.60863.11290

153570

1501591.1733529.161701000

==

++=

++++=×

∑

∑

i

ii

w

P

Pw

The crop prices have increased by 63.56% of its 2003. Changing the Base The base of an index number series is changed by taking proportions as shown below. Index A has 2000 as a base year and index B has 2003 as a base year. To convert index A to index B, each index A value was divided by 140. It can be seen that the numbers for each year are in the same proportions for both index A and index B.

Base Change

Year Index A Index B 2000 100 71.4 2001 120 85.7 2002 130 92.9 2003 140 100 2004 150 107.1

Laspeyres Indices

A Laspeyres Index is a special case of a weighted aggregate index, which always use base year weights. It can be either a price index or quantity index.

Laspeyres Price Index 10000

10 ×==∑∑

Pq

PqLp

or Laspeyres quantity index 10000

01 ×==∑∑

Pq

PqLq

Paasche Indices

A Paasche is a special case of a weighted aggregate index, which uses current year’s weights. It can either be a price index in which case:

250

Paasche price index 10010

11 ×==∑∑

qP

qPPp

or Paasche quantity index 10001

11 ×==∑∑

qP

qPPq

Example 4

Calculate the Laspeyre and Paasche indices of share prices in 2003 and based on 2000 given the data below:

Share Prices (K) No. of Shares Sold

Company 1st Jan 2000 ( )0P

1st Jan 2003 ( )

1P

1st Jan 2000 ( )0q

1st Jan 2003 ( )1q

X 150 160 6 000 10 000 Y 250 255 21 000 26 000 Z 350 395 42 000 62 000

( ) ( ) ( )

( ) ( ) ( )12.77100

00070029

00090522

100000623500002625000010150

00042395000212550006160

10000

01

=×=

×++

++=

×=∑∑

qP

qPLaspeyre

( ) ( ) ( )( ) ( ) ( )

17.11010000070029

00072032

100000623500002625000010150

000623950002625500010160

10010

11

=×=

×++++=

×=∑∑

qP

qPIndexPaasche

The Laspeyre’s price index is a reasonable measure of the change in prices over a short period of, say, two years, but if the given year is a longer period in time from the base year, the weights used tend to become out of date as spending habits change and no longer give a realistic comparison between the two years. This disadvantage maybe

251

overcome by the paasche’s price index. However, it is equally unrealistic in that it compares hypothetical past quantities with current real quantities rather than vice versa. One suggested way out of the dilemma is to calculate an average index number which is the geometric mean of the laspeyres and the paasche index numbers. This is beyond this manual. Four main considerations to be borne in mind when constructing an index number. i) The purpose of the index number unless the purpose is clearly defined the

usefulness of the final index will be suspect. The index must be designed with something in particular.

ii) Selection of items for inclusion. The items to be included must be agreed upon.

These should be relevant to the index being calculated. iii) Selection of appropriate weights. The weights should reflect the importance of

the items under consideration.

Chain Index Numbers

In a chain base index the base period progresses by one time period each time, therefore each index number is interpreted relative to the previous period.

1001/Pr

/Pr ×−

=ntimeatQuantityice

ntimeatQuantityiceIndexChain

Example 5

The table below shows the week ending share price on the stock exchange over a period of four weeks for a local company’s shares:

Week 1 2 3 4 Price (K) 150 200 250 175

Calculate and interpret a chain base index using week 1 as the base.

252

( )

( )

( )

( ) 70100250

175:4

125100200

250:3

33.133100150

200:2

100100150

150:1

=×

=×

=×

=×

weekindex

weekindex

weekindex

weekindex

At the end of the second week the share price had increased by 33.33% from the end of the first week. By the end of the third week the share price had increased again but at a slower rate (25%) when compared to week 2. In week 4 the price had gone down with a 30% decrease from week 3.

Exercise 3

1) Show changes in the price of copper between 2000 and 2001 by finding indices based on 2003 prices. Year 2000 2001 2003 2004 Average price of copper (K per tone)

10 000

10 500

10 800

11 000

2) Find the chain-based indices for the value of exports achieved by a company

during 2000 – 2004. Year 2000 2001 2002 2003 2004 Total exports (x 1 000 000) 500 530 520 650 850

3) Calculate the Laspeyre and Paasche price indices for 2003 given the data shown

below (base year = 2001).

2001 2003 Item

Price (K) Quantity

Purchased Price (K) Quantity

Purchased I 10 500 30 11 000 30 II 9 500 50 10 000 70 III 8 000 70 8 500 60 IV 12 000 110 12 500 100

4) Calculate indices based on 2000 for the volume of sugar exported given the table below.

253

Year 2000 2001 2002 2003 Total export (1 000 tonnes) 60 68 70 74

5) Find (a) a laspeyre price index, and (b) a paasche price index of the quantity

produced in 2003 based on the figures in 1993.

No. of Units Produced Cost of Units (K)

Product 1993 2003 1993 2003 A 4 000 6 000 10 000 23 000 B 6 000 7 000 5 000 9 500 C 12 000 15 000 900 1 350



1.1 Prices have been as follows (million kwacha)

Year 1990 1991 1992 1993 1994 Price 4.1 3.7 3.5 3.8 3.9

When converted to index numbers with base 1990, the index for 1994 to the nearest whole number is:

A. 95 B. 105 C. -0.2 D. 5

(Natech, 1.2 Mathematics and Statistics, December 2003) 1.2 The paasche price index is:

A. Relating the cost of buying current period quantities at current period prices to the cost of buying base period quantities at base period prices.

B. Relating the cost of buying base period quantities at current period prices

to the cost of buying base period quantities at base period prices.

C. Relating the cost of buying current period quantities at current period prices to the cost of buying current period quantities at base period prices.

254

D. Relating the cost of buying base period quantities at current period prices to the cost of buying current period quantities at base period prices.


1.3 The following table shows index numbers for the period 1994 – 1998, with 1994

as base year.

Year 1994 1995 1996 1997 1998 Index No. 100 108 111 120 125

If it is decided to commence a new series using 1998 as the base year, what are the index numbers for the years 1994 through 1998? A. 75, 83, 86, 95, 100 B. 10, 11, 11, 12, 100 C. 80, 86, 89, 96, 100 D. 75, 77, 86, 95, 100

1.4 The numbers for the years 1996, 1997 and 1998, calculated on the chain base

method, for a particular commodity, are shown below.

Year 1998 1999 2000 Chain Index No. 100 107 108

What is the 2000 index, to the nearest whole number, using 1998 as base? A. 100 B. 116 C. 108.00 D. 115.56

(Natech, 1.2 Mathematics and Statistics, December 2002) 1.5 Taking 1996 as base (1996 = 100), the price index of a certain commodity in 1998

was 118. which of the following is the price index (to the nearest whole number) of the same commodity in 1996, taking 1998 as base?

A. 82 B. 84 C. 85 D. 87

(Natech, 1.2 Mathematics and Statistics, June 2001) 1.6 An index number represents a comparison with respect to:

A. Quantity B. Price C. Value D. Any of the above can be involved.

1.7 A simple price relative can be computed by using the formula:

255

A. 10000

11 ×qP

qP B. 100

0

1 ×P

P C. 100

1

0 ×P

P

D. 10011

00 ×qP

qP

1.8 Laspeyres’ weighted aggregate price index can be computed by using the

formula:

A. 10010

11 ×∑∑

qP

qP B. 100

11

10 ×∑∑

qP

qP C. 100

00

11 ×∑∑

qP

qP

D. 10000

01 ×∑∑

qP

qP

1.9 An index number that represents a comparison over time for a group of commodities rather than for a single commodity is termed:

A. Complex B. Component C. Simple D. Composite

1.10 If the consumer price index is 150 for a given period, the purchasing power of the

kwacha in the given period as compared with the base period is:

A. K0.15 B. K0.67 C. K1.50 D. K15.00

SECTION B QUESTION ONE a) Calculate the chain-base index numbers of the following series:

Year 1995 1996 1997 1998 1999 Value 46 52 62 69 74


b) A company uses three raw materials (R, S and T) in its production process. The

following is the information about the prices (K per tonne) of the raw materials in 2001 and 2002 and the average weekly quantities (Q) used during 2002.

256

Raw Materials 2001 Prices 2002 Prices Q R S T

5 800 4 200 500

6 000 4 000 750

100 40

1 000 Calculate using 2001 as the base year:

(i) Simple price index. (ii) Weight mean of price relatives.

c) Distinguish between a simple index and a weighted index.

(Natech, 1.2 Mathematics and Statistics, December 2004) QUESTION TWO a) The following extract is taken from an article. Published in the X daily paper,

dated 21 January 2000:

The rate of inflation continued to fall last month putting the Government’s target of single-figure price increases within reach in the next month or two. Shop prices rose 0.6% in December, the same increases as in the previous month, bringing the annual rate of inflation from 14% to 13%. The index of retail prices rose to 199.5 (Jan 1998 = 100). This month could see a bigger fall in the year-on-year rate of price increases. The annual rate of inflation will fall to near 10% or even below it. Inflation over the last six months has been running at about 6.3% at an annual rate.

i) Explain why, if shop prices rose 0.6 percent in December, the annual rate

of inflation should fall from 14 percent to 13 percent. ii) What do you understand by the index of ‘199.5 (Jan 1998 = 100)’.


b) The price of tomatoes and potatoes, and the amount consumed in two years is as

follows:

257

You are required to:

i) Construct a price relative index using quantity weights 200 = 100 (base year).

ii) Interpret the result.

Natech, 1.2 Mathematics and Statistics, December 2003)

c) The following table gives the index number for different groups together with

their respective weights for the year 2000 (base year 1995).

Group Food Clothing Electricity &

Charcoal

Rent Miscellaneous

Group Index No. (I) 130 280 190 300 200 Group Weight (a) 60 5 7 9 19

i) What is the overall cost of living index number for the year 2000? ii) Suppose a person was earning K1 500 000 per month in 1995, what should

be his salary in 2000, if his standard of living in that year was the same as in 1995?


QUESTION THREE a) Calculate the Laspeyre and Paasche indices of the share prices in 1998, based on

1995, given the data tabulated below.

Share Price (K’000)

Share Price (K’000)

No. of Shares Sold

No. of Shares Sold

Company

1st Jan 1995 1st Jan 1998 1st Jan 1995 1st Jan 1998 A 0.80 0.90 5 000 10 000 B 1.80 1.75 20 000 15 000

Item 2001 2002 Units Consumed Tomatoes Potatoes

2 000 15 000

3 000 16 000

2 5

258

C 2.40 2.85 40 000 60 000

(Natech, 1.2 Mathematics and Statistics, December 1995 (Rescheduled)) b) The following relate to ADB (Z) Limited’s Sales in a 5 month period:

Month Sales (Kwacha) December 782 000 January 875 000 February 621 000 March 681 000 April 997 000

Taking January as base period, compute the index for each month.


259

CHAPTER 6

FINANCIAL MATHEMATICS

6.0 Introduction This Chapter is a continuation of the previous chapter, it introduces the types of interest

and applies geometric and arithmetic progressions to solve problems in Financial Mathematics. It concludes with annuities. 6.1 Simple Interest

Persons who rent buildings or equipment expect to pay for the use of someone else’s property. Similarly, those who borrow money must pay for the privilege of borrowing another’s money. This privilege is called interest. The amount of money that was borrowed is the principal of a loan. This interest which increases in value by the same amount each year is called Simple interest. This simple interest is given by the formula

)1.6(→××= trPI where I = interest P = Principal

r% = Interest rate

100

r

T = time in years Therefore, the total value (Amount or future value) after t years, is the principal plus interest and is given by

)2.6()1( →+=××+=

rtP

trPPAt

When the total value (Amount or future value), the interest rate and time are known, the principal (present value) may be calculated by rewriting formula (6.2) as:

)3.6(1

)1(

→+

=

+=

rt

AP

rtPA

t

t

260

Formula (6.3) is often referred to as the “present value” formula.

Example 1 K25 000 000 is invested for three years at an interest rate of 15%. a) Calculate the simple interest paid in any one year. b) Calculate the total value of savings at the end of one, two, and three years. Show

that the total value of savings at the end of successive years is an arithmetic progression.

c) Calculate the present value (principal) when the future value (total value) is K3

750 000 after three years.

a) Start by writing down any information given in the question.

P = 25 000 000; 100

15=r and t = 3 years.

Therefore:

00025011

3100

1500000025

K

trPI

=

××=

××=

b) Using formula (6.2), the total value of the savings after t years is

calculated as: After 1 year, t = 1

00050032

)3.1(00000025

))2(15.1(00000025

2,2

00028750

))1(15.1(00000025

)1(

2

1

K

A

tyearsAfter

K

A

rtPAt

==

+==

=+=

+=

261

After 3 years, t = 3

00025036

)45.1(00000025

))3(15.1(000000253

K

A

==

+=

When the total value of the investment is calculated for each year, notice that the

increments are constant, indicating that this is an arithmetic progression where the difference between any two consecutive years is K3 750 000.

c) The present value K36 250 000 earned in three years’ time may be

calculated by using (6.3), given r = .15 and t = 3.

.00000025

45.1

00025036

)3(15.01

00025036

1

K

rt

AP t

=

=+

=+

=

6.2 Compound Interest In the modern business environment, the interest on money borrowed (lent or

invested) is usually compounded. For example, if K10 000 is placed in savings Account at 20% per year interest, then I = 10 000 × .20 ×1 = K2 000 interest will be added to the account in the first year to bring the balance to K12 000. During the second year I = 12 000 × .20 × 1 = K24 000 will be paid. Interest calculated in this way is called Compound Interest.

In other words, compound interest pays interest on the principal plus any interest

accumulated in previous years. When interest is compounded in this way, the total value tA , of principal P, at 2% per annum is given by the formula

)4.6()1( →+= t

t rPA

Example 2 K1 million is invested at an interest rate of 12%. What is the value of the

investment at the end of year 9?

262

76.0787732

)773078757.2(0000001

)12.1(0000001)1(

9,100

12,0000001

99

K

rPA

yearstrP

t

=

=

+=+=

===

Example 3

a) Calculate the amount owed on a loan of K5 000 000 over 4 years at an

interest rate of 12.5% compounded annually.

20.0330098

)125.1(0000005)1(

4,125.100

5.12,0000005

44

K

rPA

yearstrP

t

==+=

====

b) Musenge places K25 000 on deposit in a bank earning 5% compound

interest per annum. Find the amount that would have accumulated:

i) After 1 year ii) After 2 years iii) After three years.

The final amount accumulated (terminal value), nrPS )1( += Where P = Principal r = Interest rate per annum n = time.

i) 000500287)15.1(00000025 KS == ii) 50030623)15.1(00000025 2 KS == iii) .87502138)15.1(00000025 3 KS ==

263

6.3 Terminal Values

Comparison of Projects.

If we were to be given the choice between two Projects A and B, the expected profits of which over the next four years are:

A: K25,000,000 at 5% inter per annum B: K30,000,000 per annum

Which would we prefer (assuming both require the same initial outlay)? On Project A, we have to compound each flow by adding on interest at 5% pa for

the number of years remaining until the end of the projects, that is the year 1 cash flow of K25,000,000 earns 3 years interest and is thus worth K107 753 125 at the end of 4 years. We have compounded the flows to produce what is termed as the Terminal value of each flow.

Project A Cashflows K

Year 1 625,940,28)05.1(00000025 3 =

Year 2 25 000 000 2)05.1( = 27,562,500

Year 3 25 000 000(1.05) = 26,250,000

ear 4 25 00 000 25,000,000 K107,753,125 =========

Project B

Year 120,000,000 = 120,000,000

264

Net Terminal Value With the calculations just carried out , we are in a position to choose between the two projects since they have the same outlay. However, we have not as yet considered whether either of them is worthwhile. This will depend on the initial outlay required to generate K107 753 125 which we could receive by investing in Project A. If we end up with a deficit, we could reject the project. Suppose in this case the projects require an initial outlay of K15 000 000 at the beginning of year 1 (refered to as year 0). We cannot compare this outlay directly to K107, 753,125 generated since this is the return at the end of four years. At that time we will have lost four year’s potential interest on the outlay of K15 000 000. To allow for this we need to calculate the terminal value of the initial outlay by adding four years’ interest at 5% The complete solution is as follows: Year Cash flow Compound factor Terminal Value 0 (15 000 000) 4)05.1( (18 232 593.75)

1 25 000 000 3)05.1( 28 940 625

2 25 000 000 2)05.1( 27 562 500

3 25 000 000 )05.1( 26 250 000 4 25 000 000 25 000 000 K89 520 531.25 ========== Note that the year column refers to the end of various years. Thus the initial outlays occur at the start of the project, i.e., the end of year 0 (which means the beginning of year 1). The first cash flow is received at the end of year 1, and so on for the subsequent cash flows. The net surplus in this case K89,520,531.25 is called the net terminal value (NTV) and since it is positive, indicating a surplus, the project is worthwhile and should be accepted. The positive net terminal value indicates that the cash and interest earned from the project exceed the value of the initial outlay plus interest. If the net terminal value is negative, indicating a deficit, the project would be rejected.

265

Example Find the terminal values of the following investment a) An initial outlay of K500 000 which will generate the following cash flows.

Year Cash flow

1 30 000 2 20 000 3 40 000 4 50 000

The annual interest rate available for deposit is 7.5%. b) An initial outlay of K5 000 000 which will generate the following cash flows: Year Cash flow

1 2 500 000

2 2 000 000 3 - 4 5 000 000

The annual interest rate available for deposits is 10%. c) An initial outlay of K3 000 000 which will generate cash flows of k120 0000 for

four years. The annual interest rate available for deposits is 8.5%.

Solution

The net terminal values are calculated as follows: a) Year Cash flow Compound factor Terminal value

00.00050000504

00.00043)075.1(000403

50.11223)075.1(000202

91.26837)075.1(000301

)57.734667()075.1()000500(0

2

3

4

×××

×

(514 353.16)

So the project would be rejected.

266

b) Year Cash flow Compound factor Terminal value

000000500000054

3

0004202)10.1(00000022

0053273)10.1(00050021

)5003207()10.1()0000005(0

2

3

4

−−−××

×

3 427 000

So the project would be accepted. c) Year Cash flow Compound factor Terminal value

00.000200100012004

00.0003021)085.1(00012003

00.6704121)085.1(00020012

95.7463251)085.1(00012001

)10.5761574()085.1()0000003(0

2

3

4

×××

×

(1 289 840.85)

========= So the project would be rejected.

267

6.4 Other Applications of the Compound Interest Formula In the Compound Interest Formula, there are four variables, .,, tandrPAt If

any three of these variables are given, the fourth may be determined. In some cases, you will require the rules for indices and logs. For example, a general expression for r may be derived as:

t

t rPA )1( +=

dividing throughout by P, we get

tt rP

A)1( +=

then take the t th root on both sides

rP

A tt +=

1

1

making r the subject of the formula gives the required formula for r .

)5.6(1

1

→−

=t

P

Ar t

Example 4

a) Find the compound interest rate required for K15 000 000 to grow to

K25 000 000 in 5 years. b) A bank pays 13.5% interest compounded annually. How long will it take

for K15 000 000 to grow to K20 000 000?

a) Calculater when given

5,00000015,000000255 === tandPA

268

Direct from the formula, (6.5)

%.8.10

108.01108.1

10000015

00000025 51

==−=

−

=

ror

r

An interest rate of 10.8% is required if this investment is to be doubled in value over a period of 5 years.

b) When %.5.1300000015,00000025 === randPAt Direct from

compound interest formula (6.4),

t

t

)135.1(667.1

)135.1(0000001500000025

=

+=

taking natural logarithms or common logarithms on both sides gives us

0355.4

12663265.0

511025603.0

)135.1ln(

)667.1ln(

)135.1ln()667.1ln(

=

=

=

=

t

t

So, at 13.5% interest, it will take over 4 years for the investment to double

in value.

