Explain about DIAGRAMMATIC Representation.

Explain about DIAGRAMMATIC Representation.

Bar Diagrams

1) Simple 'Bar diagram’: - It represents only one variable. For example sales, production,

population figures etc. for various years may be shown by simple bar charts. Since these are of the

same width and vary only in heights ( or lengths ), it becomes very easy for readers to study the

relationship. Simple bar diagrams are very popular in practice. A bar chart can be either vertical

or horizontal; vertical bars are more popular.

Illustration:- The following table gives the birth rate per thousand of different countries over a

certain period of time.

Country Birth rate Country Birth rate

India

Germany

U. K.

33

15

20

China

New

Zealand

Sweden

40

30

15

Represent the above data by a suitable diagram.

Comparing the size of bars, you can easily see that China's birth rate is the highest while Germany

and Sweden equal in the lowest positions. Such diagrams are also known as component bar

diagrams.

2) Sub - divided Bar Diagram:- While constructing such a diagram, the various components

in each bar should be kept in the same order. A common and helpful arrangement is that of

presenting each bar in the order of magnitude with the largest component at the bottom and the

smallest at the top. The components are shown with different shades or colors with a proper index.

Illustration:- During 1968 - 71, the number of students in University ' X ' are as follows. Represent

the data by a similar diagram.

Year Arts Science Law Total

1968 -

69

1969 -

70

1970 -

71

20,000

26,000

31,000

10,000

9,000

9,500

5,000

7,000

7,500

35,000

42,000

48,000

3) Multiple Bar Diagram:- This method can be used for data which is made up of two or

more components. In this method the components are shown as separate adjoining bars. The height

of each bar represents the actual value of the component. The components are shown by different

shades or colors. Where changes in actual values of component figures only are required, multiple

bar charts are used.

Illustration:- The table below gives data relating to the exports and imports of a certain country

X ( in thousands of dollars ) during the four years ending in 1930 - 31.

Year Export Import

1927 - 28 319 250

1928 - 29 339 263

1929 - 30 345 258

1930 - 31 308 206

Represent the data by a suitable diagram

4) Deviation Bar Charts:- Deviation bars are used to represent net quantities - excess or

deficit i.e. net profit, net loss, net exports or imports, swings in voting etc. Such bars have both

positive and negative values. Positive values lie above the base line and negative values lie below

it.

Illustration:-

Years Sales Net

profits

1985 - 86

1986 - 87

1987 - 88

10%

14%

12%

50%

-20

-10%

resent the above data by a suitable diagram showing the sales and net profits of

private industrial companies.

Pie Chart:

i) Geometrically it can be seen that the area of a sector of a circle taken radially, is proportional to

the angle at its center. It is therefore sufficient to draw angles at the center, proportional to the

original figures. This will make the areas of the sector proportional to the basic figures.

For example, let the total be 1000 and one of the component be 200, then the angle will be

In general, angle of sector at the center corresponding to a component

.

Tuition fees $ 6000

Books and lab. $ 2000

Clothes / cleaning $ 2000

Room and boarding $ 12000

Transportation $ 3000

Insurance $ 1000

Sundry expenses $ 4000

Total expenditure = $ 30000

Now as explained above, we calculate the angles corresponding to various items (components).

Tuition fees =

Book and lab =

Clothes / cleaning =

Room and boarding =

Transportation =

Insurance =

Sundry expenses =

Uses: - A pie diagram is useful when we want to show relative positions(proportions ) of the

figures which make the total. It is also useful when the components are many in number.

Explain about Graphical Representation with example.

A graph is a visual representation of data by a continuous curve on a squared ( graph ) paper. Like

diagrams, graphs are also attractive, and eye-catching, giving a bird's eye-view of data and

revealing their inner pattern.

Graphs of Frequency Distributions: -

The methods used to represent a grouped data are :-

1.Histogram

2.FrequencyPolygon

3.FrequencyCurve

4.Ogive or Cumulative Frequency Curve

1. Histogram :- It is defined as a pictorial representation of a grouped frequency

distribution by means of adjacent rectangles, whose areas are proportional to the

frequencies.

To construct a Histogram, the class intervals are plotted along the x-axis and corresponding

frequencies are plotted along the y - axis. The rectangles are constructed such that the

height of each rectangle is proportional to the frequency of the that class and width is equal

to the length of the class. If all the classes have equal width, then all the rectangles stand

on the equal width. In case of classes having unequal widths, rectangles too stand on

unequal widths (bases). For open-classes, Histogram is constructed after making certain

assumptions. As the rectangles are adjacent leaving no gaps, the class-intervals become of

the inclusive type, adjustment is necessary for end points only.

For example, in a book sale, you want to determine which books were most popular, the high-

priced books, the low-priced books, books most neglected etc. Let us say you sold a total 31 books

at this book-fair at the following prices.

