The data show the number of patients in a sample of six hospitals who acquired an infection while hospitalized. Find the mean.

Solution W.P(1) :

!

The mean of the number of hospital infections for the six hospitals is 64.83

Mean estimate =!

TheMean

W.P(1) : 3-1 Example (3-2/p-107) : (Hospital Infections) Crude Data

110 76 29 38 105 31* Source: Pennsylvania Health Care Cost Containment Council.

X =X∑n

= 110 + 76 + 29 + 38 +105 + 316

= 3896

= 64.83

W.P(2) : 3-1 Example : Grouped Data (Without Class Interval)

Data (X) Frequency (f) f.x

5 1 56 3 18

7 4 288 8 64

9 3 2710 5 50

Total = 24 192

X =f • X∑n

= 19224

= 8

Page � of �1 10

ROYAL COMMISSION YANBU COLLEGES AND INSTITUTES YANBU INDUSTRIAL COLLEGE

DEPARTMENT OF GENERAL STUDIES SEMESTER I, 1437-38 (H) / 2016-17 (G)

Exercises # Ch3-1 Measures of Central Tendency

Student’s ID: _______________ Student’s Name: ______________________________________________

Course: STAT-201 Section: Date: 19 – 10 – 2016

Instructor’s Name: Mr. Osama Alwusaidi Instructor’s Signature:

Using the frequency distribution for Example 2–7, find the mean. The data represent the number of miles run during one week for a sample of 20 runners.

Solution W.P(3) :

Mean estimate =!

EstimatingtheMean

W.P(3) : 3-1 Example (3-3/p-107) : (Miles Run per Week) Grouped Data (With Class Interval)

Class limits Class boundaries Midpoints (xm) Frequency (f) f . xm

6 - 10 5.5 - 10.5 8 1 8

11 - 15 10.5 - 15.5 13 2 26

16 - 20 15.5 - 20.5 18 3 54

21 - 25 20.5 - 25.5 23 5 115

26 - 30 25.5 - 30.5 28 4 112

31 - 35 30.5 - 35.5 33 3 99

36 - 40 35.5 - 40.5 38 2 76

Total = 20 490

X =f •Xm∑n

=49020

= 24.5 miles

Page � of �2 10

Find the median for the daily vehicle pass charge for five U.S. National Parks. The costs are : $25, $15, $15, $20, and $15.* Source: National Park Service.

Solution W.P(4) :

!

The median cost is : $15

The number of tornadoes that have occurred in the United States over an 8-yearperiod follows. Find the median.

Solution W.P(5) :

!

!

The median number of tornadoes is ……………..

TheMedian

W.P(4) : 3-1 Example (3-5/p-110) : (National Park Vehicle Pass Costs) Ungrouped Data [an odd number of values]

i = n+12

=5+12

=62= 3

$15 $15 $15 $20 $25

Median

W.P(5): 3-1 Example (3-6/p-110) : (Tornadoes in the United States) Ungrouped Data [an even number of values]

684 764 656 702 856 1133 1132 1303* Source: The Universal Almanac.

i = n+12

=8+12

=92= 4.5

MD =764+856

2=16202

= 810

656 684 702 764 856 1132 1133 1303

Median

Page � of �3 10

810

4.5

3

- An approximate median can be found for data that have been grouped into a frequency distribution.- First it is necessary to find the median class. This is the class that contains the median value. That is the

data value. Then it is assumed that the data values are evenly distributed throughout the median class.

- The formula is : �

where n = sum of frequenciescfb = cumulative frequency of class immediately preceding the median classw = width of median classfm = frequency of median classLm = lower boundary of median class

Solution W.P(6) :

- Find the median index : !

For our example: �

cfb = 9w = 1fm = 5Lm = 6.5

!