269

6.5 Present Value At Compound Interest

At present value of a future sum, tA , is the amount which, when put on deposit

now (i.e. Pt ,0= ), at (r%) rate of interest, will grow to the value of tKA after t

years. The present value, P, is calculated by arranging the compound interest (6.4).

)6.6()1(

)1(

→+

=

+=

tt

tt

r

AP

rPA

Example 5 If the future value of an investment is K25 000 000 invested at 12.5% compound

interest per annum for five years. Compute the present value. yearstrKA 5,125.,000000255 === . Using formula (6.6)

93.22387313

)125.1(

000000255

KP

P

=

=

6.6 Nominal and Effective Rates When interest is compounded several times per year, for example it may be compounded

daily, weekly, monthly, quarterly, semi-annually or continuously. Each period is called a conversion period or interest period. Then the amount of the future value is given by the formula

where tmn ×= = total number of conversion periods m = conversion periods per year t = number of years

)7.6(1 →

+=mt

t m

rPA

270

Example 6

K25 000 000 is invested for five years at 12.5%. Calculate the total value of the investment when compounded i) Monthly ii) Daily

Assume a year has 365 days. i) P = 25 000 000, r = 12.5% = .125, t = 5 and m = 12. Using formula

(6.7) the total value after three years with 60512 =×=×= tmn conversion periods is calculated as:

11.40255546

)01041666.1(00000025

12

125.100000025

1

60

60

5

K

P

m

rPP

mt

t

==

+=

+=

ii) 18255365,365,5%,5.12,00000025 =×=×===== tmnmtrP

00070046

)868.1(00000025

365

125.100000025

1

1825

5

K

m

rPP

tm

==

+=

+=×

When working with problems involving interest we use the term payment

period as follows: Annually Once per year Semi-annually Twice per year Quarterly 4 times per year Monthly 12 times per year

Daily 365 timer per year (some basics use 360 times per year)

271

6.7 Effective Rate Of Interest Interest rates are usually cited as nominal rates of interest expressed as per annum

figures. However, as compounding may occur several times during the year with the nominal rate, the amount owed or accumulated will be different from that calculated by compounding once a year. So a standard measure used to compare the amount earned (owed) at quoted nominal rates of interest when compounding is done several time per year is called the annual percentage rate (APR) or effective annual rate or effective rate of interest. Let us consider formula (6.7)

mt

tm

rPA

+= 1 nominal rate compounded m times per year.

t

t APRPA )1( += APR rate compounded annually. Note that m = 1 i.e, once per

year. Equating the two amounts since they are the same, we have

tmt

APRPm

rP )1(1 +=

+

making APR subject of the formula, we have

)8.6(11 →−

+=m

m

rAPR

Example 7

Interest on a savings account is payable semi-annually at a (nominal) rate of 12.5% per annum. What is the effective rate of interest? R = 0.125 and m = 2 because interest is payable twice yearly, so

1289.0

12

125.01

2

=

−

+=APR

The effective rate of interest is 12.89%

272

Example 8 a) A finance company advertises money at 25% nominal interest, but compounded

quarterly. Find the effective interest rate (APR) b) Two banks in a local town quote the following nominal interest rates. Bank X

pays interest on a saving account at 12.5% compounded monthly and bank Y pays 12.5% on a savings account compounded quarterly, which pays its savers the most interest?

a) r = .25, and m = 4 because interest is payable 4 times a year, so

274.0

14

25.01

4

=

−

+=APR

The effective rate of interest is 27.4%.

b) Bank X pays interest of 12.5% compounded monthly

%2.13

132.0112

125.1

12

=

=−

+=APR

Bank Y offers the greater effective interest rate and thus pays its savers more interest.

6.8 Investment Appraisal If an initial investment will bring in payments at future times, the payments are called

cash flows. The net present value, denoted NPV, of cash flows is defined to be the sum of the present values of the cash flows (revenue), minus the initial investment (cost). If NPV > 0, then the investment is profitable, if NPV < 0 the investment is not profitable.

273

Example 9 Suppose that you can invest K100 000 000 in a business that guarantees you cash flows at

the end of years 1, 2 and 3, as indicated in the table. Assume an interest rate of 12.5% compounded annually and find the net present value of the cash flows.

Year Cash Flow

1 5 000 000 2 400 000 3 200 000

NPV = cash inflows – cash outflows (Revenue) (cost)

79.03909995

00000010039.46614038.04931644.4444444

000000100)125.1(000200)125.1(000400)125.1(0000005 321

K−=−++=

−++= −−−

Since NPV < 0, the business venture is not profitable. If one considers the time value of

money, it would be better to invest the K100 000 000 in a bank paying 12.5%.

6.9 Internal Rate Of Return The discount rate at which a project has a net present value of zero is called the Internal

rate of return (IRR). There is no precise formula for calculating the IRR of a given project. However, it can be estimated (using linear interpolation technique) with:

a) graphically, or

b) by formula given by IRR = )9.6(12

1221 →−

×−×NPVNPV

NPVrNPVr

Both techniques need that the NPV is calculated using two different discount rates. We

explain the two methods in the following examples.

Example 10

274

A project involves an initial outlay of K100 000 000. The expected cash flow at the end of the

next four years is given as given as follows (the amounts are in million of kwachas).

1 2 3 4 Cash flows 50 125 167 182

a) Determine IRR graphically by plotting the NPV against r. For r = 0.05, 0.08, .10,

.20. b) By formula, show that the value of the IRR is slightly different when calculated

from pairs of points.

a) Graphically. Excel is ideal at this point. Since calculation of NPVs requires the repeated use of the formula t

t rANPV −+=∑ )1( where ∑ is

the symbol for the sum of several NPVs. Then we plot the curve of NPV against r .

Table 6.1 Excel Sheet for Calculating NPVs at Different Interest Rates

t Cash flow = 0.05 = 0.08 r = 0.10 r = 0.20 r =0.24 0 1 2 3 4

300− 50

125 167 182

300− 47.61905 113.3787 144.2609 149.7319 154.9905

300− 46.2963

107.1674 132.5700 133.7754 119.8091

300− 45.45455 103.3058 125.4696 124.3084

98.53835

300− 41.66667 80.0000 85.5040 74.5472

-18.2821

-300 40.32258 81.2953

87.58937 76.98119 -13.8113

275

NPV 200 150 • • 100 • 50 •• 0 • • -50 0.1 0.2 0.3 r IRR Figure 6.1 The NPV for each discount rate is plotted in Figure 6.1. The IRR is the value of

r at which this graph crosses the horizontal axis. In Figure 6.1 this point is between r = 10% and 20%, but considerably closer to 20%.

b) In Table 6.1 are several positive and negative NPVs. Therefore, the IRR given by

formula (6.9) is calculated from any such pair. Summarising points already calculated. Points A B C D E

r 0.05 24.020.010.008.0 NPV 154.9905 119.8091 98.53835 −18.2821 −13.8113

i) From points C and D

276

1843.0

82045.116

53586.21

53835.982821.18

)53835.98(20.0)2821.18(10.0

12

1221

=

−−=

−−×−−×=

−×−×=

IRR

NPVNPV

NPVrNPVrIRR

Therefore,IRR = 18.43%.

ii) From points B and E,

22346.0

6204.133

859088.29

8091.1198113.13

)8091.119(24.0)8113.13(08.0

12

1221

=

−−=

−−×−−×=

−×−×=

NPVNPV

NPVrNPVrIRR

Therefore, IRR = 22.346% The two results (i) and (ii) demonstrate that slightly different estimates are

calculated from different pairs of points.

277

6.10 Comparison of Appraisal Techniques: NPV and IRR When comparing the profitability of two or more projects, the most profitable project

would be (a) the project with the largest NPV, (b) the project with the largest IRR. The advantages of NPV method is that : i) It gives results in cash terms ii) It is practical as it discounts net cash flows. The disadvantage is that it relies on the choice of one discount rate which means that a

change in the discount rate could lead to a change in the choice of project. The advantage of the IRR is that it does not depend on external rates of interest. A major

weakness is that the method does not differentiate between the scale of projects; for example, one project might involve a cash flow in units of K5 000 000 while another involve units of K5. Not that in most cases where two or more similar project are being ranked in order of preference, the methods of NPV and IRR will generally agree on the best project but this is not a hard and fast rule.

Example 11 The net cash flow for two projects, A and B, is as follows: K’000, 000 Year 0 1 2 3 4 Project A -45.0 -13.5 18 27 36 Project B -22.5 -9 4.5 13.5 22.5

a) Use the net present value criterion to decide which project is the most profitable if a discount rate of

i) 8%, and ii) 12% is used.

b) Calculate the IRR of each project. Which project would now be considered more profitable?

278

Project A

Net Flow Discount Factor at

8% Present Value Discount factor at

12% Present Value

1.0000

0.9259

0.8573

0.7938

0.7350

12.50

1.0000

0.8929

0.7972

0.7118

0.6355

45.00

12.05

116.0

42.6

7464.0

42.6

048.06984.0

82.56.0

)6.0)(08.,10()82.5(12.

12

2112

+=

−−=

−−−=

−−−=−+−=

−×+×−=

NPVNPV

NPVrNPVrIRR

279

Project B

Net Flow Discount Factor at

8% Present Value Discount factor at

12% Present Value

1.0000

0.9259

0.8573

0.7938

0.7350

1.0000

0.8929

0.7972

0.7118

0.6355

22.50

8.04

3.59

9.61

14.30 3.04

083.0

33.3

278.0

42.6

0348.02432.0

29.004.3

)29.0(12.)04.3(08.0

12

1221

=

−=−

−−=

−−−−=

−×−×=

NPVNPV

NPVrNPVrIRR

From (a) at rate 8%, project A has the highest NPV (9.42) and thus would be chosen as best.

(Note also that project B has the highest NPV (0.29)). And from (b) project A has the highest rate of return at 11.6% and thus would be chosen as best agreeing with the choice in (a). Overall clearly project A is the best choice.

280

Exercise 1 1) K5 400 000 is invested at 9.5% simple interest. How much will have to be

accrued after 5 years? 2) Find the amount of:

a) K 2 160 000 compounded at 13.5% for 3 years.

b) K5 580 000 compounded at 8.5% for 10 years.

3) Calculate the present value of K56 million that is expected to be received in five years’ time when simple interest is 6.5%.

4) Calculate the compound interest rate required for K250 000 to grow to K450 000

in 3 years’ time.

5) Calculate the number of years it will take a sum of K450 000 to grow to K1 800 000 when invested at 4.5% interest compounded annually.

6) Calculate the APR for a 6.5% nominal rate of interest which is compounded

a) four times per year b) 12 times per year, and c) 3 times per year.

7) Two banks in a Geal town quote the following nominal interest rates: Bank X

charges interest on a loan at 9.5% compounded semi-annually and bank Y charges 9.0% on a loan compounded quarterly. Which bank charges the most interest on a loan?

8) You have a choice of two savings schemes. Scheme A offers 7.0% interest

payable semi-annually and scheme B offer 6.5% interest payable quarterly. Which bank charges the most interest on a loan?

9) Find the present value of K25 million in 6 years time if the discount rate is 13.5%

compounded semi-annually.

10) A Financial group can make investment of K240 million now and receive K264

million in two years time. What is the internal rate of return?

281

11) An investment project has the following NPV calculated for a range of discount rates. Give an approximate IRR for the project.

Discount rate (%) NPV (K’million)

5 5.5 6

6.5 7

11.25 7.02 3.42 1.2

-2.349

The company considering the project could invest an equivalent amount of money for a

similar length of time at an interest rate of 7.5%. Should they undertake the project?

12) Calculate the NPV of a project, which requires an initial outlay of K90 million

now but should return K36 million at the end of year 2 and K20 million at the end of four years. Assume a discount rate of 3.5% compounded annually. Estimate the IRR of this project.

13) Find the terminal value of the following compounded deposits. a) K15 000 000 deposited for 3 years at an annual interest of 8.5%.

b) K30 000 000 deposited for 5 years at an annual interest rate of 12%.

c) K4 800 000 deposited for 4 years at an annual interest rate of 25%.

14) Find the terminal value and compound interest payable if you deposit K24 000 000 for one year with a bank offering 5% interest per month on deposit accounts.

282

6.11 Series of Payments

6.11.1 Amount of an Annuity and their Present Values

An annuity is a sequence of payments made at fixed periods of time over a given time interval. The fixed period is called the payment period, and the given time interval is the term of the annuity. An example of an annuity is depositing of K100,000 in a savings account every 6 months for a year.

In this Section, we consider the amount accrued from a series of such payments

and also the present values of a series of such payments which are to be made.

The amount of an annuity is found using the following formula:

( ))10.6(

11 →

−+=r

rRA

n

while the present value of an annuity is found using the following formula.

)11.6()1(1

)(Pr →

+−=−

r

rRannuityofvalueesentP

n

where A is the amount of an ordinary annuity r is the interest rate per period n number of period P is the present value of an ordinary annuity.

Example 11

Find the amount of an annuity consisting of payments of K225 000 at the end of

every 3 months for 4 years at the rate of 6.5% compounded quarterly.

To find the amount of the annuity we use equation (6.10) with R =22 000,

283

n = 4(4) = 16, and r = 01625.04

065.0 =

67.8490734

)10599851.18(000225

01625.0

1)01625.1(000225

16

K

A

==

−=

Example 12

What is the terminal or future value of an annuity of K4 500 000 for five years at 10% of interest rate per annum? R = 4 500 000, r = 0.10, n = 5 and ≠n 6

00.95047227

)1051.6(0005004

10.0

1)10.1(5004

5

K

A

=

=

−=

Alternatively, we reason as follows: The first payment will be made at the end of year 1, and so at the end of 5 years, it will have

been invested for 4 years and will have a value of 4)10.1(0005004 . The next payment is made at the end of year 2, and so at the end of year 5 it will have been invested for 3 years and will have a value of .)10.1(0005004 3 It is easy to work using a table as follows:

284

End of Year Amount (K) Value of the end of Year 5

5

4

3

2

1

0005004

0005004

0005004

0005004

0005004

000

)10.1(000

)10.1(000

)10.1(000

)10.00091

2

3

4

Notice the pattern At the end of year 5 the total value of all the payments will be the total of the third column. In

reverse order, we can see that this is a geometric series with the first term a = 4 500 000, common ratio r = 1.10 and n = 5.

That is 432 )10.1(000450)10.1(00450)10.1(0004500)10.1(00045000004500 ++++

−−=1

1

r

raS

n

n using formula (5.6)

−−=110.1

1)10.1( 5

5 aS

.95047227)1051.6(0004500 K==

The present value of an annuity is the sum of the present values of all the payments. It represents

the amount that must be invested now to purchase the payments due in the future. Unless otherwise specified, we assume that each payment is made at the end of a payment period, that is called an ordinary annuity. Example 13

Find the present value of an annuity of K450 000 per month for 3 years at an interest rate of

6.5% compounded monthly.

Using equation (6.11), R = 450 000, .36)3(1212

065.0 === nandr

Thus

285

77.73168614

6371817.32(450

0054.0

)0054.1(1000450

36

K

P

=

=

−=−

Alternatively, there will be altogether 12(3) = 36 monthly payments. The interest rate is

.0054.012

065.0 = The present value of the payment is therefore:

362 )0054.1(

000450...

)0054.1(

000450

0054.1

000450 +++

This is a geometric series with

.360054.1

1,

0054.1

000450 === nandra

So the sum is

−−=1

1

r

raS

n

n using formula (5.6)

77.731686141

0054.1

1

10054.1

1

0054.1

000450

36

K=

−

−

=

6.12 Equation of Value Example 14 Suppose that Mr Chilufya owes Mr Banda two sums of money: K500 000 due in 2 years

and K300 000 due in 5 years. If Mr Chilufya wishes to pay off the total debt now by a single payment, how much should the payment be? Assume an interest of 5% compounded semiannually.

The single payment x due now must be such that it would grow and eventually

pay off the debts when they are due. That is, it must equal the sum of the present values of the future payments. As shown in the figure below, we have

286

104 )025.1(000300)025.1(000500 −− +=x This equation is called an equation of value. We find that

84.33468752.23435932.452975 Kx =+= Year 0 1 2 3 4 5 Single Payment x 500 000 300 000 Present 500 000 (1.025)-4 4 periods Value of Dents

300 000(1.025)-10 10 periods

Replacing two future payments by a single payment now for Example 13

In general, an equation of value illustrates that when one is considering two methods of paying a debt (or other transaction), at any time the value of all payments under one method must equal the value of all payments under the other method. Example 15

A debt of K15 000 000, which is due 7 years from now is instead to be paid off by three payments: K2 500 000 now, K7500 000 in 4 years, and a final payment in the 6th year. What would this payment be if an interest rate of 5% compounded annually is assumed? Let x be the final payment due in 6 years. Setting up the equation of value, we have

1

26

)05.1(00000015

)05.1(0007500)05.1(0002500−=

++ x

287

Year 0 1 2 3 4 5 6 7 2500 000 7500 000 x 15 000 000 7500 000 (1.05)2

2500 000(1.06)6

15 000 000 (1.06)-1

Time values of payment for Example 15 3546297.78 + 8268 750 + x = 14285714.29 x = K2470666.51

6.13 Perpetuities

When the present value of an annuity continue for an indefinitely long period of time, we have what we call ‘Perpetuities’ . The present value of annuity say receivable in arrears in perpetuity at a given discount rate r is given by the following formula:

PV of perpetuity = )12.6(→

=

ratediscount

cashflowsAnnual

r

A

Example 15 Find the present value of K25 000 000 receivable annually in arrears at a discount rate of

7.5% . A = 25 000 000, r = 0.075

30.333333333075.

00000025K

r

APV ===

288

Example 16 The T Company is expected to pay K11 250 every 6 months indefinitely on a share of its

preferred stock. If money is worth 6.5% compounded semi annually to X, what should he be willing to pay for a share of the stock?

A = 11 250, r = 0.0325; then

85.1533460325.0

11250K

r

APV ===

6.14 Amortization of Loans

When borrowing a sum of money from a bank or building Society for house purchase (a

mortgage), it is usual to repay it by a series of regular equal installments. The present value of the series will be the same as the amount borrowed.

Example 17 Chisha wishes to borrow a sum of money to buy a house. She wishes to repay exactly

K2 500 000 a month for 15 years at the rate of 13.5% compounded monthly,. Find how much she can borrow.

Using formula (6.11), 1801215,01125.012

135.,0005002 =×==== nrR .

80.750556192

)02270031.77(0005002

01125.0

)01125.1(10005002

180

K==

−=−

Alternatively, there will be 1801215 =× . Monthly repayments of K2 500 000. The

interest rate is 1.125%. The present value of repayments is therefore

289

( )

80.750556192

101125.1

1

101125.1

1

01125.1

0005002

180,01125.1

1,

01125.1

0005002

01125.1

0005002...

)01125.1(

0005002

)01125.1(

0005002

01125.1

0005002

180

18032

K

nra

=

−

−

=

===

++++

Example 18 Lungowe borrows K11250 000 to be paid in 4 years to buy a car. How much must she

repay per month assuming an interest rate of 7.5% a year compounded monthly. Using formula (6.11)

00625.012

075.0,48)12(4,00025011

)1(1

=====

+−=−

rnPwhere

r

rRP

n

Substituting these values in the above formula

−=−

00625.0

)00625.1(100025011

48

R

and making R the subject of the formula 70312.5 = R(0.258489819) R = K272 012.65

290

6.15 Amortization Schedule An analysis of how each payment in the loan is handed can be given in a table called an

Amortization Schedule. The amortization schedule contains i) Principal outstanding at the beginning of the period

ii) Interest for period

ii) Payment at the end of the period

iv) Principal repaid at end of period

A loan is amortized when part of each payment is used to pay interest and the remaining

part is used to reduce the outstanding principal. Since each payment reduces the outstanding principal, the interest portion of a payment decreases as time goes on. For examination purposes, a schedule would only be asked for where the period was relatively short; for example, up to 5 or 6 time periods.