$ ....2, $ 1, $ 2, $ 2, $ 3, $ 5, $ 6, $ 17, $ 17, $ 7, $ 15, $ 7, $ 7, $ 18, $ 8, $ 10, $ 10, $ 9, $ 13, $

11, $ 12, $ 12, $ 12, $ 14, $ 16, $ 18, $ 20, $ 24, $ 21, $ 22, $ 25.

The books are ranging from $1 to $25. Divide this range into number of groups, class intervals.

Typically, there should not be fewer than 5 and more than 20 class-intervals are best for a

frequency Histogram.

Our first class-interval includes the lowest price of the data and, the last-interval of course includes,

the highest price. Also make sure that overlapping is avoided, so that, no one price falls into two

class-intervals. For example, you have class intervals as 0-5, 5-10, 10-15 and so on, then the price

$10 falls in both 5-10 and 10-15. Instead if we use $1 - $5, $6=$10, the class-intervals will be

mutually exclusive.

Therefore, now we have distribution of books at a book-fair

Class-interval Frequency

$ 1- $ 5

$6 - $10

$11 - $15

$16 - $20

$21 - $25

6

8

10

3

4

Total n = fi = 31

Note that each class-interval is of equal width i.e. $5 inclusive. Now we draw the frequency

Histogram as under.

Relative Frequency Histogram:- It uses the same data. The only difference is that it

compares each class-interval with the total number of items i.e. instead of the frequency of each

class-interval, their relative frequencies are used. Naturally the vertical axis (i.e. y-axis) uses the

relative frequencies in places of frequencies.

In the above case we have,

Class-interval Frequency Relative

frequency

$ 1 - $ 5

$ 6 - $10

$11 - $15

$16 - $20

$21 - $25

6

8

10

3

4

6/31

8/31

10/31

3/31

4/31

The Histogram is same as in above case.

Construction of Histogram when class-intervals are unequal: - In a Histogram, a

rectangle is proportional to the frequency of the concern class-interval. Naturally, if the class-

intervals are of unequal widths, we have to adjust the heights of the rectangle accordingly. We

know that the area of a rectangle = l. h. Now suppose the width ( l ) of a class is double that of a

normal class interval, its height and thus the corresponding frequency must be halved. After this

precaution has been taken, the construction of the Histogram of classes of unequal intervals is the

same as before.

Illustration: - Represent the following data by means of Histogram.

Classes : 11-14 16-19 21-24 26-29 31-39 41-59 61-79

Frequencies : 7 19 27 15 12 12 8

Solution: Note that class-intervals are unequal and also they

are of inclusive type.

We have to make them equal and of the exclusive

type.

Correct factor = ( 16 - 14 ) / 2 = 1. Using it we

have

Classes : 10-15 15-20 20-25 25-30 30-40 40-60 60-80

Frequencies : 7 19 27 15 12 12 8

Adjusted Heights : 7 19 27 15 12/2 12/4 12/4

(Frequencies) = 6 = 3 = 3

2) Frequency Polygon:- Here the frequencies are plotted against the mid-points of the class-

intervals and the points thus obtained are joined by line segments.

Example Height in cm. 150 - 154 154 - 158 158 - 162 162 - 166 166 - 170

No. of children 10 15 20 12 8

The polygon is closed at the base by extending it on both its sides ( ends ) to the midpoints of two

hypothetical classes, at the extremes of the distribution, with zero frequencies.

On comparing the Histogram and a frequency polygon, you will notice that, in frequency polygons

the points replace the bars ( rectangles ). Also, when several distributions are to be compared on

the same graph paper, frequency polygons are better than Histograms.

3) Frequency Distribution (Curve):- Frequency distribution curves are like frequency

polygons. In frequency distribution, instead of using straight line segments, a smooth curve is used

to connect the points. The frequency curve for the above data is shown as:

4) Ogives or Cumulative Frequency Curves:- When frequencies are added, they are

called cumulative frequencies. The curve obtained by plotting cumulating frequencies is called

cumulative frequency curve or an ogive ( pronounced ogive ).

To construct an Ogive:-

1) Add up the progressive totals of frequencies, class by class, to get the cumulative frequencies.

2) Plot classes on the horizontal ( x-axis ) and cumulative frequencies on the vertical

( y-axis).

3) Join the points by a smooth curve. Note that Ogives start at (i) zero on the vertical axis, and

(ii) outside class limit of the last class. In most of the cases it looks like 'S'.

Note that cumulative frequencies are plotted against the 'limits' of the classes to which they refer.

(A) Less than Ogive:- To plot a less than ogive, the data is arranged in ascending order of

magnitude and the frequencies are cumulated starting from the top. It starts from zero on the y-

axis and the lower limit of the lowest class interval on the x-axis.