Anapproximatemedian

3-1 Exercises (42/p-122) : (An approximate median)

MD =n2 − cff

w( )+ Lm

W.P(6) : 3-1 Example (a) : (An approximate median) Grouped Data (Without Class Interval)

Data (x) Frequency (f) c.f

3 1 1

4 2 3

5 2 5

6 4 9

7 5 14

8 3 17

9 2 19

10 1 20

Total = 20

i =f∑ +12

= 20 +12

= 10,5

n = f∑ = 20

MD =n2 − cfbfm

w( )+ Lm =202( )−95

+6.5 = 6.7

Page � of �4 10

Using the frequency distribution for Example 2–7, Find the approximate median.The data represent the number of miles run during one week for a sample of 20 runners.

Solution W.P(7) :

- Find the median index : !

For our example: �

cfb = 6w = 10.5 - 5.5 = 5fm = 5Lm = 20.5

!

W.P(7) : 3-1 Example (b) : (An approximate median) Grouped Data (With Class Interval)

Class limits Class boundaries Frequency (f) cf6 - 10 5.5 - 10.5 1 1

11 - 15 10.5 - 15.5 2 3

16 - 20 15.5 - 20.5 3 6

21 - 25 20.5 - 25.5 5 11

26 - 30 25.5 - 30.5 4 15

31 - 35 30.5 - 35.5 3 18

36 - 40 35.5 - 40.5 2 20

Total = 20

i =f∑ +12

= 20 +12

= 10,5

n = f∑ = 20

MD =n2 − cfbfm

w( )+ Lm =202( )−65

×5⎛

⎝⎜

⎞

⎠⎟+20.5 = 24.5

Page � of �5 10

Find the mode of the signing bonuses of eight NFL players for a specific year.The bonuses in millions of dollars are

Solution W.P(8) : Since $10 million occurred 3 times—a frequency larger than any other numberthe mode is $10 million.

Find the mode for the number of branches that six banks have. 401, 344, 209, 201, 227, 353

Solution W.P(9) :

Since each value occurs only once, there is no mode.

* Note : Do not say that the mode is zero. That would be incorrect, because in some data, such as temperature, zero can be an actual value.

�

The modal class is simply the class interval of highest frequency.

The modal class is = 21 - 25

TheMode

W.P(8) : 3-1 Example (3-9/p-111) : (NFL Signing Bonuses)

18.0 14.0 34.5 10 11.3 10 12.4 10* Source: USA TODAY.

W.P(9) : 3-1 Example (3-10/p-112) : (Branches of Large Banks)

401 344 209 201 227 353

W.P(10) : 3-1 Grouped Data (Without Class Interval)

Mode =33 occurs 6 times

TheModalClass

W.P(11) : 3-1 Grouped Data (With Class Interval)

Modal class = 21 - 25

Class limits Frequency (f)6 - 10 1

11 - 15 216 - 20 321 - 25 526 - 30 431 - 35 336 - 40 2Total = 20

Page � of �6 10

In the last two winter seasons, the city of Brownsville, Minnesota, reported these numbers of water-line breaks per month. Find the midrange.

Solution W.P(12) :

!

Hence, the midrange is 4.5.

* Note : If the data set contains one extremely large value or one extremely small value, a higher or lower midrange value will result and may not be a typical description of the middle.

I. Crude Data

!

II. Grouped Data

!

The numbers 40, 70 and 65 have weights 2, 3 and 4 respectively. What is the weighted mean?

TheMidrange

W.P(12) : 3-1 Example (3-15/p-114) : (Water-Line Breaks)

2 3 6 8 4 1

MR = lowest value + highest value2

=1+8

2= 4.5

WeightedMean

W.P(13) : 3-1 Crude Data

Xw =Xiwi

i=1

n

∑

wii=1

n

∑= X1w1 + X2w2 + ...+ Xnwn

w1 +w2 + ...+wn

Xw =fixi

i=1

k

∑

fii=1

k

∑= f1x1 + f2x2 + ...+ fkxk

f1 + f2 + ...+ fk

(Xi) wi Xi wi

40 2 8070 3 21065 4 260

175 9 550

Xw =Xiwi

i=1

n

∑

wii=1

n

∑=The weighted mean is: 61.11

Page � of �7 10

Harmonic Mean: The harmonic mean is a measure of central tendency. It is the number of values divided by the sum of the reciprocals of each value.