Example 19 A debt of K22 500 000 with interest at 6% compounded semi-annually is amortized by

equal payments for the next 2 years.

a) Find monthly payment b) Construct an amortization schedule.

a) Using formula (6. 11)

52.1080536

)717098403.3(00050022

03.0

)03.1(100050022

422,2

06.0,00050022

4

KR

R

R

nrP

=

=

−=

=×===

−

The monthly repayments must be K6 053 108.52

291

b) The Amortization Schedule is shown in Table 1.0 Period Principal Outstanding

at Beginning of period

Interest for period Payment at End of Period

rincipal Repaid at End of Period

22 500 000 17 121 891.48 11 582 439.70 5 876 804.37

675 000 513 656.74 347 473.19 176 304.13 1712 434.06

6 053 180.52 6.053 108.52 6 053 108.52 6 053 108.50 24 212 434.06

5 378 108.52 539 457.78

57 05 635.33 5 876 804.37 22 500 000.00

The final payment is adjusted to offset rounding errors. Note that: 1) Principal repaid = payment made + interest paid

2) Outstanding principal at beginning of period = principal outstanding (current)

At beginning of period (previous) – principal repaid at end of period (previous)

6.16 Sinking Fund

A sinking fund is a fund into which periodic payments are made in order to satisfy a future obligation. This is the amount of an annuity as opposed to the present value of an annuity in the case of a loan. Here we use formula (6.10).

−+=r

rR

annuityanofAmountAn 1)1(

)(

Sinking funds are commonly used for the following purposes:

a) repayments of debts

b) to provide funds to purchase a new asset when the existing asset is fully depreciated.

c) To pay for future school fees or a pension.

292

Example 20 The Board of Education received permission to issue K200 000 000 I bonds to build a

new block of classrooms. The board is required to make payments, every 6 months into a sinking fund paying 12.5% compounded semi-annually. At the end of 15 years the bond obligation will be retired. What should each payment be?

The payment R required twice a year to accumulate to K200 000 in 15 years 30215( =×

payments at a rate of interest 0625.02

125. = per payment period).

Using formula (6.10),

38.5674202

)62525618.82(

0625.0

1)0625.1(000000200

1)1(

30

KR

R

R

r

rRA

n

=

=

−=

−+=

Example 21 A woman borrows K13 500 000, which will be paid back to the lender in one payment at

the end of 4 years. She agrees to pay interest semi annually at 15%. At the same time she sets up a sinking fund in order to repay the loan at the end of 4 years. She decides to make equal deposits into her sinking fund, which earns 6.5% interest compounded semi-annually.

a) What is the monthly sinking fund deposit? b) Construct a table that shows how the sinking fund grows over time.

c) How much does she need each month to be able to pay the interest on the loan and make the sinking fund deposit?

293

a) The sinking fund deposit is the value of R in the formula.

−+=r

rRA

n 1)1(

where A equal the amount to be accumulated namely,

.2

065.0,824,0050013 ==×== randnKA The sinking

fund deposit is therefore:

55.7455041

)97161647.8(00050013

2

065.

12

065.01

00050013

8

KR

R

R

=

=

−

+=

b) The table below shows the growth of the sinking fund over time. The entries for payment number 8 are obtained by using the amount of an annuity formula for a monthly payment of K1 504 745.55 made for 8 months at 6.5% compounded monthly.

79.8163186

)199259328.4(55.1504745

2

065.0

12

065.01

55.1504745

8

K

Total

=

=

−

+=

The deposit for payment number 8, the final payment, is only K1 504 745.58 because a deposit of K1 504 745.58 results in a total payment of K134 99999 97.

294

Period Interest Added Deposit Increase in Fund Amount in Fund at End of Period

0

48 904.23

97.85

151 532.51

205 361.55

260 940.03

318 324.81

377 574.59

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1 504 745.55

1504 745.55

1 553 649.78

1 604 143.40

1 656 278.06

1 710 107.10

1 765 685.58

1 823 070.36

882 320.17

1 504 745.55

3 058 395.33

4 662 538.73

6 318 816.79

8 028 923.89

9 794 609.47

11 617 679.83

13 500 000.00

c) The monthly interest payment due on the loan of K13 500 000 at 15% interest is

found using the simple interest formula.

500,012,12

1)15.0(00050013 KI =

=

Thus the woman needs to be able to pay K1 504 745.55 + K1 012 500 = K2 517 245 55 each month.

Example 22

A copper mine is expected to yield an annual net return of 900 million for the next 15 years, after

which it will be worthless. An investor wants an annual return on the investment of 18%. If she can establish a sinking fund earning 12% annually, how much should he be willing to pay for the mine?

Let x be the purchase price. Then 0.185x represents an 18.5% annual return on investment. The annual sinking fund contribution needed to obtain the amount x in 15 years is found

by solving 11)1( −

−+r

rx

n

where n = 15 and r = 0.12. The investor should be willing

to pay an amount x so that

295

Annual return + Annual sinking = Annual return on investment fund requirement

00.5268052484

000000900211824239.0

000000900026824239.0185.0

00000090012.

1)12.1(185.0

115

Kx

x

xx

xx

=

=

=+

=

−+−

A purchase price of K4 248 805 526.00 will achieve the investor’s goals.

Example 23

A machine costing K18 million new is estimated to have after 6 years of use a scrap value of K1.8million. if the depreciation fund earns 3%, use the sinking fund method to:

a) Find the annual deposit into the fund,

b) Find the amount in the fund at the end of 4 years,

c) Prepare a depreciation schedule.

a) Original cost – Scrap value = 18 – 1.8 = 17.2

Therefore:

51.4795042

194052296.0

486.0

1)03.1(

)03.0(2.17

03.0

1)03.01(2.17

6

6

=

=−

=

−+=

R

R

296

Depreciation Charge

Interest fund Increase fund Amount in fund Book value

0

2 504 479.51

2 504 479.51

2 504 479.51

2 504 479.51

2 504 479.51

2 504 479.51

0

0

152 522.80

232.232.87

314 334.24

398 898.66

486 00.00

0

2 504 479.51

2 657 002.31

2 889 235.18

3 203 569.42

3 602 468.08

4 088 468.08

0

2 504 479.51

5 084 093.41

7 741 095.72

10 477 808.1

13 296 621.85

16 200 000.02

18 000 000

15 495 520.49

12 915 906.59

10 258 904.28

7 522 191.90

4 703 378.15

1 799 999.98

The error of K0.02 in the final book value is due to rounding off all entries to 2 decimal places.

Exercise 2

1) Find the present value of the given future payment at the specified interest rate.

a) K24 million due in 15 years at 5% compounded annually.

b) K15.75 million due in 7 years at 6% effective

c) K18 million due in 12 years at 7.5% compounded semi-annually

d) K11.25 million due in 15 months at 6.5% compounded quarterly.

e) K9 million due in 2½ years at 18% compounded monthly.

2) Mulenga wishes to borrow a sum of money to buy a house. She wishes to repay exactly K2 000 000 a month for 15 years, starting at the end of the present. How much can she borrow assuming an interest rate of 13.5% payable monthly?

3) Construct amortization schedules for the indicated debts. Adjust the final payment, if necessary.

a) K25 million repaid by four equal yearly payment interest at 15%

compounded annually.

297

b) K36 million repaid by six equal semi annual payments with interest rate of

8% compounded semi annually.

c) K4050 000 repaid by five equal quarterly payments with interest at 10% compounded quarterly.

4) A person borrows K9million and will pay off the loan by equal payments at the end of each month for 6 years. If interest is a the rate of 18.5% compounded monthly, how much is each payment?

5) Find the monthly payment of a 6-year loan for K31.5million if interest is at

12.24% compounded monthly. 6) What sum must a parent invest now in order to obtain school fees of K9 million

payable 3 times a year for 10 years starting in 4 months’ time, when the nominal interest rate is 6.5% compounded 3 times a year?

7) Chibuye is saving for his daughter’s wedding in 3 years’ time. He decides to save

either

i) K67 500 a month starting at the end of the current month at a nominal rate of 7.5% a year compounded monthly or

ii) K 2 070 000 a quarter starting at the end of the first 3 months at a nominal

rate of 6.5% a year compounded quarterly. How much does he save?

8) A debt of K500 000 000 compounded at 18.5% is to be discharged over 4 years using a sinking fund method. Find the annual payment (based on an ordinary annuity) if the fund earns 12.5%. Draw up a schedule showing both the position of the debt and the fund each year.

9) A debt of K2 000 000 due in 4 years and K2 000 000 due in 5 years is to be repaid by a single payment now. Find how much the payment is if an interest rate of 10% compounded quarterly is assumed.

10) A debt of K5 000 000 due in 5 years is to be repaid by a payment of K3 000 000

now and a second payment at the ends of 6 years. How much should the second payment be if the interest rate is 5% compounded quarterly?

298


MULTIPLE CHOICE QUESTIONS

(SECTION A)

1.1) Today Chisenga purchase an annuity of K6 750 000 per year for 15 years from an insurance company which was 3% compounded annually. If the first payment is due in one year, what did the annuity cost Chisenga?

a) K80 581 061.25 b) K3 037 500 c) K125 542 668.70 d) K9 787 500

(Natech, 1.2 Mathematics & Statistics, November/December 2000) 1.2) A certain machine costs K5 400 000. The depreciation for a month at the end of

any month is estimated to be 5% of the value of the beginning of the month. At what value is the machine cared after 24 months of use?

a) K1 659 727.08 b) K1 576 740.73 c) K11 610 000 d) K90 990 000

(Natech, 1.2 Mathematics & Statistics, November/December 2000)

1.3) Find the Net present value of a project which requires an initial outlay of K50 000 and guarantees you a cash flow of K30 000 per annum for the next three years with an interest sale of 10%.

a) K25 608 b) K24 606 c) K50 000

d) K25 000 (Natech, 1.2 Mathematics & Statistics, June 2002)

1.4) In how many years will K1 000 000 amount to K3 207 000 at 6% per annum compound interest? (Give your answer to the nearest whole number).

a) 17 b) 18 c) 19 d) 20


1.5) Calculate the annual effective rate of interest of 5% compounded monthly to two

decimal places. a) 6% b) 5.12% c) 5% d) 4.91%


299

1.6) What is the present value of an annuity that pays K400 000 a month for the next five years if money is worth 12% compounded monthly.

a) K24 000 000 b) K17 982 015 c) K28 800 000 d) K2 808 000


1.7) What is the amount for an ordinary annuity of K10 000 a year for 4 years at 8% compounded annually?

a) K45 061.12 b) K48 500.50 c) K13 300 d) K13 604.89 (Natech, 1.2 Mathematics & Statistics, December 1998)

1.8) If a boy undertakes to deposit K100 on September 1, K200 on September 2, K400 on September 3, K800 on September 4, and so, how much will be deposit from September 1 to September 15, inclusive?

a) K32 767.00 b) K327 670.00 c) K3 276 700.00 d) K3 276.70 (Natech, 1.2 Mathematics & Statistics, December 1998)

1.9) Find the compound interest on K800 000 for 2 years at 6% per annum, interest to be added half yearly.

a) K96 200.00 b) K192 400.50 c) K100 407.05 d) None of these


1.10) If K40 000 000 invested for 5 years yields a simple interest of K3 800 000, what will be the interest on K24 000 000 invested at the same rate for 7½years?

a) K380 000 b) K240 000 c) K342 000 d) K760 000


300

MULTIPLE CHOICE QUESTIONS (SECTION B)

1.1) What sum will earn K15 750.00 simple interest in 146 days at 4.5% per annum? a) K15 778.35 b) K284 112.30 c) K157 500.00 d) K875 000.00

(Natech, 1.2 Mathematics & Statistics, December 2002) 1.2) If 1 500 000 is deposited at simple interest of 3% per year, what amount of money

would be in the account at the end of 12 years? a) K180 000 b) K518 000 c) K699 000 d) K18 000


1.3) What is the discountable value of a bill for K475 000 drawn on 4th march at 3 months and discounted on 10th may at 6%?

a) K472 813.70 b) K28 500.00 c) K427 500.00 d) 2 186.30 (Natech, 1.2 Mathematics & Statistics, December 2002) 1.4) The terms for a five year lease agreement are that, K10 million must be paid at the

beginning of the first year, to be followed by four-equal payments at the beginning of years two, three, four and five at a discount rate of 8%. If the present value of the four equal payment is K26 496 000, the total amount to be paid during the lease period is close to?

a) K32 million b) K40 million c) K42 million d) K44 million

1.5) A machine assumed to depreciate at a fixed rate of 12% per annum, will have a book value of K9 288 080 in six years time. Its purchase value to the nearest ten is:

a) K15 million b) 21.5 million c) K25 million d) K20 million.

301

1.6) A government bond of K1 million is advertised to become K1.57 million after 5

years. The effective annual rate of interest to the decimal place is:

a) 7.5% b) 9.0% c) 10.8% d) 11.4%

1.7) The net present value of an investment at 20% is K12 400 000 at 12% is a loss of K8 000 000. What is the internal rate of return of this investment?

a) 18% b) 17% c) 16% d) 15% (Natech, 1.2 Mathematics & Statistics, December 2003) 1.8) How long will it take for K4 275 000 to amount to K4 446 000 at 8% simple

interest rate giving your anser in months? a) 5 months b) 6 months c) 5.8 months d) 3.0 months


1.9) A bank accounts pays 12% annual interest compounded monthly. How much

must be deposited now so that the account contains exactly K45 000 000 at the end of the year?

a) K39 935 215.14 b) K50 707 126.36 c) K450 000.00 d) K40 000 000.00


1.10) Rearranging the compound interest formula nrPS )1( += , where P is the original

principal and S is the compound amount at the end of n interest periods at the periodic rate of r. To make n the subject of the formula result in

a) S

rPn

)1( += b)

+=

)1(ln

rP

Sn c)

)1ln(

ln

rP

S

n+

=

d) )1(ln

)ln(

rP

Sn

+=


302

SECTION B QUESTION ONE

a) A borrower receives K4 000 000 today agreeing to repay the lender a total of K4 800 000 at the end of 12 months. What annual simple interest rate is being charged?

b) Find the present value of K15 000 in 9 months’ time at a simple interest rate of

6%.

c) What lump sum would ha e to be invested at 14% , compounded quarterly, to provide an annuity of K1 250 000 a quarter for four years?

(Natech, 1.2 Mathematics & Statistics, December 1999 – (rescheduled))

QUESTION TWO

a) An initial investment of K90 000 000 in a business guarantees the following cash flow:

Cash Flow K28 800 000 K36 000 000 K50 400 000

Assume an interest rate of 5% compounded Semi-annually.

i) Find the net present value of the cash flows. ii) Is the investment profitable?

b) Chipasha Mulenga recently purchased a computer for K5 400 000 and agreed to

pay it off by monthly payments of K270 000. if the store charges interest at the rate of 12% compounded monthly, how many months will it take to pay off the debt?

c) i) Find the amount of an annuity consisting of payments of K180,000 at the

end of every 3 months for 3 years at the rate of 6% compounded quarterly.

iii) Also find the compounded interest. iv)


d) Miss Mwalilino has an obligation of K1 350 000 due five years from now. If interest is assumed to be 7 percent and is compounded yearly, what is the present value of the obligation?


303

QUESTION THREE a) Moonga borrows K500 000 now at an interest rate of 5% per annum. The loan

has the be repaid through five equal installments at the end of each year for the next five years. Calculate the annual repayment.

b) Mr Sokonjo has just received his gratuity amount to K60 million. He wishes to

invest K50 million of the gratuity. He is now faced with a choice between two investment opportunities, A and B. capital outlay for each is K50 million.

A is estimated to yield an annuity of K20 million at the end of each year

receivable every 5 years. B yields K11 million receivable at the end in perpetuity. If the discounting rate is

estimated at 20% for Mr Sokonjo,

i) Evaluate the two (2) investment opportunities using Net Present value (NPV) method.

ii) Recommend with a reason which one of the two (2) investment

opportunities Mr Sokonjo should choose. (Natech, 1.2 Mathematics & Statistics, December 2004)

QUESTION FOUR a) Supremo Stores advertises goods at K700 00 deposit and three further equal

annual payments of K500 000 for the next three years. If the discount rate is 7.5%, calculate the present value of the goods.

b) i) For how many years must I invest K20 000 if I want it to have a value of

at least K500 000 and the interest rate is 6%, payable annually?

ii) How does your answer change if interest is payable every 6 months? (Natech, 1.2 Mathematics & Statistics, June 2003)

c) Mr Musole, a sole proprietor, is paying K500 000 each quarter into a fund which pay 12% per year interest, compounded quarterly. How much will have accumulated in the fund by the end of the fifth year?


d) M-net offers a decorder for K540 000 and K67 500 per month for the next 12

months. If interest is charged at 9% compounded monthly, find the equivalent cash value to be paid now.


304

CHAPTER 7

7.0 Business Mathematics Techniques 7.1 Explanation of Usefulness of Computer Packages and Relations with

Business Decision Making

Computer packages such as computer Stat, MINITAB, Microsoft Excel or other statistical software packages, or a graphing calculator such as the TI-83 have made life easy to solve problems or complete projects for students. Computers are able to perform millions of calculations with speed and accuracy, often summarizing the solutions in interesting, colourful charts and tables. Students who have studied computer languages can write their own programs to solve many of the exercises in this book or they may prefer to use one or more available software packages which provide ready made programs as mentioned above. In either case the student benefits from an interesting and useful experience. It may be noted, however, that the use of the computer is not essential to the reader. Almost all the exercises on this manual can be solved with pencil and paper, but the use of an inexpensive hand-held calculator will reduce the amount of labour involved in many calculations. Some exercises are provided in which the use of a computer is recommended, and these are clearly marked.

7.2 Matrix Algebra This section introduces matrices as structures in which data can be stored and

manipulated. The rules for adding and multiplying matrices are given. Also the manipulation of simultaneous equations using matrix algebra is explained through examples. 7.2.1 Matrices: Definition

Matrices are rectangular arrays of numbers or symbols. The dimension of a matrix is stated as the number of rows by number of columns. The following are examples of matrices with their dimensions:

( )

312322

143

11

02

13

22

13

×××

=

=

−=

Dimension

CBA

305

A business might have the number of employees in each of its departments classified as to their sex for example.

Male Female Accounts 16 5 Ledgers 10 3 Purchasing 5 8

The actual matrix is a framework, which holds the data values; the sections Accounts, ledgers, purchasing, male and female in our example are there simply to identify precisely what the data is. Thus, if we call the whole matrix set A, for convenience, we have:

23:

85

310

516

×

=

Dimension

A

Where it is understood that the first row is Accounts, the second row refers to Ledgers and third row refers to purchasing. Similarly, the first column is the number of male employees, the second female employees. It is important to keep each element in its correct position in the matrix. The null matrix is a matrix of any dimension in which every element is zero, such as:

( )

411322

0000,

0

0

0

,00

00

×××

Dimension

The unit matrix or identity matrix is any square matrix in which every element is zero except the elements on the main diagonal, each of which has the value 1, such as:

=

=

100

010

001

,10

01II

306

The unit or identity matrix is represented by the symbol I. Matrices are equal if they are of the same dimension and the corresponding elements are identical:

.

23

23

12

,221

332

51

32,

51

32equalnotarematricesthebutequalare

−

−

The elements are identical but the dimensions are not. The transpose of a matrix is the matrix obtained by writing the rows of any matrix as columns or vice versa as follows:

=

43

21

42

31

2

1T

row

row

where the superscript T indicates that the matrix is to be transposed.