(B) Greater than Ogive:- To plot this ogive, the data are arranged in the ascending order of

magnitude and frequencies are cumulated from the bottom. This curve ends at zero on the the y-

axis and the upper limit of the highest class interval on the x-axis.

Illustrations:- On a graph paper, draw the two ogives for the data given below of

the I.Q. of 160 students.

Class -intervals :60 - 70 70 – 80 80 – 90 90 - 100 100 - 110

No. of students : 2 7 12 28 42

Class -intervals 110 - 120 120 - 130 130 – 140 140 - 150 150 - 160

No. of students 36 18 10 4 1

Uses :- Certain values like median, quartiles, deciles, quartile deviation, coefficient of skewness

etc. can be located using ogives. it can be used to find the percentage of items having values less

than or greater than certain value. Ogives are helpful in the comparison of the two distributions.

Measures of central tendency:

A measure of central tendency gives a single representative value for a set of usually

unequal values. The single value is the point of location around which the individual values of the

set cluster. the measures of central tendency are hence known as 'measures of location'. they are

called averages. Various measures of central tendency are the following:

1. Arithmetic mean

2. Median

3. Mode

4. Geometric Mean

5. Harmonic Mean

Arithmetic mean

Series Direct method Short cut method Step deviation

method

Individual series �̅� =

∑𝑋

𝑁 �̅� = 𝐴 +

∑𝑑

𝑁

d = X-A

�̅� = 𝐴 +𝑐 ∑𝑑′

𝑁

d' = 𝑋−𝐴

𝑐

Discrete series �̅� =

∑𝑓𝑋

𝑁 �̅� = 𝐴 +

∑𝑓𝑑

𝑁

d = X-A

�̅� = 𝐴 +𝑐∑𝑓𝑑′

𝑁

d' = 𝑋−𝐴

𝑐

Continuous series �̅� =

∑𝑓𝑚

𝑁 �̅� = 𝐴 +

∑𝑓𝑑

𝑁

d= m-A

�̅� = 𝐴 +𝑐∑𝑓𝑑′

𝑁

d' = 𝑚−𝐴

𝑐

Median

Series position value

Individual series 𝑁 + 1

2 -

Discrete series 𝑁 + 1

2 -

Continuous series 𝑁

2 M = L+

𝑖(𝑁

2−𝑐𝑓)

𝑓

f-frequency

cf- cumulative frequency

Mode

Series position value

Individual series Highest frequency -

Discrete series Highest frequency -

Continuous series Highest frequency Z = L+ 𝑖(𝑓1−𝑓0)

2𝑓1−𝑓0−𝑓2

Geometric Mean

Series Geometric Mean

Individual series Antilog(∑ log𝑋

𝑁)

Discrete series Antilog(∑𝑓 log𝑋

𝑁)

Continuous series Antilog(∑𝑓 log𝑚

𝑁)

Harmonic Mean

Series Harmonic Mean

Individual series 𝑁

∑(1𝑋)

Discrete series 𝑁

∑(𝑓𝑋)

Continuous series 𝑁

∑(𝑓𝑚)

WEASURES RES OF CENTRAL VALUE

aleulation

,lon of Simple Arithmetic MeanIndividual Observations

ncess of computing mean in case of individual observations (Le

181

The frequencies are not given) is very simple. Add together the various phere

of the variable and divide the total by the number of items. alues

Symbolicallp

X1+g+Xg*tX or = or -N N

Here X = Arithmetic Mean, X = Sum of all the values of the variable

ie. X1, Xa, A3...Ån : N = Number of observations. Steps. The formula involves two steps in calculating mean:

Add together all the values of the variable X and obtain the total,

Le., ZX.

i Divide this total by the number of the observations., i.e., N.

lustration 1. The following table gives the monthly income of 10 employees in an office:

h:ome (Rs.) 1,780 1,760 1,690 1,750 1,840 1,920 1,100 1,810 1,050 1,950

Calculate the arithmetic mean of incomes. Let income be denoted by the symbol X.

Solution.

CALCULATION OF ARITHMETIC MEAN

Monthly Income (Rs.) 1,920

Employee Employee Monthly Income

(Rs.) 1,780 1,760 ,690

1, ,100 ,810

1,050 1,950

EX = 16,650

2

3 4 1750

10 1,840 N= 10

X = EX = 16,650, N= 10

= 16,65 = 1,665. 10

Hence ne average income is Rs. 1,665.

Short-cut method. The arithmetic mean can be calculated by using 1s known as an arbftrary m as an arbitrary orlgin. When deviations are taken from an

Origin, the formula for calculating arithmetic mean is

X=A+ N

metiEuc mean of a sample is designated by the symbol which X is read X- and the arithmetic Sman of a population is designated by the Greek Letter u pronounced as 'mu.

Tmathema is the letter capital sigma of the Greek alphabet and is used in

to denote the sum of the values. ymbol E

NETHOTSR STATSTICAL MET 182

where A is the assumed mean and d is the deviation of iten assumed mean, ie., d = (X - A).