7.2.2 Definition of Matrix Addition and Subtraction

To add matrices, add the corresponding elements, for example,

=

++++

=

+

23

32

2012

0311

21

01

02

31

To subtract matrices, subtract the corresponding elements, for example,

−=

−−−−

=

−

21

30

2012

0311

21

01

02

31

In general, two matrices can be ADDED (or one matrix, subtracted from another) only if

they have identical sizes. That is, the number of rows in each of the two matrices must be the same and the number of columns in the two matrices must be the same.

307

Example 1 Given the following matrices

=

=

=

631

421,

104

83,

21

03CBA

a) Calculate i) A + B ii) A - B

b) Why is it not possible to calculate B + C or B – C? Hence, state the restrictions on matrix addition and subtraction.

a) i)

=

++++

=

+

125

86

10241

8033

104

83

21

03

ii)

−−−

=

−−−−

=

−

83

80

10241

8033

104

83

21

03

b) When attempting to add the corresponding elements of the matrices B and C, it is found that there is no third column in matrix A, therefore it is not possible to add pairs of corresponding elements in the two matrices. Matrix addition is not possible.

++++++

=

+

6?3211

4?2013

631

421

21

03

No corresponding element to add.

The same problem arises when attempting to subtract matrices B and C: there is no third column in matrix B from which to subtract the elements in column three of matrix C.

308

7.2.3 Multiplying a Matrix by a Number

An ordinary number: 1, 2, -3, 4, etc., is called a Scalar hence matrix multiplication by a number is sometimes called Scalar multiplication. And when scalar multiplication is performed, each element in the matrix is multiplied by the scalar:

Example 2 a) Given the matrix

=

52

13B calculate 3B.

b) Calculate 2I, where I is the 2 x 2 unit matrix.

a)

=

=

156

39

52

1333B

b)

=

=

20

02

10

0122I

7.2.4 Matrix Multiplication

Matrix multiplication AB is possible if the number of elements in the rows of the first matrix (A) is the same as the number of elements in the columns of the second matrix (A). This condition for matrix multiplication can be established quickly by writing down the dimensions of the matrices to be multiplied, in order

A × B = product

Dimensions: ( ) ( ) ( )424332 ×=×××

The ‘inside’ numbers are the same, therefore multiplication is possible. The ‘outside’ numbers give us the dimension of the product.

Dimension of product 2 × 4

The same, so multiplication is possible

309

Example 3 Given the matrices

=

=

=

210

113,

11

01,

53

12CBA

Calculate: (a) AC (b) CA (c) AB (d) BA

Compare your answers for (a) and (b)

a)

=

210

113

53

12AC

Dimension: 2 × 2 × 2 × 3 = 2 × 3

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

=

++++++

=

++++++

=

1389

436

1035309

221206

251315130533

211211120132

b)

=

53

12

210

113CA

2 × 3 × 2 × 2

c)

=

11

01

53

12AB

2 × 2 × 2 × 2 = 2 × 2

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

=

++++

=

++++

=58

13

5053

1012

15031513

11021112AB

Same, multiplication possible

Not same, multiplication not possible

Same multiplication possible

310

d)

=

53

12

11

01BA

2 × 2 2 × 2

( ) ( ) ( ) ( )( ) ( ) ( ) ( )

=

++++

=65

12

51113121

50113021BA

In the above example,

=

=

65

12

58

13BAandAB therefore

.BAAB ≠ (Since two matrices are equal only if all corresponding elements are identical). So in matrix multiplication, the order of multiplication is important. In general, in matrix multiplication .BAAB ≠ .

7.2.5 Find the Inverse Solving AwhenbAx = is a square matrix. The dimension of the matrix x is an

1×r . If this was the equation bandxawherebax ,,= are numbers, and not

matrices, we could solve for x by dividing both sides by a to give a

bx = . We

can’t do the same thing (divide by A) to solve bAX = because we have not defined a way of dividing by a matrix.

Since A is a square matrix, it has an inverse called 1−A such that

IAAAA ==⋅ −− 11 where I is the identity matrix.

Several methods exists for finding 1−A given .A Here two methods will be given through examples.

Same, multiplication possible

311

Example 4

Find the inverse of

=

32

21A

Using the following elementary row operations:

1. The interchange of any two rows of a matrix. 2. The replacement of any row of a matrix by a non-zero constant multiple of

itself.

3. The replacement of any row of a matrix by the sum of itself and a constant multiple of some other row.

Which will lead to an equivalent matrix.

10

01

original identity matrix matrix Step 1 The main idea is to create an identity matrix in the original matrix. The matrix obtained in the position of the Identity Matrix is the inverse of the original matrix. Multiply row 1 by –2 and add the resulting row to row 2 to get the new row 2.

-2 -4 -2 0 2 3 0 1 0 -1 -2 1

−10

21

− 12

01

32

21

312

Step 2

Multiply row by –1 to get the new row 2.

10

21

−12

01

Step 3 We make the entry '' 12a zero by multiplying row by –2 and adding this resulting row 2 to get the new row 1.

1 2 1 0 0 -2 -4 2 1 0 -3 2

10

01

−−

12

23

We have managed to get the identity matrix in the original position. Hence

−−

=−

12

231A

Check:

=

−+−−+−

=

−−

=−

10

01

3466

2243

12

23

32

21

1 IAA

Method 2 An alternative method of finding and inverse was determinants. A determinant of matrix A is denoted by A or Det A and is defined as follows: (for a 2 × 2

matrix)

313

If

=

dc

baA then bcadA −=

and

−−

−=−

ac

bd

acadA

11

Applying this method to our matrix

=

32

21A

( ) ( )

−−

=

−−

−=∴

−=−=−=

−

12

23

12

23

1

1

1432231

1A

A

which is the same result as obtained by using the elementary row operations.

7.2.6 Solving Simultaneous Equations by Matrix Algebra

Matrix algebra can be useful for solving simultaneous equations.

Example 5 Solve, using matrix algebra, the following simultaneous equations.

a) 193

2223

=+=+

yx

yx

b) qP

qP

311

222

+=−=

a) Write the system as:

=

19

22

31

23

y

x

314

Method 1

Using elementary row operations, we have

31

23

19

22

matrix of matrix of constants coefficients

Step 1 Interchange row 1 and row 2 to get

23

31

22

19

Step 2 The new row 2 is obtained by multiplying row 1 by –3 and adding the resulting row to row 2. ( )212 3 rrr +−→ where 1 stands for new row 2.

− 70

31

− 35

19

Step 3

Divide row 2 by –7 (i.e., 22 2

1rr −→ ) to get

10

31

5

19

315

Step 4 To get the new row 1, multiply row 2 by –3 and adding this resulting row to row 1 ( )122 2 rrr +−→ to get

10

01

5

4

Therefore, .5,4 == yx

Check: ( ) ( )

( ) 19154534

2210125243

=+=+=+=+

Method 2

.5

4

35

28

7

1

5722

3866

7

1

19

22

31

23

7

1

31

23

7

1,729

19

22,,

31

23

1

1

=

=

+−−

=

−−

=

=

−−

==−=

=

=

=

−

−

y

x

y

x

y

x

bAXThen

AA

by

xXA

316

b) Rewrite the given equations in the form.

133

222

−=−=+

qp

qp

−− 13

22

31

21

Step 1

212 rrr +−→

−

− 35

22

50

21

Step 2

−→

7

22

10

21

5

122 rr

Step 3

78

7

8

10

012 121

==∴

+−→

qandP

rrr

Method 2

317

.78

,7

8

35

40

5

1

1322

2666

5

1

13

22

11

23

5

1

11

23

5

1

523

13

22,

31

21

1

1

==∴

=

−−

−=

−−+−

−=

−

−−−

−=

=

−−−

−=

−=−−=

−=

=

−=

−

−

qandp

bAXTherefore

A

AThen

bandq

pXA

Example 5 A distributor records the weekly sales of television sets in three retail outlets in different parts of the country (See Table 7.1). Table 7.1 Number of Television Sets Sold in Each Shop

Sharp Sony Phillips Shop X 50 300 150 Shop Y 85 425 213 Shop Z 90 30 28

The cost price of each model is: Sharp K900,000, Sony K1.0 million, Phillips K1.2 million. The retail price of each model in each of the three shops is given in Table 7.2.

Table 7.2 Selling Price of Television Sets in Each Shop

Sharp (000’000)

Sony (000’000)

Phillips (000’000)

Shop X 1.2 1.85 2.10

318

Shop Y 1.5 1.80 1.85 Shop Z 1.1 1.95 2.00

Use matrix multiplication to calculate:

a) The total weekly cost of television sets to each shop.

b) The total weekly revenue for each model for each shop.

c) The total weekly profit for each shop.

a) The numbers sold from Table 7.1 may be written as a matrix, Q:

Television Sets Sharp Sony Phillips

=28

213

150

30

425

300

90

85

50

ZShop

YShop

XShop

Q

Write the cost of each type of television set as a column matrix:

Phillips

Sony

Sharp

C

=2.1

0.1

9.0

Total cost of television sets to each shop:

CostTotalCQ =⋅

Dimension 3 × 3 3 × 1 = 3 × 1

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

=

++++++

=

6.144

1.757

525

2.1280.1309.090

2.12130.14259.085

2.1150103000950

2.1

0.1

9.0

283090

21342585

15030050

Cost to Shop X = K525 million, Shop Y = K757.1 million, Shop Z = K144.6 million.

319

b) The total revenue = price × quantity. The quantities are given by the matrix Q, for the data in Table 7.1. The prices are obtained from the data in Table 7.2. Matrix multiplication is carried out by multiplying rows by columns. Therefore to multiply quantity × price for each television set, we get the transpose matrix of the data in Table 7.2 that’s:

Phillipsofprice

Sonyofprice

Sharpofprice

PT

−−−

=00.285.110.2

95.180.185.1

1.15.12.1

The required total required is obtained by multiplying.

Sharp Sony Phillips

=×28

213

150

30

425

300

90

85

50

ZShop

YShop

XShop

PQ T

00.285.110.2

95.180.185.1

1.15.12.1

phillipsofprice

sonyofprice

sharpofprice

X

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )

++++

++=

00.22895.1301.190

85.121380.14255.185

10.215085.13002.150

XX

XX

X

The off diagonal entries marked X don’t apply. Total revenue for shops X, Y and Z are summarized as follows: TR;

320

=

++++++

=

5.213

55.1286

930

565.5899

05.3947655.127

31555560

c) Profit = TCTR −

=

−

=

9.68

45.529

405

6.144

1.757

525

5.213

55.1286

930

Example 6 A manufacturer produces two products A and B. For each unit of A sold the profit is K50,000, and for each unit of B sold the profit is K68,750. From past experience it has been found that 12.5% more of A can be sold than of B. Next year the manufacturer desires a total profit of K262.5 million. How many units of each product must be sold?

Let x be the number of units of A to be sold and y the number of units of B to be sold. Then yx 125.1= and .0005002627506800050 =+ yx

We solve the two equations

( )( )20005002627506800050

10125.1

=+=−

yx

yx

−000500262

0

7506800050

125.11

321

+→

−→

−+−→

2100

5.2362

10

01125.1

2100

0

10

125.11

000125

1

000500262

0

0001250

125.1100050

121

22

212

rrr

rr

rrr

Thus ,210050.2362 == yandx so approximately 2362.50 units of A and 2100 units of B must be sold.

Exercise 1

1) Taking

=

=527

951

323

322

923

452

BA

Calculate: (a) A + B (b) A – B (c) AB

2) Find

=−

047

110

3411 AwhenA

3) a) State the general rule for two matrices X and Y to be multiplied.

b)

=

=

=

=40

54,

552

311,

04

15

04

,

025

203

112

DCBA

Which of the following matrix multiplication are possible? DCCDCACBCDBBBAADACAB Indicate the dimension of the product matrix where possible.

4) Given

−=

−=

535

403

23

52BandA

322

Find: a) AB and b) BA.

5) Given ( )

−==

052

21325 BandA

Find: a) AB b) BA. 6) Compute the following if possible:

a)

−−+

−125

304

551

232

b)

−−

−131

501

650

412

c)

+

−432

210

42

35

d)

−−−

+

−3324

2163

1114

5231

7) Let ( )cr × denote a matrix with shape ,cr × i.e, r denotes the number of rows

and c the number of columns.

a) What is the dimension of the product of two matrices ( ) ( ).xwcr ×××

b) Find the dimensions of the product of the matrices with the following dimensions.

i) ( ) ( )6445 ×× ii) ( ) ( )5113 ×× iii) ( ) ( )3525 ×× iv) ( ) ( )3736 ×× v) ( ) ( )155 ××

323

vi) ( ) ( )5556 ×× 8) a) Solve the following equations for yandx using matrix algebra.

623016

1152545

=+=+

yx

yx

b) If

=

=

66

51

25

70BandA

Find (i) AB and (ii) BA. 9) A fast food chain has three shops, X, Y and Z. The average daily sales and profit

in each shop is given in the following table.

Units Sold Units Profit (Kwacha) Shop X Shop Y Shop Z Shop X Shop Y Shop Z

Chips 900 500 600 3500 3000 2500 Chicken 750 400 500 1500 2000 3500 Beef 400 1100 800 2000 2500 1500

Use matrix multiplication to determine. a) The profit for each product. b) The profit for each shop. 10) The percentage of voters who will vote for party candidates P, Q and R is given in

the following table:

P Q R Number of Votes Area X 35% 15% 50% 110 000 Area Y 65% 25% 10% 80 000 Area Z 45% 36% 19% 75 000

Use matrix multiplication to calculate the total number of votes for each candidates.

324



Answer all sub-questions 1.1 The inverse of the following matrix

25

28 is given by

A.

3

1

6

53

1

3

4

B.

−

−

3

4

6

53

1

3

1

C.

−−85

22

D.

6

2

6

56

2

6

8

(Natech, 1.2 Mathematics and Statistics, Nov/Dec 2000)

1.2 If

−−−−−

=

=421

421

421

963

642

321

YandX

What is XY?

A.

000

000

000

B.

100

010

001

C.

−−−−−−−−−

963

642

321

D.

−

1282

221

100

325

(Natech, 1.2 Mathematics and Statistics, June 2003).

1.3 Find the product AB of the following set of matrices:

=53

41

22

A

=

20

21B

A.

68

105

84

B.

163

101

82

C.

1084

9106

D.

169

810

46


1.4 Find the inverse matrix, 1−A , of matrix A given as follows:

−42

31

A.

−−−

43

21 B.

− 10151

10352 C.

−24

13

D.

10

01

1.5 If thenABCandBA ,34212

56

15

23−=

−=

−−

=

A. 1512 =C B. 7422211211 =+++ CCCC C.. 0312 =+C D. 331211 =+CC

326

1.6 The Inverse of the matrix A if A =

02

18 is

A.

−−82

10 B.

−

−

412

10 C.

012

14

D.

−

−

412

10

1.7 If A = isBAB +

=

,

210

321,

54

31

A.

64

52 B.

44

10 C.

74

62

D. None of these.

1.8 Find the product of BA of the following matrices

=

=

21

11,

02

51BA

A.

22

116 B.

21

11 C.

23

62

D. None of these.

1.9 If A = Afind,

312

234

112

−−

A. −8 B −8 C. −40 D. 8

327

1.10 Evaluate the determinant 52

13.

A. 17 B. 13 C.

−−32

15 D. None of these.

SECTION B

ATTEMPT ANY FIVE QUESTIONS IN THIS SECTION.

QUESTION ONE a) Using the matrices P and Q below

−−

=

−=

43

13;

12

24QP

Solve the following equations: i) QPX =+ ii) PQY 4=+ b) A square matrix A is the inverse of the square matrix B if and only if

BAIAB == where I is the identity (or unit matrix). In each of the following show that matrix A is the inverse of matrix B.

i)

−−

=

=

72

31;

12

37BA

ii)

−=

−=

4.03

12;

4.06.0

2.02.0BA

iii)

−−

=

=

352

75;

52

73BA

(Natech, 1.2 Mathematics and Statistics, June 1997).

328

QUESTION TWO a) Zam Protect Insurance Company has four salesmen working in the Midlands area.

The number of policies they sold during the last month is given in matrix A, as follows:

Mumba Daka Besa Moonga

If ( )1111=S , find the product SA and interpret its elements.

b) Given that

=

=

13

24

33

21BandA

Find C, the product of the two matrices, A and B.


QUESTION THREE

a)

−=

−=

=22

03

61

,012

310,

5

3

1

CBA

Determine each of the following, if possible i) TTTTT ABvBAivBAiiiCBiiCA )))()( ++

Household

RiotFire

Life

Vehicle

A/

3

0

5

8

1

2

11

6

2

3

9

7

0

4

6

8

=

329

b) A fast food chain has 3 shops, X, Y and Z. The average daily sales and profit (in

thousands of kwacha) in each shop is given in the following table.

Sales Shop X Shop Y Shop Z Unit Profit ‘000’

Chicken Chips Drinks

1000 1150 700

600 800

1400

700 900

1100

2 3 6

5 6 5

7 9 3

Use matrix multiplication to determine

i) the profit for each product ii) the profit for each shop

QUESTION FOUR a) Given for the following matrices

.210

431,

8

6

2

,30

12,

33

51

=

=

−=

−= DCBA

i) Show, that BAAB ≠ b) Determine the following if possible DCivDCCiiiACiiADi T ))))

330

c) The percentage of voters who will vote for party candidate A, B, and C is given in the following table.

% votes for party candidates A B C Number of Votes Area X Area Y Area Z

35% 65% 28%

15% 25% 45%

50% 10% 27%

105 000 80 000 75 000

i) Use matrix multiplication to calculate the total number of votes for each candidate.

ii) On polling day if the turnout is 65% in Area X, 35% in Area Y and 25% in Area Z, use matrix multiplication to calculate the total votes for each candidate.

QUESTION FIVE a) Which of the following matrices do not have an inverse? Give reasons.

=

=

=

231

012

101

),510

123),

23

51) CiiiBiiAi

b) Use the INVERSE MATRIX method to solve the following equations.

364410

2342

1125

321

321

21

=++=++=+

QQQ

QQQ

QQ

331

QUESTION SIX

a) If

−=

=

16

15

21

32BandA , find

.)))) ABivBAiiiBiiAi +

b) Use any matrix method to solve for ,21 PandP given the following equations.

.68212

6195

21

21

=+=+

PP

PP

QUESTION SEVEN

a)

=

=

65

49,

31

33BA

i) Find A + B

ii) A −B and B − A. Which general rule does this prove? ii) Find AB and BA. Which general rule does this apply?

b) Solve the following equations for yandx using matrix algebra:

2752515

5255025

=+=+

yx

yx

332

QUESTION EIGHT a) The ABC’s polling organization asked a sample f voters for their preference for

mayor of the city of Kitwe in the coming election. The organization found the following table:

MMD Candidate

UPND Candidate

Other Candidates

Undecided

MMD UPND Independents

550 200 50

100 450 100

60 80 15

8 3

100

i) Write the corresponding matrix.

ii) How many MMDs were interviewed?

iii) How many voters preferred the UPND candidate?

iv) How many independents were undecided?

b) ( ) .

4

1

1

3

,2102,06

24

−=−=

−= CBA Find

.),2

1)6),) BCivAiiiAiiABi

333

7.3 Basic Linear Programming

7.3.1 Introduction In this section we find the optimal value of linear functions (cost, profit, revenue,

output, etc) subject to several constraints. Industrial production, flow of resources, investments, nutrition, etc all involve complex interrelationships, among numerous activities. A common feature of many of these problems is to formulate a course of action that will minimize or maximize some essential quantity.

7.3.2 Feasible Regions

Several business production constraints can be graphically demonstrated using their separate respective inequalities to produce a feasible region of operation.

In a linear programming problem, the function to be maximized or minimized is called the objective function. Although there are usually infinitely many solutions to the system of constraints (these are called feasible solutions or feasible points), the aim is to find one such solution that is an optimum solution (that is, one that gives the maximum or minimum value of the objective function).

Example 1

A factory can produce two products, X and Y. The contribution that can be

obtained from the products are,

A contribution of K10 000 per unit B contributes K15 000 per unit and it is required to maximize its contribution. The objective function for this factory can be expressed as as

producedYofunitsofnumbery

producedXofunitsofnumberxwhere

yxMaximize

==

+ 0001500010

Note that X and Y produced are sometimes called decision variables. A factory requires three types of processing. The length of time for

processing each unit is given in the table below.

334

Product X (hr/unit)

Product Y (hr/unit)

Maximum process capacity per day

(hrs) Process I

Process II

Process III

12 3 8

12 6 4

840

300

480

10 000 15 000

The limitations or constraints are circumstances which always exist and these govern the achievement of the objective. The constrains in any given problem must be clearly identified, quantified, and expressed mathematically. The constraints as regard to processes can be stated as follows:

48048Pr

30063Pr

8401212Pr

≤+

≤+

≤+

yxIIIocess

yxIIocess

yxIocess

In addition, a general limitation applicable to all maximizing problems is that it is not possible to make negative quantities of a product, i.e

.0,0 ≥≥ yx These non-negativity constraints exclude any possibility of negative production levels which have no physical counterpart. Together they include the x -axis and the y -axis as possible boundaries of the feasible area.

335

y 120 Process III constraint 48048 ≤+ yx

70 A(0, 50) B Process II constraint 30063 =+ yx

2

)80,

3

140( (40, 30)

Feasible Process I constraint Region 8401212 =+ yx

C D 60 70 100 x The optimum solution can be found by looking at the corner points and substituting these values in the objective function. Corner Point Objective value function yx 0001500010 + A(0, 50) 10 000(0) + 15 000(50) = 750 000

B

3

80,

3

140 10 000

3

140 + 15 000 67.666866

3

0002600

3

80 ==

C(60, 0) 10 000 (60) + 15 000(0) = 600 000 D(0, 0) 10 000(0) + 15 000(0) = 0

336

The maximum contribution occurs at point B. Therefore 3

140 units of X and

3

80 units of

Y should be produced to realize a profit of K866 666.67. We classify the constraints of a linear program as either binding or non binding. A binding constraint must pass through the optimum solution point. If it does not, it is non binding. In our example, only process II and III constraints are binding. When a constraint is binding, we may regard it as a scare resource since it has been used completely.

4803

1440

3

320

3

1120

3

804

3

1408Pr

3001601403

806

3

1403Pr

==+=

+

=+=

+

IIocess

IIocess

On the other hand, a nonbonding constraint represents an abundant resource. A redundant constraint is one which can be deleted from the solution space without affecting the solution space. In our example, process I constraint is a redundant constraint. Example 2 The linear programming problem is now Maximize yxZ 0001500010 +=

s.t

.0,0

48048

30063

780132

≥≥≤+≤+≤+

yx

yx

yx

yx

337

y 120 Process III (0, 60) Process I

B

11

420,

11

260

A(0, 50)

C

7

60,

7

390

feasible Process II region 0 D(60, 0) (65, 0) (100,0) x

Point B is the point of intersection between Process I and process II is

30063,780131211

420,

11

260 =+=+

yxyx . Point C is the point of intersection between

process Iii and process I .48048,78013127

60,

7

390 =+=+

yxyx

338

Corner Objective value function yx 0001500010 +

00.000600)0,60(

29.7146857

60,

7

390

91.0908097

60,

11

260

00.000750)50,0(

KD

KC

KB

KA

The optimal solution is to produce 11

260 units of X and

11

420 units of Y and the maximum

profit is K809 090.91. Constraint I and II are binding while constraint III is non binding as shown in the table below. Process I Process II Process III

11

260 units of X 12

11

2608

11

2603

11

260

11

420 units of Y

11

4204

11

4206

11

42013

Hours used 780 300 Maximum hours available 780 300 480

Unused hours 0 0 138.18 or 11

2138

Process I and Process II have used all their hours completely while Process III has 11

2138

hours un used.

339

Example 3 Maximize the objective function yxZ 25 += subject to the constraints.

.0

,0

,1232

,82

≥≥≤+

≤+

y

x

yx

yx

y

8 82 =+ yx D feasible

region 1232 =+ yx A B x

Figure 7.1 A, B, C and D are corner points of Feasible Region.

In Figure 7.1 the feasible region is non-empty and bounded. We have shaded the unwanted region. Thus Z is a maximum at one of the four corner points. It can be shown that: A linear function defined on a non-empty bounded feasible region has a maximum (minimum) value, and this value can be found at a corner point. The coordinates of A, B D are obvious on inspection. To find C we solve the equations

123282 =+=+ yxandyx simultaneously, which gives .2,3 == yx Thus,

( ) ( ) ( ) ( ).4,0,2,3,0,4,0,0 ==== DCBA

4

C

340

Evaluating Z at these points, we obtain

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) 84205

192235

200245

00205

=+=

=+=

=+=

=+=

DZ

CZ

BZ

AZ

Hence the maximum value of Z, subject to the constraints, is 12 and it occurs when

04 == yandx .

Example 5 Minimize yxZ 3020 += Subject to

.0,

,5678

,2443

,102

≥≥+≤+

≤+

yx

yx

yx

yx

y 10 8 102 =+ yx 5678 =+ yx 6

• A 2443 =+ yx 5 6 8 Figure 7.2: The feasible region is empty.

x C

341

As can be seen above, the feasible region is empty, so there is no optimum solution.

Example 6 Minimize yxC += 5

Subject to

.0,

,22

,634

,33

≥≥+≥+

≥+

yx

yx

yx

yx

y 3 • 2 feasible region • B • C D

33 =+ yx • 0 1 1.5 2 x 634 =+ yx 22 =+ yx Figure 7.3 A, B, C and D are corner points of feasible region.

342

The coordinates of A, and D are obvious on inspection. To find B, we solve the equations 63433 =+=+ yxandyx . Similarly to find C, we solve the equations

22034 =+=+ yxandyx . Thus,

( ) ( )0,2,5

2,

5

6,

5

6,

5

3,3,0 =

=

== DCBA

Evaluating C at these points, we obtain

( ) ( )

( )

( )

( ) ( ) 10025

4.65

2

5

65

2.45

6

5

35

3305

=+=

=+

=

=+

=

=+=

DC

CC

BC

AC

Thus C has a minimum value of 3.0 when .30 == yandx Example 7

A production line can be set up to produce either product A or product B. The following table gives the breakdown for each product.

Product Labour

(Hours) Material

(Kg) Testing (Hours)

A 2 4

10

1

60

6 =

B

2

1

8

15

2

60

8 =

In any one week only 60 hours of labour and 560 kgs of material is available. Due to cost and availability of the test equipment, it must be used for at least 8 hours. Also because of existing orders, at least 10 of product A must be produced. The profit from each unit of A produced is K50 000 and from each unit of B is K70 000. Find the weekly production that will maximize profit and what is this maximum profit.

343

Let yandx be the number of products of A and B produced in a week. Then maximize yxP 0007000050 +=

salesx

Testingyx

materialsyx

labouryxtS

10

815

2

10

1

56084

602

12

≥

≥+

≤+

≤+⋅

Since yandx are numbers of products of A and B, then

.0

0

≥≥

y

x

y 120 1204 =+ yx 10=x 70 • A •B 1402 =+ yx feasible 60 •D region • C 0 10 30 80 140 x

815

2

10

1 ≥+ yx

344

The coordinates for A are given by the intersection of .102802 ==+ xandyx the coordinates for D are given by the intersection of

.10815

2

10

1 ==+ xandyx B is given by the intersection of

−=+=+ 2821204 yxandyx while coordinates for C are found by the

intersection of .1204815

2

10

1 =+=+ yxandyx

Thus,

( ) ( )5.52,10,13

600,

13

240,

7

440,

7

100,65,10 DCBA

( ) ( )

( ) 00.000,175,45

105000701000050

15.846,153,413

60000070

13

24000050

71.285,114,57

44000070

7

10000050

00.000,050,5000701000050

KPD

KPC

KPB

KPA

=

+=

=

+

=

=

+

=

=+=

The maximum number of units is 7

440

7

100 == yandx giving a maximum

profit of K5,114,285.71.

345

Exercise 2

1. Maximize yzxz 2420 +=

Subject to

.0,

,02

,60

≥≤−

≤+

yx

yx

yx

2. Maximize yxP 57 +=

Subject to

.0,

,21032

,22023

≥≤+≤+

yx

yx

yx

3. Minimize yxC 23 +=

Subject to

.0,

,10025

,15023

,802

≥≥+≥+

≥+

yx

yx

yx

yx

4. Minimize yxC 715 +=

Subject to

.0,

,44

,634

,33

≥≥+≥+

≥+

yx

xy

yx

yx

346

5. A company extracts minerals from ore. The numbers of kilograms of minerals A and B that can be extracted from each ton of ores I and II are given in the table below together with the costs per ton of the ores. If the company must produce at least 3500kg of A and 3000kg of B, how many kilograms of each ore should be processed in order to minimize cost? What is the minimum cost?

Ore I Ore II Mineral A Mineral B Cost per kg

100kg 200kg K200 000

200kg 50kg K240 000

6. A firm manufactures two products A and B. Each product requires machine time and finishing time as given in the table below. The number of hours of machine time and finishing time available per day is 24 hours respectively. The unit profit on A and B is K13 500 and K18 000 respectively. What is the maximum profit per day that can be obtained?

Machine Time

Finishing Time

A 120 minutes 240 minutes B 240 minutes 120 minutes

7. A farmer raises chickens and fowls on her farm. She wants to rise no more than

80 birds altogether. The number of chickens should not be less than 50. She spends K10, 000 to raise a chicken and K12, 000 to raise a fowl. She has K1, 000,000 available for this purpose. Chickens sell for K15, 000 and fowls for K18, 000 each respectively.

a) How many of each kind should be raised for maximum profit.

b) What is the maximum profit?

8. A company operates two types of airplanes, the ZA106 and the ZA108. The

ZA106 is capable of carrying 18 passengers and 14 tons of cargo, whereas the ZA108 is capable of carrying 25 passengers and 7 tons of cargo. The company is contracted to carryat least 207 passengers and 84 tons of cargo each day. If the cost per journey is K300 000 for a ZA106 and K400 000 for a ZA108, what choice of airplane will minimize cost?

347

9. The XYZ company is planning production of two products: forks and spoons, for a certain period of time. Each product requires three types of processing. The length of time required for processing each unit is given in the following table:

Forks

(hr/unit) Spoons

(hr/unit) Maximum

Process Capacity

(hr) Process I Process II Process III

4 1 1

2 1 0

14 5 4

Profit per unit K10 000 K13 000 Given these circumstance, you are required to: a) State the constraints of the model. b) Determine the coordinates of the corner points of the feasible region.

c) Find the optimal strategy by evaluating the objective function at each corner point.

10. A firm is comparing the cost of advertising in two medias, Newspaper and radio.

For every kwacha’s worth of advertising, the table below gives the number of people, by income group, reached by these media. The firm wants to reach out at least 32000 persons earning under K100, 000 000 and at least 9000 earning over K100,000,00. Find the amounts that the firm should spend on newspaper and radio advertising so as to reach these numbers of people at a minimum total advertising cost. What is the minimum total advertising cost?

__________________________________________________ Under Over K100 000 000 K100 000 000 Newspaper 200 50 Radio 160 180 __________________________________________________

348

EXAMINATION QUESTIONS SECTION B QUESTION ONE A company produces two qualities of maheu; these are, type A 940% sweet and in standard bottles) which is intended for the home market, and type B (60% sweet and in large bottles) which is for export. After maturing in vats, both types require two stages of processing; blending and bottling. The process times for a standard batch of each type of maheu are:

Process Time (Hours) Blending Bottling Type A 1½ 1 Type B 2 3

There are 2,400 hours available for each process but because of the steady but limited demand for type A, the number of batches of that type must not exceed 1,200; apart from this, all stocks produced can be sold. The contribution to profit and fixed overheads per batch is K2,000 for type A and K3,000 for type B. You are required to: i) State the objective function; and ii) State the constraining factors for this situation with a view to deciding which

processing mix will maximize the total contribution to profit. (NB: No graphing required).


QUESTION TWO In a factory, two products Alpha and Omega are produced using four different materials. The table below indicates the material requirement in kilograms per unit of each product.

Material Requirement

Alpha Omega

Aeto 6 kgs 4 kgs Beto 5 kgs 3 kgs Ce to 5 kgs 10 kgs Deto 16 kgs 8 kgs

349

During a normal working month, there are 2400 kgs of Aeto, 1500 kgs of Beto and 2000 kgs of Ceto. Currently there is no restriction on the availability of Deto. The products earn contribution per unit of K22, 500 and K30, 000 for Alpha and Omega respectively. Company’s policy is to earn maximum profit by maximizing contribution. i) Determine the linear programme model using the above details. ii) Graphically, estimate the optimum production mix and hence calculate the

maximum contribution earned. iii) State the binding constraints and the redundant constraint(s).

(Natech, 1.2 Mathematics and Statistics, December 2004) QUESTION THREE For the following linear programming problem: i) Graph the constraints; ii) Find the coordinates of each corner point of the region of feasible solutions; and iii) Determine the optimal strategy by evaluating the objective function at each corner

point. Minimize yxZ 43 +=

Subject to:

902

15032

605.1

≥+≥+≥+

yx

yx

yx

and .0, ≥yx


350

QUESTION FOUR Simba Lawn-Care Services contracts with homeowners and business firms for a package of service including planting, fertilizing, weed control, and maintenance of grass lawns. Mr. Zimba, owner of Simba Lawn-Care Services mixes his own lawn treatment formulae to meet the special needs of the individual account. Currently, a treatment solution is needed which contains at least 14 measures of chemical A, at least 5 measures of chemical B, and at least 12 measures of chemical C. Two preparations containing these chemicals are sold commercially. Each canister of solu-X contains 4 measures of chemical A, 1 measure of chemical B, and 2 measures of chemical C. Each canister of Phos-Pho-Gen contains 2 measures of chemical A, 1 measure of chemical B, and 3 measures of chemical C. Each canister of solu-X costs K4,000, while each canister of Phos-Pho-Gen costs K3,000. The total revenue Simba Lawn-Care Services receives from an account is fixed by contract; thus the company would like to perform the required services at a minimum cost. Mr. Zimba, therefore, needs to know how to combine the two products, Solu-X and Phos-Pho-Gen, to obtain a lawn treatment containing the required quantities of each chemical. The chemical requirement per container of each of the two products, solu-X and Phos-Pho-Gen, are given in the following table:

CHEMICAL REQUIREMENTS AND AVAILABILITY

Chemicals Required

Quantities of Chemical Per

Container (Measures)

Minimum Amount Required in Treatment

(Measures) Solu-X Phos-Pho-Gen Chemical A 4 2 14 Chemical B 1 1 5 Chemical C 2 3 12 Cost of container K4 000 K3 000

Given these circumstances, you are required to: i) State constraints of the model. ii) Determine the co-ordinates of the corner points of the feasible region. iii) Find the optimal strategy by evaluating the objective function at each corner

point.


351

QUESTION FIVE A small Copperbelt-based biscuit-making factory makes two brands of biscuits. To make a batch of standard biscuits takes 20 kg of flour and 2 kg of butter, whereas a batch of special biscuits requires 10 kg of flour and 5 kg of butter. The firm makes a profit of K20,000 on a batch of standard biscuits and a profit of K60,000 on a batch of special biscuits. The factory has at most 200 kg of flour and 40 kg of butter available each day. You are required to: i) Formulate this as a linear programming problem. ii) Determine by graphical means how many batches standard biscuits and special

biscuits the factory should produce each day in order to maximize profit.

(Natech, 1.2 Mathematics and Statistics, June 2003) QUESTION SIX Suppose that a company is planning production for a period of one week. It is making two products, X and Y, each of which requires certain foundry, machinery, and finishing capacity as shown in the following table:

Product Foundry Machining Finishing X 6 hr/unit 3 hr/unit 4 hr/unit Y 6 hr/unit 6 hr/unit 2 hr/unit

The following number of hours are available in each area during the week being planned.

Foundry 420 hours Machining 300 hours Finishing 240 hours

Given these circumstances, you are required to: i) State and graph the constraints of the model. ii) Determine the coordinates of the corner points of the feasible region. iii) State the objective function and determine how many units of each product should

be produced in order to achieve maximum profit, given that each unit of product X produces a profit of K3,000 and each unit of product Y produces a profit of K2,000.


352

7.4 INTRODUCTION TO DIFFERENTIAL CALCULUS

1. Introduction

This Chapter covers elementary calculus. The field of calculus is a large and complex field of mathematics with applications in many different fields. This manual concentrates only on simple differentiation and integration which is likely to be relevant in accounting and business. The rules for differentiation, its integration and its practical use in the finding of the maximum and minimum values of functions or turning points on curves are covered. Finally, the rules of integration are given.

Contrary to (often popular) belief, that calculus is a difficult area of study, NO, it is not especially when one follows:

i) a few simple rules which will be given; ii) an understanding of some relevant applications.

Differentiation could be used to find the maximum or minimum points of the curves of certain business functions. For example, cost, revenue and profit functions. The aim of any business is minimize cost and maximize profit. If one is able to determine the production function of his or her business, the level at which the maximum profit will occur can be determined. Differentiation can also be used to find the “rate of change” of cost or revenue functions. This rate of change is called marginal. Hence the rate of change of the cost function is the marginal cost and that of the revenue function is the marginal revenue etc.

Integration is the process of reversing differentiation. It can be used to find revenue or cost function from the marginal revenue or marginal cost function respectively.

1) Simple Functions

The most commonly used functions in business are the linear and quadratic functions. For example a simple linear function for total cost might have the form:

bxay += where y = total cost, the dependent variable. x = output or activity, the independent variable and a and b are constants, representing fixed cost and variable (or marginal) cost respectively.

353

Such a function is shown in Figure 1.0.

y

b = the slope of the line which represents the marginal cost

Costs • a = fixed cost

activities x Graph of a simple linear cost function. Figure 1

2) Quadratic Functions

The other common function found in business and accounting is a quadratic function. This is of the form cbxaxy ++= 2 where y = total cost or revenue etc, the dependent variable, x = output activity, the independent variable, a , b and c are constants. ‘a , b ’will form part of the variable cost and c is the fixed cost.

Rule and Notation For Differentiation

Differentiation can be thought of as a process, which transforms one function into

a different one. The new function is known as the derivative of the original one.

354

Derivative of a Simple Function The simple function nKxy = (where K and n are any numbers) can be differentiated to give the new function:

1−= nKnxdx

dy

where dx

dy is read as “the Derivative of y with respect to x”.

dx

dy is also denoted

by y′ . In particular,

a) ,Kxy = then Kdx

dy =

b) If ,Ky = then 0=dx

dy

Thus, for example,

If 48xy = , then ;32)4(8 314 xxdx

dy == −

If ,5xy = then ;5)1(5 11 == −xdx

dy

If ,20=y then 0=dx

dy

Example 1 Using the above rule find the following derivatives.

a) If 325xy = , then 313 75)3(25 xxdx

dy == −

b) If 7165xy −= , then 617 1155)7(165 xxdx

dy −=−= −

c) If ,9 100xy = then 991100 900)100(9 xxdx

dy == −

355

Derivatives where the function is a sum and a difference The Derivative of the sum of two (or more) simple functions is the sum of the separate derivatives of the functions. Example 2

a) If ,20123 34 ++−= xxxy then 13612 3 +−= xxdx

dy.

b) If ,5375317 2132

3+−+=+−+= −− xxxx

xxy then xxx

dx

dy621 24 −−= −− .

c) If ,5 56 xxy += then .256 45 xxdx

dy += .