Steps.

items iron

(1) Take an assumed mean.

(2) Take the deviations of items from the assumed mean and .

these deviations by d.

(3) Obtain the sum of these deviations, i.e., Zd.

denor

(4) Apply the formula : X=A+

From Ilustration 1 calculate arithmetic mean by taking 1,800 as t N

assumed mean.

Solution.


(X-1800) -20

Employee Income

1,780 2 1,760 -40

1,690

1,750 3 -110

4 -50

5 1,840 +40

1,920 +120

7 1,100 -700

8 1,810 10

1,050 -750 10 1,950 +150

N= 10 2d =- 1350

= A N

A 1800, 2d= - 1350, N = 10

X = 180o - 1,350 10

= 1,800 135 1,665.

Hence the average income is Rs. 1,665. Note. The reader will find that the calculations here are more than s

we used the formula

We had

N This is true for ungrouped data. But for grouped data consideraC possible by adoping the short-cut method.

saving in

Calculation of Arlthmetic Mean-Discrete Serles

In discrete serles arlthmetic mean may be computed by app (1) Direct method, or

(il) Short-cut method.

ing

Any value whether existing in the data or not can be taken as esU and the final anSwer would be the same. However, the nearer the to the actual mean, ue lesser are the calculations.

Ie asstumed DSUmed e

183 MEASURES OF CENTRAL VALUE

Direct Method

rhe formula for computing mean is

N

where, f= Frequency: X = The variable in question; N* = Total number of observations, ie., f. Steps: ( Multiply the frequency of each row with the variable and obtain the total y.

(ti) Divide the total obtained by step () by the number of observations, Le., total frequency.

lustration 2. From the following data of the marks obtained by 60 students of a class, calculate the arithmetic mean:

Marks No. of Students Marks No. of Students 20 8 50 10 30 12 60 6 40 20 70

Let the marks be denoted by X and the number of students by Solution.


Marks No. of students fX X

8 160 20

12 360 30

0 800 40

10 500 50

360 60

4 280 70

N = 60 fX = 2,460

-22460 60

41

Short-cut Method According to this method.

N Henc the average marks 41.

=A+ 2 A = Assumed mean; d = (X - A}; N = Total number of observations.

e., 2f. where A =

total der should note carefully that n diserete and continuous frequency distributions the

number Der of observatlons, Le. N= the sum ot Irequeincy or N= X

STATSTICAL METH 184

Steps:

(i Take the deviations of the variable X from the assumed

denote the deviations by d.

(i) Multiply these deviations with the respective frequency and tal

total fd

(iw) Divide the total obtained in third step by the total frequencu

(9 Take an assumed mean.

Tmean and e te

lustration 3. Calculate arithmetic mean by the short-cut method using fr distribution of illustration 2.

frequecy

Solution.


fd No. of Students f

(X 40) Marks X

20 8 -20 -160

30 12 -10 -120

40 20 0

50 10 +10 +100

60 6 +20 +120

70 +30 +120

N 60 2fd = 60

60 X= A+= 40 + 40+1 = 41 60

Calculation of Arlthmetic Mean-Continuous Series

a In continuous series, arithmetic mean may be computed by appy of the following methods: (0 Direct method, (0 Short-cut method.

Direct Method When direct method is used

N where m = mid-point of varlous classes; f = the frequeney N the total frequency. ency

of each e

the reader should be famillar with the terms 'groupe erles

ereas

USced

the

data also. Ungrouped dala refer to the Individual observations W*

tta' and

Note. For the sake ol clarity and ease of understanding we naghoul individua observatlons, discrete serles and continuous series

the

However, foupe.

refer to coninuous SCies and the discrete serles.

Mid polnt=Ower limit + Upper limit 2

185 MEASU

OF CENTRAL VALUE

Steps:

Obtain the mid-point of each class and denote it by m. these mid-points by the respective frequency of each class

tMultiply

and obtain the total Yfm.

aDivide the total obtained in step (0 by the sum of the frequency, ie., N.

lustratio ehration 4. From the following data compute arithmetic mean by direct method:

0-10 10-20 20-30 30-40 40-50 50-60 Marks

10 5 10 25 30 20 No. of students

Solution.

CALCULATION OF ARITHMETIC MEAN BY DIRECT METHOD

Marks Mid-point No. of Students tm m

5 25 0-10 5

15 10 150 10-20

25 625 20-30 25

35 30 1,050 30-40

20 900 40-50 45

10 550 50-60 55

N 100 fm = 3,300

Xim 3,300 33. N 100

omput by applying the following formula: Short-cut Method When short-cut method is used, arithmetiç mean is

K=A+ N

assumed mean; d= deviatlons of mid-points from assumed Stee, (m - A); N = total number of observatlons.