Derivatives where the function is a product.

If ),2)(5( 2 ++= xxxy find dx

dy.

Consider y as a product of two functions

)2()5( 2 ++− xxxy first second

then )((sec)(sec)( firstdx

dondond

dx

dfirst

dx

dy ++=

)5()2()2()5( 22 xxdx

dxx

dx

dxx +++++=

first second Derivative Derivative of first Of second

1014

)52)(2()1)(5(2

2

++=++++=

xx

xxxx

The derivative of the product of two functions is the first function times the derivative of the second, plus the second function times the derivative of the first.

356

Example 3

a) If );76)(35( 42 ++= xxy then )10)(76()24)(35( 432 xxxxdx

dy +++=

xxx 7072180 35 ++=

b) If );7)(53( 21

3 ++= xxy then )9)(5()2

1)(23( 23 2

121

xxxxdx

dy +++=

25492

325

21

27

xxxx +++=

Derivates where the function is a quotient

If ,2

5

++=

x

xy find

dx

dy

Consider y as a quotient provided the denominator is not zero. Then:

2)min(

)minmin)(

atordeno

atordenodx

dnumerator

dx

datordeno

quotientdx

d −=

.)2(

3

)2(

52

)2(

)1)(5()1)(2(

)2(

)2()5()5()2(

22

2

2

+−=

+−−+=

++−+=

+

++−++=

xx

xx

x

xx

x

xdx

dxx

dx

d

xdx

dy

That is, the derivative of the quotient of two functions is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

357

Example 4

a) If 2)2(

)2()3()3()2(;

2

3

−

−+−+−=

−+=

x

xdx

dxx

dx

dx

dx

dy

x

xy

.)2(

52−

−=xdx

dy

b) If ;53

32 −

=x

xy

22

22

)58(

)58(3)3()58(

−

−−−=

x

xdx

dxx

dx

dx

dx

dy

.)58(

241522

2

−−−=

x

x

dx

dy

Derivatives Functions of A Function Consider the function 4)52( −= xy where the expression in the brackets is a differentiable function say u, i.e 52 −= xu , the whole expression can be written as

.4Uy = In such cases the rule for differentiation is

dx

du

du

dy

dx

dy ×= which is known as the chain rule. Thus to differentiate .)52( 4−= xy We

let 52 −= xu then .4uy = And 34udu

dy = and .2=dx

du

Therefore 33 82.4 uudx

dy == . But the original function is in x. Hence, 3)52(8 −= xdx

dy.

358

Example 5 Using the Chain Rule

a) If dx

dyfindxy 3)52( −= .

dx

du

du

dy

dx

dy.=

Let 3,52 uythenxu =−=

235 udu

dyand

dx

du =−=

Therefore .15)3(5 2uudx

dy −=−= We can write our answer in terms of x alone by

replacing u by .52 x−

.)52(15 2xdx

dy −−=

b) If ,21 xy −= find .dx

dy Here, we let .,21 2

1

uythenxu =−= Then by chain rule

dx

du

du

dy

dx

dy.= = )2.(

2

121

−u

21

21

)21( xudx

dy −−=−=−

Second Derivative of A function Differentiation can be repeated as many times as necessary on any given function. The second derivative is important for the syllabuses covered by this manual as it is used to determine whether a point is a maximum or minimum.

The second derivation of any function y is written as yordx

yd ′′2

2

and is obtained by

differentiating a given function y twice.

359

Example 6

a) If ,5232

+−= xxy find 2

2

dx

yd

Then 6)26(2

2

=−= xdx

d

dx

yd

b) If ,2635 345 +−++= xxxxy find 2

2

dx

yd

Then 631225 2342

2

−++= xxxdx

yd

xxxdx

yd636100 23

2

2

++=

The Practical use of Differentiation Now that the idea of differentiation has been explained and the rules given for differentiating common functions, it is time to look at practical examples. Example 7 A firm has analyzed their operating conditions, prices and costs. The Cost Accountant have developed the following functions. Revenue 25350)( QQRK −= and cost 6020)( 2 ++= QQCK where Q is the number of units sold. The firm wishes to maximize profit and wishes to know a) What quantity should be sold? b) At what price c) What will be the amount of profit?

360

a) Profit = Total revenue – Total Cost

.603306

)6020(53502

22

−+−=

++−−=

QQ

QQQQ

dQ

profitdequaiting

QdQ

profitd

)(

33012)( +=

to 0 implies 033012 =+− Q 5.27=Q

012)(

2

2

<−=dQ

profitd implies maximum profit when 5.27=Q . Therefore

the firm should sell 27.5 units to maximize profit.

Alternatively From basic economic theory it will be recalled that profit is maximized when Marginal Revenue = Marginal Cost.

202)(

6020

10350)(

5350

2

2

+==

++=

−==

−=

QdQ

CdMC

QQC

QdQ

RDMR

PQR

point of profit maximization is when

5.27

12330

20210350.

)()(

==

+=−

==

Q

Q

QQei

dQ

Cd

dQ

RdorMCMR

361

b) Total Revenue

75.5843

25.37819625

)5.27(5)5.27(350 2

=−=

−=

27.5 units will be sold for 5843.75 kwachas.

The price will be 50.2125.27

75.5843K=

c) Total Profit = revenue - Cost revenue from above = K5 843.75 from above and Cost 60)5.27(20)5.27( 2 ++=

25.1366

6055025.756

K=++=

Therefore profit = K5 843.75 – K1 366.25 = K4 477.5. Procedure for identifying the turning points of a curve. Using differentiation techniques, turning points of curves can be identified using three steps. We shall show this by means of an example. Example 8

Suppose the function 1524

1 2 +−= xxy is a production function where y is the cost (in

million of kwachas) of manufacturing x (in hundreds) items for some process. The procedure is as follows:

Step 1 : Obtain dx

dy for the given function 2

2

1 −= xdx

dy

362

Step 2: Solve the equation 0=dx

dy, which will give the x-coordinates of any

turning points that exists.

Since 22

1 −= xdx

dy the equation 02

2

1 =−x needs to solved. .4=x

Step 3: Evaluate 2

2

dx

yd at which (and any) x value found in Step 2.

02

2

>dx

yd signifies a Minimum point;

02

2

<dx

yd signifies a Maximum point

In our example, .2

1,2

2

12

2

=−=dx

ydthereforex

dx

dy

Thus the turning point at 4=x is a minimum since .02

1 > Hence, 400 items must be

manufactured in order to minimize total costs. The minimum total cost at this value of x

is 11158415)4(2)4(4

1 2 =+−=+−=y (million kwacha).

Example 9 Determine the coordinates and nature of any turning points on the curve represented by

the function ,3

218.

3

218

33 xxySolve

xxy −=−=

2218 xdx

dy −= and 02

2

=dx

yd gives

3

9

02182

2

±==

=−

x

x

x

The turning points exist on the curve .333

218 3 −=−= andxatxxy

363

Note that when xdx

yd4

2

2

−= and 012)3(4,32

2

<−=−==dx

ydx Signifying a maximum.

When ,012)3(4,32

2

>=−−=−−dx

yd signifying a minimum.

Substituting x = 3 into the original equation will give the y-coordinate of the maximum

as: .361854)3(3

2)3(18 3 =−=−=y Hence, the maximum point for the curve is at

).36,3(363 oryandx ==

Similarly, substituting 3−=x into the original equation will give the y-coordinate of the

minimum as: 361854)3(3

2)3(18 3 −=+−=−−−=y . Hence, the minimum point for the

curve is at .)36,3(363 −−−=−= oryandx

Example 10 Find the rate of change of 6xy = with respect to x and evaluate it when 2=x and when

1−=x . Interpret your results.

56xdx

dy =

When 192)2(6,2 5 ===dx

dythenx . This means that if x increases by a small amount,

then y increases approximately 192 times as much. More simply we say that y is

increasing 192 time as fast as x decreases. When .6)1(6,1 5 −=−==dx

dythenx The

importance of the minus sign is thaty is decreasing 6 times as fast as x increases. Integration As mentioned earlier, integration can be regarded as the reverse process to differentiation. For example, since differentiating 34 205 xgivesx , integrating 320x should give .5 4x The problem occurs when differentiating a constant. This is because it always gives zero. Example, 95;35;25 222 +−+ xxx are the same function when differentiated with respect to x . Hence the derivative is 10x . Therefore, the best that can be done is to

364

integrate 10x to give ,5 2 Cx + where C is some arbitrary constant. If conditions are given about the function under integration, we can find the particular value of C.

The rule for integrating a simple function Cn

xdxxisx

nnn +=

+

∫1

where ∫ represents

‘the integral of’ and x is the variable of integration. Note that C is the constant of integration. Example 11 Find the integrals of functions

a) CxxCxxCxx

dxxx ++=++=++

++

=+++

∫2424

11133

2

52

2

5

4

8

11

5

13

8)58(

b) ∫ ++−=+− Cxxxdxxx 53

7)527( 222

c) Cxxdxx ++=+∫ 25)210( 2

d) Cxxdxx ++=+∫ 4

3

3

5)

4

35.2( 2

321

e) If 55 =+= yandxdx

dy when ,0=x find y in terms of x .

Now integrating dx

dy will give y . That is, ∫ ++=+= .5)5( 2 Cxxdxxy But

5)0(50.05 2 =++== Cxwheny giving C = 5. Thus .552 ++= xxy

365

Example 12 The total revenue obtained (in million of kwacha) from selling x hundred items in a particular month is given by R which is a function of variable x .

Given that xdx

dR250−=

a) Determine the total revenue function R; b) Find the number of items sold on one day that will maximize the total revenue

and evaluate this total revenue.

a ) Revenue = ∫ ∫ +−=−= Cxxdxxdxdx

dR 250)250(

When no items are sold revenue will be zero. That is 0)0()0(50 2 =+−= CR

So that, 250 xxR −= . b) The value of x that maximizes revenue R is found by solving the equation

.0=dx

dR That is 0250 =− x

02

25

2

2

<−=

=

dx

Rd

x

confirms that when 25=x , the total revenue is maximized. Hence 2 500 items

should be sold per month to maximize revenue. This maximum revenue is .000000625)25()25(50 2 KR =−=

366

Exercise 3 Differentiate each of the following functions 1) 5=y 2) xy 35+= 3) 63xy =

4) )5()( 33 += sssf 5) xxf 5)( = 6) x

xxy

3

5 25 +=

7) 1

1)(

2

2

−+=

z

zzf

8) The cost function is 500152 +−= qqc where q = quantity of units produced.

Find the point of minimum cost. 9) If 00012604.0005.0 23 ++−= qqqc us a total cost function, find the marginal

cost when q = 50. 10) A process has a total cost function give by xc 1050+= and a revenue function

given by ,888 2xxR −= where x is the level of activity (in hundreds). The cost(C) and revenue (R) are both in units of a million kwacha.

a) Derive an expression for the total profit (P).

b) Calculate the level of activity that maximizes profit and the amount of

profit at this level.

c) What is the profit situation when 500 units are produced.

367


MULTIPLE CHOICE QUESTIONS

SECTION A

1.1. The derivation of the function, xx

xf 56

)(2

−= is equal to:

a) 21

56 2 xx − b) 2

512 23−

+− x

x c)

xx

2

512 −−

d) xx 2

5123

−

(Natech, Mathematics and Statistics; December 2000)

1.2. Find the gradient of 18423 22 =−++= xatxxxy . a) 17 b) 1 c) 27 d) -17

(Natech, Mathematics and Statistics; December 2004) 1.3. A firm’s output is given by ,2120 42 LLQ −= where L is the number of labour

hours in thousands. How many labour hors maximize output? a) 30.0 b) 8.0 c) 24.0 d) 5.5

(Natech, 1.2 Mathematics and Statistics; June 2003) 1.4 The demand equation for a certain product is xxP 0125.0500)( −= where )(xP

is the price per unit, in kwacha and x is the quantity demanded. For what value of x is revenue a maximum.

a) 20 000 b) 500 c) 0.0125 d) -0.025

(Natech, 1.2 Mathematics and Statistics; December 2002)

1.5 What is dx

dy of the following function? 32 /)1( xxy +=

a) 232

xx b) 3

2)3(x

x−− c) 4

23(x

x−−

d) xxx 2)1(3 2 ++−

(Natech, 1.2/B1 Mathematics and Statistics; December 1999 (Rescheduled))

368

1.6. Find the derivative of the function 23

)4( 2 −= xY

A. 23

)2(2 x B. 21

)4(3 2 −xx C. )4(2

3 2 −x

D. . None of these. (Natech, 1.2 Mathematics and Statistics; December 1998)

1.7 Find the differential coefficient, ,dx

dy of the following function .)154( 823 xxy −=

A. 723 )154(8 xx − B. 823 )154( xx − C. )8)(3012( 2 xx − D. 7232 )154)(3012(8 xxxx −−


1.8 If ,312 25

21

xxy −−= then which one of the following statements is true?

A. 2

3

2

1

2

156

xx

dx

dy −−=−

B. 2

3

2

1

312 xxdx

dy −−=−

C. 21

)2

5(3)

2

1(12 2

3−

−−= xxxdx

dy D. none of these

1.9 The total profit from selling x pencils is 35720)( 2 −−= xxxP . The marginal

average profit function is equal to:

A. xx 720 − B. x

x 720 − C.

2

2 3520

x

x +

D. x

xx 35720 2 −−

1.10 If )(),1)(23()( xfthenxxxf ′−+= using the product rule, is equal to:

A. )1(2)1(3 −+− xxx B. 16 −x C. 23 2 −− xx D. (3)(1)

369

SECTION B

1) Kango Pottery Ltd, a manufacturer of ornamental china plates estimates its profit function to be 20002000060)( xxxP −+= where x is the amount of money, in kwacha, spent on advertising.

i) What amount spent on advertising will yield maximum profit? ii) What is the maximum profit?

(Natech, 1.2 Mathematics and Statistics; June 2001) 2) Find the relative maximum and minimum values of the following function 593)( 23 +++−= xxxxf


3) a) Differentiate the function 1002 )12( +−= xxy with respect to x and simplify your answer as much as possible.

b) A manufacturer knows that if x (in hundreds) products are demanded in a

particular week, the total cost function is x314+ and the total revenue function is .219 2xx −

i) What is the total profit function? ii) Find the profit break-even points

iii) Calculate the level of demand that maximizes profits and the amount of profit obtained.

4) Find the point(s) where the gradients of the following equations are zero. i) 231820 xxy −+=

ii) 723632 23 +−−= xxxy (Natech, 1.2 Mathematics and Statistics; June 2002)

370

5) a) A company has examined its cost structure and its revenue structure and has determined that the following functions approximately describe its costs and revenues ,015.0100 2xC += where C = total cost and x = number of units produced. , where R =x2 total revenue and x = number of units produced and sold. Find, by calculus method, the output which will maximize profits for this company.

b) If ,15629)( 2 −−= xxxR find: i) the values of R(0) and R(3)

ii) the value of x, correct to nearest whole number, for which R(x) is maximum.


6) a) The revenue and cost function of product x have been identified as: Total revenue (K’000) = xx 202 +− Total Cost (K’000) = 216 +x were x = sales in units.

i) Determine the profit-maximizing price and quantity.

ii) Compute the maximum profit based on the results in (i) above. b) Find the first derivative of:

i) x

xxy

6

202 +−=

ii) 52 )104( +−= xxy


7) a) Find the 1st derivative of the following 12

3

2

1 23 −++= tttT .

b) Find the points where the gradients of the following equations are zero. 723632 23 +−−= xxxy


371

8) a) In a certain office, examination and analysis of past records show that there is a relationship between the number of clerks employed and the average cost of processing an order for new business. If q is the number

of clerks employed, average cost is given by qq

qC10

11000)( +=

What value of q will minimize this expression?

b) Examine the following expression for maximum and minimum values using the second derivation method 2)12( xxy −= .

(Natech, 1.2 Mathematics and Statistics; November/December 2000)

9) Differentiate the function given by 23

54 +

=x

y

(Natech, 1.2/B1 Mathematics and Statistics; December 1999 (Rescheduled))

372

CHAPTER 8

SOLUTIONS

CHAPTER 1

Exercise 1 1) 30 2) 17 3) 22 4) 6 5) 105 6) 48

7) 19 8) -19 9) 3

53 10) 25 11)

3

31− 12) 88

13) 134 14) 4

7− 15) 240

Exercise 2

1) a) 1

2 b)

5

6 c)

3

2 d)

1

3

2) 500

1 3) 416 4) K336 000 5) K86 400

Exercise 3 1) K91 200 2) 22 500 3) K825 4) K1 350 000 5) 400km 6) 416 7) K712 500 8) $91 9) 15 hours 10) K6 000 000

Exercise 4 1) 4 hours 2) 24 days 3) 4 sweets 4) 24 days

5) 7

22 days

373

Exercise 5 1) 11.51 SAR 2) 147.77E 3) 2,228,571.43 SAR 4) K2 765 000 5) £123.68 6) 1 SAR = K750 7) a) $111.11 b) K2362 500 c) K2 344 440

Exercise 6 1) 1500m 2) 0.55kg 3) 55.65kg 4) 513kg 5) 35cm 6) 45.6m 7) 0.135kilo litre 8) 2902500m 9) 167cm 10) 25.02kg 11) 33735cm

CHAPTER 2

Exercise 1 1) a) 4 b) 31 c) 9− d) 12− e) 12− f) 15 g) 5 h) 15− .

Exercise 2 1) a) 11a b) 15a c) 5a d) 3a e) 3a

f) 363− g) 243− h) 20

36

b

a− .

2) a) 27.621 b) 105.382 c) 12.2497 d) 20.967 3) a) 1.682 b) 0.1108 c) 0.6487 d) 5.213 e) 12.1825 f) 145.413 g) 0.9802 h) 2.8629

Exercise 3

1) 2,2 == yx 2) 5,8 == yx 3) 5,2 == yx 4) 43 == yx 5) 5,8 == pq 6) 3,2,1 === zyx

7) 3,5 == pq 8) 56,15,59 321 ==−= PPP

374

9) 50,45 == yx 10) 2

7,

4

27 == yx

EXAMINATIONS QUESTIONS WITH ANSWERS

Multiple Choice

1.1 B 1.2. A 1.3 B 1.4 B 1.5 A 1.6 D 1.7 A 1.8 A 1.9 A 1.10 C

SECTION B

Solution One

a) 532

52

3 =++− q

q

q

Multiply both sides by )52( −qq

71.054.38

173578.118

129

8

17

64

129

8

17

4

10

64

289

8

17

04

10

4

17

010174

25101086

25101510643

)52(5)32)(52(3

2

2

2

2

22

22

orq

q

q

q

q

qq

qq

qqqq

qqqqqq

qqqqq

=

+±=

±=−

=

−

−=

−

=+−

=+−

−=−−−−−−++

−=+−+

b) At equilibrium sdds QQandPP ==

375

12

1089

4925200

=−=−

+=−

Q

Q

QQ

when 14060200)12(5200,12 =−=−== PQ c) )1(53 →=+− zyx

)3(113

)2(4322

→=−+→=++

zyx

zyx

Equation (1) add to reaction (3) to eliminate z. )4(1624 →=+ yx Equation (2) and 3 × equation (3) to get equation (5) 37115 =+ yx .

Multiply equation (4) by –5 and equation (5) by 4 and add the two new equations to get the value of .y

1484420

801020

=+−=−−

yx

yx

2

6834

==

y

y

Substituting this value of y in equation (4),

3

124

1644

==

=+

x

x

x

Finally, substituting 2,3 == yx in equation (1), 22)3(35 −=+−=z . x

Solution Two i)

1 2 3 4 5 6 7 8

622 ++= xxC 9 14 21 30 41 54 69 86

x

376

ii)

C 18 • 12 • 622 ++= xxC • 6 • 0 1 2 3 4 5 iii) 6225 2 ++= xx

tonnesx

x

x

cba

xx

347

472.32

94.82

2

)19)(1(422

19,2,1

0192

2

2

≅

=

±−=+±−

=

−====−+

377

CHAPTER 3

Exercise 1 2) Score Tally Frequency

6 I 1 7 I 1 8 IIII IIII 10 9 IIII IIII IIII 14 10 IIII I 6 11 IIII 4 12 IIII 4 ∑ = 40f

3) 30 26 22 18 Number of stores 14 10 6 2 100 150 185 200 250 277 price (Kwacha/g)

378

4) 28 24 21 18 Number of Accounts 15 12 9 6 3 0.5 10.5 20.5 30.5 40.5 50.5 60.5 price (Kwacha/g) 5) 200 150 Number of people 100 50 0 2 5 8 11 14

379

Age (years) 6) i) ‘less than’ distribution f F 0 – 10 26 26 10 – 20 44 70 20 – 30 36 106 30 – 40 30 136 40 – 50 8 144 50 – 60 6 150 150 125 • • • 100 • Cumulative 75 frequency • 50 • 25 10 20 30 40 50 60 70

Monthly Bonus (Kwacha)

380

ii) “greater than” distribution f F 0 – 10 26 150 10 – 20 44 124 20 – 30 36 80 30 – 40 30 44 40 – 50 8 14 50 – 60 6 6 150 • 125 100 Cumulative 75 frequency • 50 • 25 • • 10 20 30 40 50 60 70

Monthly Bonus (Kwacha)

381

7) others Competitions TV Newspaper Radio

o

o

o

o

o

Others

nsCompetitio

Newspaper

Radio

TV

14846.13360260

10

2869.27360260

20

6923.69360260

50

42538.41360260

30

20869.207360260

150

≅=×

≅=×

≅=×

≅=×

≅=×

382

8) 300 No. of 200 Employees 100 50 A B C D Factory 9) 300 No. of

200 Employees 100 50 X Y Z W Factory Unskilled Semi-skilled Skilled

383

1000 900 8 700 600 Frequency 500 400 300 200 100 Companies Company X Company Y Company Z

Exercise 2 1) a) 592.5 b) 9.083 c) 14.6 d) 8

e) −8.8 2) a) 24.5625 b) 4.918 c) 47.385 d) 28.058

e) 45.894 3) K345 833.33 4) i) 9.681 ii) 1.55 iii) 1.2

384

5) i) 1 ii) 1 6) a) 169.643 b) i) 186.765 ii) 176.667 7) i) a) 5.21875 median = 5.333 Mode = 5.167 ii) a) 241.216 b) 21.071 c) 21.667 8) Mean = 155.857 median = 156.429 Mode = 184.375 9) Mean = 3.267 median = 3 Mode = 3 and 4 10) a) mean b) mode c) mean d) mean e) mode f) mode

Exercise 3 2) range = 6 692.15.15,2 31 ==== σQDQQ

3) a) 99.517 b) 10.286 c) 9 d) 10.34% 4) a) 64 b) 35 c) 30 d) 17 e) 50 f) 16.5 g) 98.163 h) 0.061 5) i) 6.148 ii) 6 iii) 5 iv) 7.308 6) Year 1 6.85554.115.13 === cvσµ Year 2 4.76308..105.13 === cvσµ There is less variability in the number of rooms per dwelling in Zambia in the

second year than the first year.

385


1.1 B 1.2 D 1.3 C 1.4 C 1.5 C 1.6 B 1.7 B 1.8 C 1.9 C 1.10 C

SECTION B Solution One a) Arrange the data in decreasing order 2, 4, 5, 7, 8, 11, 13, 18 Since 8=n , and

5.22

7

2

411

11;6)2(4

3

4;2)8(4

1

3

1

==−=∴

==

==

QD

Qposition

Qposition

b) ooDRCongo 1363609.156

3.59 =×=

oo

oo

oo

oo

Zambia

Kenya

Tanzania

BrazavilleCongo

233609.156

9.9

243609.156

3.10

363609.156

8.15

1413609.156

6.61

=×=

=×=

=×=

=×=

386

Zambia 23 Kenya 24 Congo DR 36 136

Tanzania 141 Congo Brazaville

Pie chart for five distant geographical market. Solution Two a) For category x

intmidpox − f xf fx2 6.95 10.95 14.95 18.95 22.95 26.95

3 5 7 6 3 1

∑ = 25f

20.85 54.75 104.65 113.7 68.85 26.95

∑ = 75.389xf

144.9075 5.995125 1564.5175 2154.615 1580.1075 726.3025

∑ = 9625.67692 fx

387

Thus, the standard deviation is

( )

( )

)2(27.57504.27

2525

75.3899625.6769

2

2

2

placesdecimalS

S

f

f

fxfx

S

==

−=

−=∑ ∑

For category y,

intmidpox − f xf fx2 6.95 10.95 14.95 18.95 22.95 26.95

4 8 8 3 3 4

∑ = 30f

27.8 87.6 119.6 56.85 68.85 107.8

∑ = 5.468xf

193.21 959.22 1788.02

1077.3075 1580.1075 2905.21

∑ = 075.85032 fx

Thus, the standard deviation is:

388

( )

( )

29.6

5556.39

3030

5.468075.8503

2

2

2

≅

=

−=

−=∑ ∑

f

f

fxfx

S

b)

Complaints per week (x) Number of weeks (f) Cumulative frequency (F) 0 1 2 3 4

5 12 7 2 1

5 17 24 26 27

Position of the median is given by 5.13)27(2

1 =

The 1.5th observation will be found where cumulative frequency is 17. hence, the

median is 1. c)

f x fx xxf −

*0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 500 – 600 600 – 700*

3 6 11 15 12 7 6

∑ = 60f

50 150 250 350 450 550 650

150 900 2750 5250 5400 3850 3900

∑ = 22200fx

960 1320 1320 300 960 1260 1680

∑ =− 7800xx

* Approximate lower boundary and upper boundary.

389

13060

7800

37060

22200

==

−=

===

∑∑

∑∑

f

xxfdeviationMean

f

fxx

Solution Three

x f xf fx2 32.5 37.5 42.5 47.5 52.5 57.5

22 27 21 31 21 18

∑ = 140f

715 1012.5 892.5 1472.5 1102.5 1035

∑ = 6230xf

23237.50 37968.75 37931.25 69943.75 57881.25 59512.50

∑ = 2864752 fx

i) Mean = 5.44140

6230 ==∑∑

f

xf

ii) Standard deviation

390

( )

( )

15.8475.66

139140

6230286475

1

2

2

2

≅=

−=

−

−=

∑

∑∑∑

f

f

xffx

σ

iii) Since the data is grouped, and thus the original access times are not known, both

the measures above are estimates, i.e. approximations of the actual values. Solution Four i) Class Interval Tally Frequency(f) 35 – 39 l 1

40 – 44 IIII lll 8

45 – 49 IIII IIII I 11

50 – 54 III 3

55 – 59 II 2

60 – 64 II 2

65 – 69 II 2

70 – 74 1 1

30

ii) The modal class frequency is 45 – 49 .

391

Solution Five

a) i) The mean class size of the college can be calculated as follows:

English History

44 × 152 + 26 × 192

= 668.8 + 499.2

= 1168

Therefore, mean class size of college

= 7.166857.1670

1168 ==

ii) From the frequency distribution it is found that 4 classes were of size 1 – 6 students and 15 + 3 classes were of size 7 – 12 students. Therefore, a total of 22 classes would not be run. Thus, 70 – 22 = 48 classes will remain. The 1168 students would, therefore, be distributed over 48 classes, giving a mean class size

of 3.2448

1168=

iii) Given that the students attending classes of size 1 – 6 and 7 – 12 are not admitted, the mean class size of the college can be calculated from the modified frequency.

Size Mid-point x No. of classes (f ) fx 13 – 18 19 – 24 25 – 30 31−36

15.5 21.5 27.5 33.5

21 16 9 2

48

325.5 344.0 247.5 67.0

984

Therefore, mean = 48

984=∑∑

f

fx

= 20.5

392

Solution Six

a)

Freq

150

100

50

Mode = 26.7

0 10 20 30 40 value

b) i)

x f fx fx2 5 15 25 35 45 60 80

16 30 34 22 10 5 3

∑ = 120f

80 450 850 770 450 300 240

∑ = 3140xf

400 6750 21250 26950 20250 18000 19200

∑ = 1128002fx

.978.15

)107.26(120

112800

167.26120

3140

2

mK

mKx

=

−=

==

σ

σ

393

ii) The mean salary five years ago was K18.95m whereas toady this has increased to K20,166m, unfortunately, the variation around the mean has also increased from K10.6m to K15.978m, clearly indicating increased variability in salary.

Solution Seven

IQ No. of children

( f )

x Mid point

xf fx2 xxf −

50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109 110 – 119 120 – 129 130 – 139

1 2 8 18 23 21 15 9 3

∑ = 100f

54.5 64.5 74.5 84.5 94.5 104.5 114.5 124,5 134.5

54.5 129.0 596.0 1521.0 2173.5 2194.5 1717.5 1120.5 403.5

9910=∑ xf

297.25 8320.5 44402.0 128524.5 205395.75 229325.25 196653.75 139502.25 54270.75

∑ = 10093652 fx

44.6 69.2 196.8 262.8 105.8 113.4 231.0 228.6 106.2

4.1358=−∑ xxf

i) 1.99100

9910 ==x

584.13100

4.1358 ==

−=

∑∑

f

xxfdeviationMean

ii)

394

( )

( )

52.1684.2772

100100

99101009365

2

2

2

==

−=

−=∑ ∑

σ

σf

f

xffx

c) The standard deviation is greater than the mean deviation. This is because the standard deviation gives more prominence to extreme values. The mean deviation, on the other hand gives equal weight to extreme items and items whose deviation from the mean is small, so that the existence of several extreme items is not adequately reflected.

Solution Eight

Value x No. of workers f xf 100 000 150 000 200 000 250 000

165 190 105 92

∑ = 552f

16500 000 28500 000 21000 000 23000 000

∑ = 00089000xf

i) Mean = 552

00000089==∑∑

f

xfx

K161 231.88

ii) The modal value per order K150 since it has the highest frequency.

b)

x f fx fx2

395

0 1 2 3

4 or more

23 14 3 2 0

∑ = 42f

0 14 6 6 0

∑ = 26xf

0 14 12 18 0

∑ = 442 fx

i) Mean = 619.042

26 ==∑∑

f

xf

ii)

( )

( )

664.0

42

9047619.27

4242

2644

2

2

2

2

=

=

−=

−=

∑

∑∑∑

f

f

xffx

σ

iii) 8151.06644.0 ==σ c)

x f xf fx2 4 - 5 5- 6 6.7 7- 8 8- 9 9- 10 10 - 11 11- 12 12- 13

4.5 5.5 6.5 7.5 8.5 9.5 10.5 11.5 12.5

3 7 2 4 6 10 8 4 0

13.5 38.5 13.0 30.0 51.0 95.0 84.0 46.0

0

60.75 211.75 84.50 225.0 435.5 902.5 882.0 529.0

0

396

13 – 14 13.5

8

∑ = 52f

108

∑ = 479xf

1458.0

∑ = 47872 fx

i) Arithmetic mean = 212.952

479 ===∑∑

f

xfx

ii) Modal sales = 9.5

iii) Standard deviation

( )

71.2

34653092.7

5152

)479(4787

1

22

2

=

=

−=

−

−=

∑

∑∑∑

S

f

f

fxfx

iv) S

medianmeanSK

)(3

−=

106.0

71.2

)5.9212.9(3

≅

−=

CHAPTER 4

Exercise 1

1. a) 0.14 b) 0.06 c) 0.30 2. a) 0.16 b) 0.58 c) 0.3913 d) not independent

3. 221

1 4.

12

1

397

5. a) 0.2975 b) 0.2479 c) 0.4959 d) 0.5041

6. a) 0.2097 b) 0.0000128 7. a) 0.0498 b) 0.5882 c) 0.36199 Exercise 2 1. i) 54322345 510105 babbababaa +++++ ii) 765243342567 7213535217 babbabababaaa −+−+−+− iii) 432234 6251500135018081 babbabaa +−+− iv) tsttststss −+−+− 432345 1040808032 v) .126016024019264 12103866941221518 yyxyxyxyxyxx +−+−+− 2. i) 54322345 243810108072024032 dcddcdcdcc +++++ ii) 432234 625200024001280256 srssrsrr ++++ iii) 32331 xxx −+− iv) 15129263345 32802000500062503125 babbababaa +++++

v) 8524 2

3927

5481ccc

cc++++

3. a) 0.0467 b) 0.31104 c) 0.8208 d) 0.4557 e) 0.13824 f) 0.95904 4. a) 0.0302 b) 0.3209 c) 0.1359 d) 0.6791 5. a) 0.6983 b) 0.2573 c) 0.0444 d) 0.3017 6. a) 0.0000759 b) 0.9999 7. a) 0.4219 b) 0.0469 c) 0.9961 8. P(0) = 0.178, P(1) = 0.356, P(2) = 0.2966, P(3)=0.1318 P(4) = 0.033, P(5) = 0.0044, P(6) = 0.0002 9. 0.9991 10) 0.1404 11. a) 0.0821 b) 0.9179 c) 0.2566 d) 0.1336 12. 0.9972

398

Exercise 3 1. a) 0.3085 b) 0.4602 c) 0.6179 d) 0.1859 e) 0.44 2. a) i) 0.2946 ii) 0.3859 iii) 0.346 b) 2.4655± 3. i) 0.0869 ii) 0.5299 4. 0.1711 5) a) 0.9738 b) 0.509 c) 0.2879 6. a) 0.492 b) 0.1685 7. a) 0.9969 b) 65.23 c) 0.6597 d) 67.1675

EXAMINATION QUESTIONS

Multiple Choice Questions (Section 1) 1.1 A 1.2 C 1.3 A 1.4 A 1.5 A 1.6 C 1.7 C 1.8 B 1.9 A 1.10 A

Multiple Choice Questions (Section 2) 1.1 B 1.2 C 1.3 A 1.4 B 1.5 C 1.6 A 1.7 A 1.8 B 1.9 C 1.10 C 1.11 D 1.12 A

SECTION B (1)

Solution One a) 100,150 ≅= σµ Let X be the normal random variable standing for the number of items

399

9772.0

47725.0

)2(10

)150170()170(

=+=

<=

−<=>

ZP

ZPXP

2 b) nPEV = (assuming the distribution follows a binomial distribution). The number

of defectives is 0.02(400) = 80 defectives. c) i) In a permutation there is attention give to the order of arrangement. In a

combination there is no attention given to the order of arrangement.

ii) !)!(

!

xxn

nCx

n

−=

ways

C

21

!5!2

!5)6(7

!5)!57(

!77 5

=

=

−=

Solution Two a) Let X be the duration (days) required to shift operations completely from Mongu

to Lusaka. Then X is approximately normally distributed with mean 300 days and standard

deviation 9 days.

400

i)

−<=<9

300280)280( ZPXP

014.0

)22.2(

=−<= ZP

ii)

−>=>9

300310)310( ZPXP

134.0

)11.1(

=>= ZP

iii)

−<<−=<<9

300310

9

300280)310280( ZPXP

853.0

)11.122.2(

=<<−= ZP

b) i) Let X = number of orders rejected.

X is poisson with 4=λ orders

16.0

156.0!5

4)5(

45

≅=

==−e

XP

ii) X is binomial with n = 8 orders

[ ]

45.0

4483.0

)2965.01976.00576.0(1

)2()1()0(1

)3()3(

3.0

≅=

++−==+=+=−=

≥==

XPXPXP

XPrejectedleastatP

P

c) This is a binomial question and we let P be the probability that the drug is

effective, i.e. xAndnandqP .645.0,55.0 === the number of patients.

i) 65 )55.0(6

6)45.0()55.0(

5

6)5(

+

=≥xP

= 0.1359 + 0.02768

401

= 0.16358

ii) )45.0(0

6)0(

==xP

= 0.45

iii) 335 )45(.)55.0(3

6)45.0()55.0(

1

631(

+

== orxP

= 0.06105 + 0.30318 = 0.3642

Solution Three

a) i)

−<<−02.0

8035.8

02.0

8985.7ZP

7333.0

4599.02734.0

)75.175.0(

=+

<<− ZP

Therefore probability = 1− 0.7334 = 0.2667.

ii) 50% area under the normal curve minus 2% = 48% = 0.48 the Z value is 2.05

02.0

035.805.2..,

µσ

µ −=−= eiX

Z

Therefore, mean = 8.035 – 2.05 (0.02)

= 7.994 b)

Defective 0.03 0.05 0.08 Non Defective 0.52 0.40 0.92 .55 .45 1.00

402

423.0

039.0

0165.0

0225.00165.0

0165.0

)45)(.05.0()55(.03.0

)55)(.03.0(

)()/()()/(

)()/()/(

=

=

+=

+=

+=

BPBDPAPADP

APADPDAP

c) Let X be the number of rejects. Assuming X is a binomial vr. where

.102.0,08.0 === nandqP

960.0

148.0378.0434.0

)2.0()8.0()2.0()8.0()2.0()8.0(

)2()1()0()2(82

21091

110100

010

=++=

++=

=+=+==≤

CCC

XPXPXPXP

Solution Four a) This is a binomial distribution problem where .5,6.0,4.0 === nqP . Let X

be a binomial vr. Then i) 07776.0)6.0()4.0()0( 50

05 == CP

403

ii) 92224.007776.01)(1)1( =−=−=≥ nonePXP iii) )2()1()0()2( =+=+==≤ XPXPXPXP

68256.0

3456.02592.007776.0

)6.0()4.0()6.0()4.0()6.0()4.0( 322

5411

5500

5

=++=

++= CCC

b) 20,190 == σµ Let X be a normal Vr. with

i)

−<=<20

190160)160( ZPXP

0668.0

4332.05000.0

)5.1(

=−=−<= ZP

-1.5

ii)

−>=>20

190210)210( ZPXP

1587.0

3413.05.0

)1(

=−=

>= ZP

1

404

The required number of journeys = 150 (0.1587)

= 23.805 24≅ iii) 3 hours 35 minutes is 215353180 =+× minutes. Hence

8944.0

3944.5.

)25.1(

20

190215215(

=+=

<=

−<=<

ZP

ZPXP

The true percentage figure is 89.44% 1.25 The required number of journeys = 150 (0.1587)

c) !