Steps: Take an assumed mean.

ig Tom the mid-point of each class deduct the assumed mean. ti Multtply Pro

any the respective frequencles of each class by these devlations and obtain the total 2jd.

Apply the formula: X= A+ fd N Calct

ationarithmetic mean by the short-cut method from the data of

4.

MEASURES

OF CENTRAL VALUE

ulation of Median-Individual Observations

197

Sreps

Arrange the data in ascending or descending order of magnitude. (Both arrangements would give the same answer.)

h In a group composed of an odd number of values such as 7, add l1 to the total number of values and divide by 2. Thus, 7 + 1 would be 8 which divided by 2 gives 4-the number of the value starting

at either end of the numerically arranged groups will be the median value. In a large group the same method may be followed. In a group of 199 items the middle value would be 100th value.

199+ This would be determined by In the form of formula 2

Med. Size of h item.

lustration 11. From the following data of the wages of 7 workers compute the median

wage

1200 1160 1400 Wages (in Rs.) 1100 1150 1080 1120

Solution:

CALCULATION OF MEDIAN

Wages arranged inn ascending order

SI. No. Wages arranged in ascending order

SI. No.

1160 1080

6 1200 1100 7 1400 3 1120

1150

Size ofth item = = 4th tem. of 4th item =

150 S find that median is the middlemost item : 3 persons get a wage less than Rs. We t qual number, i.e., 3, get more than Rs. 1150.

1150. Hence the median wage = Rs. 1150

The Iterocedure for determining the median of an even-numbered group

ues1s not as obvious as above. If there were, for instance, different th a and group, the median is really not determinable since both the

bth values are in the centre. In practice, the median value for a hetposed of an even number of items is estimated by fînding the

YOup arthme he mean of the two middle values-that is, adding the two values ontdle and dividing by two. Expressed in the form of formula, it nOunts to

Median = Size of th item

The bbreviation Med. represents median.

TSTICAL ME 198

Thus we find that it is both when N is odd as well

(one) has to be added to determine median value. as even tha

Illustration 12. Obtain the value of median from the following data:

777 753 2.488 1 591 407 672 522 391 384

Solution.


Data arranged in ascending order

SI. No. Data arranged in ascending order

X

SI. No.

X

384 6 672

2 391 7 753

777 405

522 9 1,490

5 591 10 2,488

Median Size of th item == 5.5th item

5th item + 6th item 591+672 1,263 631.5 Size of 5.5th item = 2 2 2

Computation of Median-Discrete Series

Steps (0 Arrange the data in ascending or descending order of mag (ii Find out the cumulative frequencies.

gnitude

Apply the formula: Median = Size of 2 (iu) Now look at the cumulative frequency column and na that

which is either equal to or next higher to that and de N+1 2

detern

the value of the variable corresponding to it. That give the

of median.

llustration 13. From the following data find the value of median: Income (Rs.) 1000 1500 800 2000 2500 1800

No. of persons 24 26 16 30 20 6

CENTRAL VALUE 199 MEASU OF

Solution.


No. of persons C.f. C.f Income arranged in ascending

order

No. of persons f

ncome

aranged in

ascending

order

800 16 16 1800 30 96

1000 24 40 2000 20 116

1500 26 66 2500 6 122

Median Size of th item =

2 122 +1 61.5th item. 2

Size of 61.5th item = 1500.

Calculation of Median-Continuous Series

Steps. Determine the particular class in which the value of median lies.

Use N/2 as the rank of the median and not (N + 1)/2. Some writers have

suggested that while calculating median in continuous series l should be

added to total frequency if. it is odd (say 99) and should not be added if

it is even figure individual and discrete series because specific items and individual values

are involved. In a continuous frequency distribution all the frequencies

ose their individuality. The effort now is not to fînd the value of one

Specific item but to find a particular point on a curve-that one value Which will have 50 per cent of frequencies on one side of it and 50 per

Cent of the frequencies on the other. It will be wrong to use the above

rule. Hence it is N/2 which will divide the area of curve into two equal

parts and as such we should use N/2 instead of (N + 1)/2, in continuous

Ees. After ascertaining the class in which median lies, the following

nula is used for determining the exact value of median.

(say, 100). However, 1 is to be added in case of

Median = L+ N/2-Cxi

Lower 1limit of the median class, Le., the class in which the

middle item of the distribution lies. L

cf. = Cumulative frequency of the class preceding the median class

the of the frequencies of all classes lower than

or sum

median class.

JSimple frequency of the median class.

It Should be remember

i= The class interval of the medlan class.

ered that while lnterpolating the median value in a

thatcy distribution it 1s assumed that the variable is continuous and

ere is an orderly and even distributlon of iten1s within each class. quen

STATSTIC CAL METHOT 200

Illustration 14. Calculate the median for the following frequency distribution:

No. of Students Marks No. of Students Marks

45-50 10 20-25 31 40-45 15 15-20 24 35-40 26 10-15 15

30-35 30 5-10

25-30 42

[B.Com., Madras Univ, 1

Solution. First arrange the data in ascending order and then find out median.