)(r

erP

x λλ −

=

3614.0)3012.0(2.1!1

2.1)1(

3012.0!0

2.1)0(

2.11

2.10

===

==

−

−

eP

eP

2169.0)3012.0(72.0!2

2.1)2(

2.12

===−e

P

0867.0)3012.0(288.0!3

2.1)3(

2.13

===−e

P

0260.0)3012.0(0864.0!4

2.1)4(

2.14

===−e

P

405

0062.0)3012.0(020736.0!5

2.1)5(

2.15

===−e

P

0012.0)3012.0(0041472.0!6

2.1)6(

2.16

===−e

P

0002.0)3012.0)(0007109.0(!7

2.1)7(

2.17

===−e

P

Solution Five

a) Using a tree diagram, where D stands for defective and G not defective P(D) = 0.05 0.45(0.05) = 0.0225 P(Old) = 0.45 P(G) = 0.95 0.45(0.95) = 0.4275 New P(D) = 0.03 0.55(0.03) = 0.0165

P (New) = 0.55

P(G) = 0.97 0.55(0.97) = 0000.1

5335.0

Figure 1.0 Therefore, from Figure 1.0;

406

.5769.0039.0

0225.0

)(

)()/(

039.00165.00225..0)(

==∩=

=+=

DP

DoldPDoldP

DP

b) This is a normal distribution problem with XLetand .45.618 == σµ be a

normal ,.vr then ;

i) 2676.0)62.0(45.6

1822)22( =>=

−≥=≥ ZPZPXP

ii) 7238.02238.05.0)93.0(45.6

1824)24( =+=≤=

−≤=≤ ZPZPXP

c) This is a binomial distribution problem with .4

3

4

1,8 === qandPn Let X

be a binomial random variable i.e. standing for the number of errors. i) )2(1)2( ≤−=> XPXP

[ ]

[ ]

3214.0

3115.0267.01001.0

4

3

4

1

4

3

4

1

4

3

4

11

)2()1()0(1

62

28

71

18

80

08

=

++=

+

+

−=

=+=+=−=

CCC

XPXPXP

ii) )1()0()2( =+==< XPXPXP = 0.1001 + 0.267 = 0.3671

Solution Six a) 10R

5W 14

9)( =RP

407

210

90)( =RRP

15

10)( =RP

14

5)( =WP

210

50)( =RWP

14

10)( =RP

210

50)( =WRP

15

5)( =WP

14

4)( =WP

210

20)( =WWP

The required probability is 21

10

210

100

210

50

210

50)()( ==+=+ WRPRWP

b) nnP .140,02.0 == is large and P is small. We use the Poisson distribution

with .8.2)140(02.0 === npλ

,...2,1,0,!

)( ===−

xwherex

exXP

x λλ where x is a Poisson r.v.

i) 2384.0!2

)8.2()2(

8.22

===−e

XP

ii) )1()0()2( =+==< XPXPXP

408

2311.0

1703.00608.0!1

)8.2(

!0

)8.2( 8.28.20

=+=

+=−− ee

c) i) 240

111027569815019018814170180 +++++++++++==∑∑

f

xfx

4

240

959

≅

=

ii) ...2,1,0,!

)( ===−

xwherex

exXP

x λλ

0183.0!0

4)0(

4

==−e

P

0732.0)0183.0(4!1

4)1(

4

===−e

P

0366.0)0183.0(2!2

4)2(

42

===−e

P

1952.0!3

4)3(

43

==−e

P

3904.0!4

4)4(

44

==−e

P

1562.0!5

4)5(

45

==−e

P

1042.0!6

4)6(

46

==−e

P

0595.0!7

4)7(

47

==−e

P

0297.0!8

4)8(

48

==−e

P

0132.0!9

4)9(

49

==−e

P

409

0053.0!10

4)10(

410

==−e

P

0192.0!11

4)11(

411

==−e

P

Exercise 4 1. K492,299.43 to K507,700.57 2. 91.31 to 96.69 3. 0.42 to 0.62 4. (a) 0.0358, 00358 (b) 0.108 to 0.292 (c) 0.128 to 0.272 5. 280=n 6. 267=n 7. 106.57 to 133.43 8. 214.37 to 249.63 Exercise 5 1. 0;2.0 HAcceptZc −= 2. 0Re;44.2 HjectZc =

3. 0Re;81.3 HjectZc = 4. 0;5199.0 HAccepttc =

5. oc Hjectt Re;301.2= 6. oc HAcceptZ ;37.1−=

7. oc HAcceptt ;6097.0−=

Exercise 6 1. oc HjectZ Re;25.2−= 2 oc HjectZ Re;699.3=

410

3. oc HjectZ Re;116.25−= 4. oc HjectZ Re;45.2−=

5. oc HjectZ Re;04.2=

EXAMINATION QUESTION WITH ANSWERS


1.1 D 1.2 A 1.3 C 1.4 B 1.5 B 1.6 B 1.7 B 1.8 A 1.9 B 1.10 C

SECTION B

Solution One a) 00,25:;000,25: 2 >= µµ HHo

b) )(1800:,1800: 2 testtailedoneHHo −>= µµ

50100,1800,1850 ==== nandX σµ

Standard error, 150

100. ==

nxofes

σ

The test statistic, ( )

54.3

50

10018001850 =−=−=

X

XZ

σµ

.33.254.3 >=cZ

Hence, we reject the null hypothesis at 1% or 0.01 level of significance and conclude that the claim is justified.

c) Null hypothesis: oH ‘the trial has not produced stronger struts’.

411

The test statistic: 62.1

25

18512501310 =−=Z

We use a one tailed test since we are only interested in improved results. Z is not significant at the 5% level. Thus, the statistical evidence as has been produced is not convincing enough that the new material is producing stronger struts.

Solution Two a) A point estimate is a single number which estimates a population parameter e.g.,

the sample mean may be used as a point estimate of the population mean. While an interval estimate is a range within which we can be confident at a given level of probability that the value of the population parameter lies. Example, the average height of a male NATech student is 1.76m is a point estimate while the average height of a male NATech student lies between 1.5 to 2.1 metres is an interval estimate.

b) i) Let =π population proportion of students who knew the intitute’s

programmes through ZACB hour

( )500

37.0500

185

sizesamplen

proportionsampleP

=

==

95% confidence interval is given by ( )

)412.0,328.0(

)02159.0(96.137.0),02159.0(96.137.0(

500

)63.0(37.096.137.0,

500

)63.0(37.096.137.0

96.137.0

+−

+−

−

There is a 95% probability that the population proportion of students who come to know about the institute’s programmes lies between 3.2% and 41.2%

ii) Sample Size

412

( ) ( )( )( )

students

n

n

d

PqZn

8955

7696.8954

01.0

63.037.096.12

2

2

2

2

≅=

=

=α

Solution Three

a) i) cmandXnX

5.775.168,100 === σ

The 99% confidence interval is given by

( )

935.175.168100

5.758.275.168

2

±

±

±n

ZX xZ

σ

(166.815, 170.685). There is a 99% probability that the mean height of all students is between 166.8 and 170.7cm.

95.1

025.02

05.0

2,95.01,01.0,05.0)

005.

2

==

===−<==

ZZ

derrorii

σ

αασ

( ) ( )

( ) ( )

( ) ( )

.97

04.96

01.0

05.096.1

01.0

05.096.1

05.096.101.0

2

≅>

>

>

>

>

n

n

n

n

n

n

Z

d Z

σα

b) mmmmXnX

06.0,38.4,10 === σ

413

( )

( )44.4,32.4

06.038.416.3

06.025.338.4

10

06.025.338.4

25.399.01

1,2

9,005.0

±

±

±

±

==−

− ntX

t

X

n

σα

α

c) %1.4500000,49,7,500,49 ====== αµ sandXnMean

143.3000,42:

000,42:

6,01.0 =>=

tH

H

a

o

µµ

41.484.1700

7500

7

4500000,42500,49 ==−=−=

n

sX

tc

µ

Reject oH if .143.3>ct Therefore, we reject oH since .143.341.4 > We can

conclude that there has been a significant increase in the weekly turnover.

000,032,1:

000,032,1:)

KH

KHd

a

o

≠=

µµ

Reject oH if .032.4032.4 5,005.05,005.0 −=−<=> ttortt cc

35.04308.370686

129900

6

61.90799210320001161900

6145.907992,1161900

6971400,1022223226.1,6 132

==−=−=

==

=×== ∑∑

n

SX

t

x

xxn

c

µ

Accept oH and conclude that there is sufficient evidence to say that the mean

bank balance is K1,032,000.

414

Solution Four a) i) The confidence limits are given by

nZx

σα2

±

The error must not exceed 20 kilometres, thus

( )

( )

93

544.92

62.9

20

13048.1

13048.120

2

>>>

>

>

n

n

n

n

n

ii) If ,n the sample size, is too large it can be reduced by either increasing the

allowed error or decreasing the confidence interval or both. b) i) You wish to detect a situation leading to a loss when it occurs. Thus, we

want to detect when .0.3<µ So that the null and alternative hypotheses are:

0.3:;0.3: 1 <= µµ HHo

ii) You wish to detect a situation leading to a profit when it occurs. Thus, the

null and alternative hypotheses are: 0.3:;0.3: 1 >= µµ HHo

c) Matched pairs

0:

0:

<−=−

BAa

BAo

H

H

µµµµ

Where AB and µµ are the average scores before and after the course respectively. Hence we compute the differences.

415

Student d d

1 -26 676 2 -6 36 3 0 0 4 16 256 5 1 1 6 -22 484 7 -26 676 8 1 1 9 2 4 10 15 225 11 9 81 ∑ −= 32d 24402 =∑d

Solution Five a) i) The company wishes to determine whether absenteeism has declined and

they want to detect the decline if it has occurred. Hence, we have

98:

98:

1 <=

µµ

H

Ho

ii) In this situation, you would be more interested in whether or not your sales

are leading to a financial disaster, and if it is true, you want to detect that. Hence, the hypotheses are:

20:

20:

1 <=

µµ

H

Ho

iii) Since, the process is out of control if and only if the mean diameter of the

machine bearings is different from 1.27cm, the hypotheses are:

27.1:

27.1:

1 ≠=

µµ

H

Ho

416

( ) ( )

62999.0

11

32.1591.2

32.1510

11

322440

1

91.211

32

22

2

−=−==

=

−−=

−

−=

−=−==

∑ ∑

∑

n

Sdd

t

nn

dd

Sd

n

dd

c

Reject .895.110,05.0 −=−< ttifH co Since –0.62999 is greater than –1.895, we

accept .oH There is no sufficient evidence at the 5% level of significance to

conclude that the course has produced some learning.

Solution Six a) The sample proportion is given by

80.02000

1600 ===n

x

sampledNumber

sampleinsuccessesofNumber

80% of the population favour more strict measures.

b) The null hypothesis, 200: =µoH

The alternative hypothesis, 200: ≠µaH

This is a two-tailed test.

19.26.1

5.3

100

162005.203 ==−=−=

n

XZc σ

µ

Reject .58.258.2005.0 −<=> coo ZorZZifH Hence accept oH

417

c) 6.5:

6.5:

≠=

µµ

a

o

H

H

Reject 65.165.1 −<> cco ZorZifH

35.2

30

4.16.55 −=−=−=

n

xZc σ

µ

Since ,65.135.2 −<−=cZ we regret .oH Hence there is a significant

improvement. Exercise 7

a) y

16 14

12

10

8 • •

6 • •

4 • •

2 •

0 x

1 2 3 4 5 6 7 8 9 10 11 12 14 16

418

b) y 10 9 8 7 6 5 • 4 3 2 • 1 2 6 8 10 12 14 16 18 20 22 24 26 x c)

5.35.3

1414

58

46

35

22

== yxmeansTotal

yx

25.85.12

3350

1016

913

811

610

== yx

yx

419

y

10

9 •

8

7

6

5

4

3 •

2

1

1 2 3 4 5 6 7 8 9 10 12 13 x 2. a)

Sales 14

12 •

10 • •

8 •

6 4

2 100 200 300 400

Advertising expenditure

•

•

420

b)

x (Advertising expenditure) y (Sales) K’000s K’ms

230 280 310 350 400 430

∑ = ,2000x ∑ = 7.65y

,6948002 =∑ x 43.7222 =∑ y

∑ == 6,22093 nxy

000,300,12.30.12,530^^

Keiyy ==

3. xy 714.0238.0^

+= 4. a) Communication 100

•

75 •

• •

50

25 (x) 25 50 75 100 Mathematics

(y)

• •

•

•

421

b) xy 696.016.13^

+=

Where x is mathematics and .^

ionCommunicaty =

c) 7232.72)85(696.016.13^

≅=+=y

d) .7548.74

696.016.1365

≅=+=

x

x

5. a) 16

14

12

10

8

6

4

2 1999 2000 2001 2002 2003 2004 2005

•

• •

• •

•

•

Birth Rate

Year

422

b)

Year ( )x Code Birth Rate ( )y 1999 1 14.6 2000 2 14.5 2001 3 13.8 2002 4 13.4 2003 5 13.6 2004 6 12.8 2005 7 12.6

140

282 =

=

∑∑

x

x

72.22115

7

8.225

2 =

==

=

∑∑

∑

y

nxy

y

xy 664.9914.70^

−=

c) ( ) 39.3511664.9914.70,11^

−=−== yx Exercise 8

1. a) 1=r b) 1−=r c) 055.0=r 2. 987.0=r 3. 714.0=r

4. a) 754.0=r b) 741.0=r 5. .227.0=r The coefficient of correlation is too low to be a reasonable

indictor for the price of company’s share.

423

EXAMINATION QUESTIONS WITH ANSWERS.


1.1 D 1.2 A 1.3 C 1.4 C 1.5 A 1.6 A 1.7 B 1.8 D 1.9 B 1.10 C

SECTION B Solution One a) The product moment correlation coefficient is given by

( )[ ] ( )[ ]∑ ∑∑ ∑∑ ∑ ∑

−−

−=

2222 xxnyyn

yxxynr

Where

11,103661798633044013467,11 22 ======= ∑∑ ∑ ∑ ∑ nyxyxxyn

( ) ( )

( ) ( )[ ] ( ) ( )[ ] 63.0440117986113301036611

330440134671122

=−−

−=r

b)

10

1.1

32

=

=

=

∑∑

n

x

x

∑∑∑

=

=

=

4.137

17548

3802

xy

y

y

The required equation is

.1179.2^

xy +=

424

c) Note: Profits depend on the amount spent on advertising therefore it is the

dependent variable .X Advertising expenditure is the independent variable .X

i) A scatter diagram is used to show that there is a relationship between two variables.

Another method is to find the product moment correlation coefficient .r

( )[ ] ( )[ ]∑ ∑∑ ∑∑ ∑ ∑

−−

−=

2222 xxnyyn

yxxynr

∑ ∑

∑ ∑∑==

====

278.54,91.3

,5819.2,3.82,87.1142,6 22

xyx

xyyn

( ) ( ) ( )( ) ( )[ ] ( ) ( )[ ]

.deg,938.0

91.35819.263.8287.11426

91.33.82278.54622

ncorrelatioofreehighverya

r

=

−−

−=

ii) xy 061.19296.1^

+= K800,000 is taken to be 8.0=x

( ).800,544,16

5448.168.006.19296.1^

^

Ky

y

=

=+=∴

425

Solution Two a) i) Regression coefficient is the number which quantities the relationship

between the dependant variable and explanatory variable, e.g., in the simple linear regression model,

;bxay += a and b are the regression coefficients. b is the slope which indicates the amount by which y changes for a given unit change in the value of x and a is the intercept which indicates the value of .0=xwheny

ii) The explanatory/independent variable is the variable in regression model

assumed to cause the dependent variable to change. I.e., in the model ,bxay += the explanatory variable is .x

b) Let =y imports and =x prices. Then

i) .608580,2995372,1322,2699 2 ====∑ ∑ ∑ ∑ xxyyx

xy 28.106.398 −= When ( ) 06.7825028.106.398,250 =−== yx Total imports – 78,060 tonnes ii) A correlation coefficient of –0.95 implies a strong negative linear

relationship between price and imports. The higher the price of apples, the less the number of apples bought.

Solution Three

a)

x y 10log x y10log

18 62 1.16 1.79 27 48 1.43 1.68 36 37 1.56 1.57 45 31 1.65 1.49 54 27 1.37 1.43 72 22 11.86 1.34 90 18 1.95 1.26

426

b) Let x = 10log x and y = y10log

∑ ∑ ∑ ∑ ∑ ===== 989.161396.160436.1956.1044.11 22 xyyxyx &

50.143662595

10

775.2log

,log

775.2775.0

775.2

10

==

=

=−=

a

a

soaoftheisA

AandB

775.050.143662595 −= xy c) If x = 64 800 000, then 775.0)00080064(50.143662595 −=y %05.526=x to 2 decimal places. Solution Four a) i) Let 2000 to 2004 be years 0 to 4

x y xy 2x 2y 0 20 0 0 400 1 18 18 1 324 2 15 30 4 225 3 14 42 9 196 4 11 44 16 121

5

10

=

=∑n

x ∑ = 78y ∑ = 134xy ∑ = 30x ∑ = 12662y

( )[ ] ( )[ ]

[ ][ ]

992.0

)7()1266(5)10()30(5

)78(10)134(522

2222

−=

−−−=

−−

−=

∑ ∑∑ ∑∑ ∑ ∑

yynxxn

yxxynr

427

ii) There is an extremely strong negative correlation between the year of sale

and units sold. The value of r is close to –1, therefore a high degree of correlation exists. This means that there is a clear downward trend in sales.

iii) Since .98.0,992.0 2 =−= rthenr That is 98% of the changes in sales can

be explained by the charge in the number of years. 2% of the changes are unexplained.

b) i) .2.220ˆ xy −= Sales for 2005 implies .5=x Therefore

,9)5(2.220 =−=y i.e. 9000 units.

ii) )1(

61

2

2

−−= ∑

nn

dr

7465.0)1144(12

)50.72(61 =

−−=r

c) i) Using the product moment coefficient of correlation.

.9339.0

49.145

,8378.20,5.1018,18.11,78 22

=

=

====

∑

∑ ∑ ∑∑

r

xy

yxyx

ii)

x y rx ry d 2d

14.0 14.0 13.5 12.5 12.0 12.0

1.90 1.91 1.86 1.84 1.84 1.83

1.5 1.5 3 4

5.5 5.5

2 1 3

4.5 1

-0.5

-0.5 0.5 0

-0.5 1

-0.5

0.25 0.25

0 0.25

1 0.25

∑ = 22d

428

775.0

2

2

50.143662595

.9429.0)136(6

)2(61

)1(

61

−=

=−

−=−

−= ∑

xy

nn

drs

Solution Five Let TC = Total cost and Q = Units produced. a) TC = 444.44 + 14.07Q TC is in thousands b) When Q = 0, TC = fixed costs = loss of K444, 444.44.

c) ( )[ ] ( )[ ]∑ ∑∑ ∑∑ ∑ ∑

−−

−=

2222 xxnyyn

yxxynr

where ∑ ∑ == 420,187000 xxy

∑

∑∑

=

==

.000187

2800550262

xy

yx

Thus, we would expect about 19 defective surgical needles in a box inspected by

a worker with 6 weeks worth of experience. Solution Six a) ii) ∑ ∑ ∑ ∑ ∑ ===== .1069,2156,732,128,72 22 xyyxyx

.1.87−=r There is a strong negative linear relationship between the of

weeks experience and number of rejections.

iii) xy 988.0892.24ˆ −=

904.23)1(988.0892.24ˆ1 =−=y

429

b) i) Correlation is to do with the strength of the relationship between two or more quantities such that a change in one of the quantities is accompanied by a predictable charge in the other. Regression is to do with describing mathematically the relationship between two or more quantities.

ii) ( )( )

( ) 86.01648

1261

1

61

2

2

=−

=−

−= ∑nn

dr

There is a strong, positive linear correlation between writing and reading ranks.

c) ∑ ∑ ∑ ∑ ==== 24895,204,5088,36 2 xyxyx

The required equation of regression is .60.4780.327 xy +−=

Solution Seven

a) i) 822.0)199(7

)10(61 =

−−=r

ii) A value of 0.822 indicates a strong positive correlation. Generally, with

an increase in quality, the price goes up. b) The required regression line is .20.6594.63 xy −=

Solution Eight a) Using the formula

[ ] ( )[ ][ ][ ]

8714.0

)864)(672(

664

)128()2156(8)72()732(8

)128(72)1069(8

)(

22

2222

=

−=

−−−=

−−

−=

∑ ∑∑ ∑∑ ∑ ∑

r

xxnyyn

yxxynr

There is a strong negative linear relationship between the number of defectives and the number of weeks worth of experience.

b) The required least squares equation is .988.089.24 xy −= c) 6=x in the model in (c), we have .962.18)6(988.089.24 =−=y

ZICA T3 - Business Mathematics & Statistics

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