C.f. Marks cf. Marks

30-35 30 149 5-10

15 22 35-40 26 175 10-15

24 46 40-45 15 190 15-20

10 200 20-25 31 77 45-50

25-30 42 119

Med. = Size ofitem 200 100th item

Median lies in the class 25-30

Med. = L+2C ki

L = 25, N2 = 100, c.f. = 77, f = 42, i = 5

5+ 5 25+2.74 27.74. 42

Ilustration 15. Calculate the median from the following data:

No. of Weight (in gms.)

No. of Weight (in gms.) Apples Apples

45 410-419 14 450-459 420-429 20 18 460-469 430-439 42 470-479 440-449 54

B.Com, A.P. Unv

he exclus

ion. Since we are given inclusive class intervals, we should convert it to educting 0.5 from the lower limits and adding 0.5 to the upper limits.

201 CENTRAL ALUE

EASURES OF

f C.f. Weight 14 14 409.5-419.5

20 34 419.5-429.5

42 76 429.5-439.5

54 130 439.5-449.5

45 175 449.5-459.5

18 193 459.5-469.5

7 200 469.5-479.5

N 200

Med.Size of th item = 200

= 100th item 2

Median lies in the class 439.5 449.5

Med = L+2- C.f xi

L 439.5, N/2 = 100, c.f. = 76, f = 54, i = 10

Med. 439.5 +0076 x 10 439.5 + 4.44 443.94. 54

lustration 16. From the following data calculate median:

Marks Less than 5

Marks Less than 5

No. of Students

No. of Students

29 30 644 10 224 "35 650

" 15

465 "40 653 "20 582 45 655 "25 634

(B:Com., Madras Univ., 1998) Solution. Since we are given cumulative frequencies; first find simple frequencies and then aculate median.

Marks No. of Students C.f

0-5 29 29 5-1 195 224 10-15 241 465 15-2 117 582 20-25 52 634 25-30 10 644 30-35 6

650 35-40 3 653 45 2 655

Med. = Size ofth item = 655 327.5th item 2

CENTRAL VAL 213 EASURES OF

Solution.

CALCULATION OF MODE

Number of times it Size of item Number of times it Size of item

OCCurs OCCurs

20 10 24

12 27 3

15 30 1

18 Total 10

Since the item 27 occurs the maximum number of times, i.e., 3, hence the modal marks

27.

Note. Thus the process of determining mode in case of individual observations

essentially involves grouping of data.

values having the same maximum

When there are

frequency, one cannot say which is the modal value and hence mode is

said to be ill-defined. Such a series is also known as bimodal or

multimodal. For example, observe the following data of income:

two or more

140 130 120 130 140 110 120 130 120 110

Income (in Rs.) Let us find out how many times each value occurs:

120 130 140 110 Size of item 3 2

2 3 No. of times it occurs

nce 120 and 130 have the same maximum frequency, ie., 3, mode is

-defined in this case.

alelation of Mode-Discrete Series

iCrete series quite often mode can be determined just by inspection,

to that value of the variable around which n disc Le, by looking most heavil

h the items are

Concentrated. For example, observe the following data

29 31 32 33

30 ze of garment No. of persons wearing

28 40 65 50 15 20

above data we can clearly say that the nodal size is 31

Lust by inspection, an error of

10

From the because 5. H he value 31 has occurred the maximum number of times, i.e.,

owever, w naximu very small and where thé mode is

determined

and the items are heavily concentrated on elther side. In such

irable to prepare a grouping table and an analysis table.

us in ascertaining the modal class.

Irequency and the frequency precediing it or succeeding it is

help

S in or put in a circle; in column 2 frequencies are grouped in

o's umn 3 leave the first frequency and then group the remaining

wo' s Mark g table has six columns. In column I the maximum frequency

lirst column 4 group the frequencies in three's; i column 5 leave

the e the quency and group

the frequencies in three's; and in column 65

rst two frequencies and then group the remaiing

in three's. In

eave

STATISTICAL ME 216

the use of non-homogeneous data. In instances where a do

bimodal and nothing can be done to change it, th

used as a measure of central tendency.

Where mode is ill-defined, its value may be ascertained by the

formula based upon the relationship between mean, median and

rbuton the mode shou not

Mode = 3 Median - 2 Mean

This measure is called the empirical mode.

llustration 25. Calculate mode from the following data:

No. of students Marks No. of studenta Marks

80 Avove 600 28 Above 0

77 Above 70 18 Above 10



Above 40 55 Above 100 0

Above 50 43

(B. Com., Andhra Univ., Nagarjun Univ., Kerala Univ, 19

Soluton. Since this is cumulative frequency distribution, we first convert it ito a Si

frequency distribution.

Marks No. of students Marks No. of students

0-10 3 50-60 15

10-20 60-70 12

20-30 70-80

30-40 10 80-90 2

40-50 12 90-100 8

By inspection the modal class is 50-60.

Mo L+ x i A1+A2

50; A, = l15-121=3, Az = l15- 121 3;i= 10

Mo=50+ x10 = 50 +5 55. 3+3

lluetration 26. Find the value of mode from the data given below: Welght (kg) No, students

Welght (kg No. of studeni

-

93-97 2 14 113-117

98-102 118-122

103-107 12 3 123-127

108-112 17 128-132

(B.A. Hons. Econ., Do

, Dolhl U

STATISTIO METHON 224

1.1761 250 15 2.3979

1.8751 40 75 1.6021

2.6990 36 1.5563

log X = 17.6373 500

G.M. = A.L.| A (17.6373= AL. (1.7637) 58.03 10

Illustration 31. Calculate geometric mean from the following data:

0.08 125 1462 38 0.22 12.75 0.5 (B.Com., Mysore Univ, 1968 Solution.

CALCULATION OF G.M.

X log X

125 2.0969

1462 3.1650 38 1.5798

0.8451 0.22 1.3424 0.08 2.9031

12.75 1.1055 0.5 1.6990

log X= 10.7368

(10.7368 G.M. = A.L. 2109X

= A.L. (1.8421) = 6.952.

Calculation of Geometric Mean-Discrete Series

G.M. = Antilog Ef log X N

Steps: (0 Find the logarithms of the variable X. () Multiply these logarithnms with the respective frequencies and the total f log X.

obtain

(tit) Divide f log X by the total frequency and take the antilog value so obtained. he

G.M. = A.L, log X N

Proof

Let x1 X2 ¥n be data having frequencies fi. .. . fa respectivey Then G.M. = S S2

X1 X2

MEAS .ctRES OF CENTRAL VALUE

225

Log Jn Log G.M. =

X1

Logxh+Log h.. . +Log x Ni Log x1 +Ja Log *2 +.. .+fn Log

2Si Log x= N i= 1

T.2f log X G.M. A.L. N Hence

Calculation of Geometric Mean-Continuous Seriees

G.M. Antilog/log X N

Steps: (0 Find out the mid-points of the classes and take their logarithms. ) Multiply these logarithms with the respective frequencies of each

class and obtain the total f log m.

(i Divide the total obtained in step (1i) by the total frequency and take the antilog of the value so obtained.

llustration 32. Find the Geometric mean for the data given below:

Marks Frequency Marks Frequency 4-8 6 24-28 12 8-12 10 28-32 10 12-16 18 32-36 16-20 30 36-40 2

20-24 15

(B.A., Madurai-Kamaraj Univ.) Solution.

CALCULATION OF GEOMETRIc MEAN

log m fx log m Marks

m.p. m 4-8

0.7782 4.6692 10.0000

20.6298

8-12 10 10 1.0000 12-16 16-20

14 18 1.1461 18 30 1.2553 37.6590 20-24 22 15 1.3424 20.1360 24-28 26 12 1.4150 16.9800 28-32 10 1.4771 14.7710 30 32-36

1.5315 9.1890 34 36-40 3.1596

Ef x log m = 137.1936

38 2 1.5798 N= 109

STATISTICAL METHOn ODS

226

G.M.= A.L 20g AL (137.1936A.L. 1.2587 18.14. 109

Uses of Geometric Mean

Geometric mean is specially useful in the following cases:

increase ser sales, production, population or other economic or busines

For example, from 1996 to 1998 prices increased by 5, 10 e .The geometric mean is used to lind the average per cent incre.

per cent respectively. The average annual increase is not 11 Der.. Cent

5+10+ 18- 11| as given by the arithmetic average but 10.9 per 3

cent as obtained'by the geometric mean.

.Geometric mean is theoretically considered t be the best average the construction of index numbers. t satisties the time reversal test

and gives equal weight to equal ratio of change.

This average is most suitable when large weights have to be given t small items and small weights to large items, situations which we usually come across in social and economic fields.

The following examples illustrate the use of geometric mean

n

llustration 33. The price of a commodity increased by 5% from 1995 to 1996, 8% from 1996 to 1997 and 77% from 1997 to 1998. The average increase from 1996 to 1998 is quoted as 26% and not 30%. Explain and verify the result.

Solution. The appropriate average here is the geometric mean and not the aithmete mean. The arithmetic mean of 5, 8, 77 is 30 but this is not the correct answer. Corect answer shall be obtained if we calculate geometric mean.

% Rise2 X Price at the end of the year

taking preceding year as 100

logX

5 105 2.0212 8 108 2.0334

77 177 2.2480

2 log X = 6.3026

G.M. A.LogX=AL = A.L. (2.1009) 126.2

The average increase from 1996 to 1998 = 126.2 100 = 26.2% or approx. 2070. Verification. When the average rise is 30%. Year

Rate of change Total change Price at the end

of each year

130.0 year 30% on 100

30 year 30% on 130

39 169.0 year 30% on 169

When the average rise is 26%: 50.7 219.7

CENTRAL VALUE 233 4SURES OF

N H.M. = In discrete seriesS,

N Ef/m)

N H.M. = In continuous series,

colculation of Harmonic Mean--Individual Observatlons

individual serles harmonic mean is computed by applylng the following

formula:

N H.M. =

X, Xo, Xg. etc. refer to the various items of the variable.

lustration 41. (a) Find the harmonic mean from the following:

2574 475 75 5 0.8 0.08 0.005 0.0009.

(B.Com., GND Univ; B.Com., Calicut, 1994) Solution.

CALCULATION OF HARMONIC MEAN

(1/X) X (1/) 2574 0.0004 0.8 1.2500

475 0.0021 0.08 12.5000

75 0.0133 0.005 200.0000

5 0.2000 0.0009 1111.1111

z(1/X) = 1325.0769

N H.M. 5(1/X) 1325.0769 = 0.006.

(b) Calculate the harmonic mean from the following data:

3834 382 63 8 0.4 0.03 0.009 0.005

Solution. (B.Com., Mysore Univ., 1995)

CALCULATION OF HARMONIG MEAN

X (1/X) X (1/X) 3834 0.0003 0.4 2.5000 382 0.0027 0.03 33.3333

63 0.0159 0.009 111.1111 8 0.1250 0.0005 2000.0000

(1/)=2147.0883 8 N

(1/X) 2147.0883 0.003726. H.M.

STATISTICAL METHOD 234

Calculation of Harmonic Mean-Discrete Series

In discrete series, harmonic mean is computed by applying tho

formula e folowing

N N H.M. = 1f/0

fx

Steps: Take the reciprocal of the various items of the variable X.

(0 Multiply the reciprocal by frequencies and obtain the total zf

(ii Substitute the values of N and 2fx in the above formula.

Note. Instead of first finding out the reciprocals and then multiplvina them by frequencies it will be far more easier to divide each frequency b the respective value of the variable.

Mlustration 42. From the following data compute the value of harmonic mean

Marks 10 20 25 40 50 No. of students 20 30 50 15 5

(B.Com., Bangalore, 1990; B.Com., Madras, 199 Solution.


Marks X (IX) 10 20 2.000 20 30 1.500

25 50 2.000 40 15 0.375 50

5 0.100

N = 120 2 (f/X) = 5.975

N 2(/X) 5.975 20.08. H.M. = 120

Calculation of Harmonic Mean--Continuous Series

For calculating harmonic mean in continuous series the procchere same as applied to discrete serles. The only difference 1s

take the reciprocal of the mid-points.

cedure is the here we

llustration 43. From the following data compute the value of harmonic me Class interval

50-60

10-20 20-30 30-40 40-50

3 Frequency 6 10 7

235 SURES OF CENTRAL VALUE

Solution.


f/m Mid-points Frequency f

Class interval

m

0.267 15 4 10-20

0.240 25 20-30

0.286 35 10 30-40

0.156 45 40-50

3 0.055 55 50-60

N = 30 Z(fm) = 1.004

H.M. 2(f/m)1.00429.88. N 30 29.88.

Uses of Harmonic Mean'

The harmonic mean is restricted in its field of usefulness. It is useful for computing the average rate of increase in profits of a concern or average speed at which a journey has been performed or the average price at which an article has been sold. The rate usually indicates the relation Detween two different types of measuring units that can be expressed reciprocally. For example if a man walked 20 km. in 5 hours the rate of is walking speed be expressed as

20 km. 4 km. per hour

5 hours

re the unit of the first term is a km. and the unit of the second term san hour. Or reciprocally,

5 km. hours per km. 20 hours

T ne unit of the first term is an hour and the unit of the secondH Term is km. where

VGraeOn 44. An automobile driver travels from plain to hill station 100 km. distance at an speed of

Mustrati Kn. per ur. What is

30 km. per hour. He then makes the return trip at average speed of 20 his average speed over the entire distance (200 km.)?

Solution. an of the problem is given to a layman he is most likely to compute the arithmetic *anl of two speeds, 1.e, 30 km. + 20 Kl = 25 km. p.h. X =

2 But this is not ation. Ha not the correct average. Hamonic mean would be more suitable in this *narmonic mean of 30 and 20 IS

2x 120 10

2 2 24 km. p.h. 1 10 20 30

H.M. =

120

onic mean is a measure of central terndeney lor data expressed as rates Kms. per hour, kms. per itre, hours per semester tonnes per month ete.

Th harmonic mean Such

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