�

W.P(14) : Example1 (p-121): the harmonic mean of 1, 4, 5, and 2 is

�

This mean is useful for finding the average speed. Suppose a person drove 100 miles at 40 miles per hour and returned driving 50 miles per hour. The average miles per hour is not 45 miles per hour, which is found by adding 40 and 50 and dividing by 2. The average is found as shown.

�

W.P(15) : 3-1 Exercises 3-1 (39.b/p-121) Using the harmonic mean, find each of these.- A bus driver drives the 50 miles to West Chester at 40 miles per hour and returns driving 25 miles per hour. Find the average miles per hour.

�

TheHarmonicmean(HM)

3-1 Exercises 3-1 (39/p-121) : (Extending the Concepts)

1ΗM

=1n

1xn( )∑ = 1

n1x1+ 1

x2+ 1

x3+ ...+ 1

xn( ) or ΗM =n

1xn( )∑

=n

1x1+ 1

x2+ 1

x3+ ...+ 1

xn( )

ΗM =4

11 + 1

4 + 15 + 1

2( )= 2.05

trip1 trip2 Total HM

Distance (n) miles 100 100 200 miles

Rate (xi) mph 40 50 _________ = 44.44 mph

Time 2.5 2 4.5 hours

ΗM =n

1x1+ 1

x2( )=

20010040 + 100

50( )=

2004.5

= 44.44 mph

trip1 trip2 Total HM

Distance (n) miles 50 50 100 miles

Rate (xi) mph 40 25 _________ = 30.769 mph

Time 1.25 2 3.25

ΗM =n

1x1+ 1

x2( )=

1005040 + 50

25( )=

1003.25

= 30.769 mph

Page � of �8 10

The geometric mean is a measure of central tendency. It is the nth root of the product of n values. The formula is:

�

W.P(16) : Example1 (p-121): - The geometric mean of 4 and 16 is

�

W.P(17) : Example2 (p-121):- The geometric mean of 1, 3, and 9 is

�

W.P(18) : Example3 (p-121): - The geometric mean is useful in finding the average of percentages, ratios, indexes, or growth rates.

For example, if a person receives a 20% raise after 1 year of service and a 10% raise after the second year of service, the average percentage raise per year is not 15 but 14.89%, as shown.

His salary is 120% at the end of the first year and 110% at the end of the second year.

�

since 114.89% - 100% = 14.89%.This is equivalent to an average of 14.89%

W.P(19) : Exercises 3-1 (40.b/p-121)Find the geometric mean of each of these.- A person received these percentage raises in salary over a 4-year period: 8, 6, 4, and 5%.

�

since 105.7% - 100% = 5.7%This is equivalent to an average of 5.7%

TheGeometricmean(GM)

3-1 Exercises 3-1 (40/p-121) : (Extending the Concepts)

GM = xii=1

n

∏n = x1( ) x2( ) x3( )... xn( )n = x1 ⋅ x2 ⋅ x3 ⋅ ...⋅ xn( )1n = xi

i=1

n

∏⎛

⎝⎜

⎞

⎠⎟

1n⇒ log GM= 1

nn log xi

i=1

n

∑

GM = 4( ) 16( )2 = 64 = 8

GM = 1( ) 3( ) 9( )3 = 273 = 3

GM = 120( ) 110( )2 = 13200 =114.89

or

GM = 1.2( ) 1.1( )2 = 1.32 =1.489

GM = 108( ) 106( ) 104( ) 105( )4 = 105.74 ≈105.7or

GM = 1.08( ) 1.06( ) 1.04( ) 1.05( )4 = 1.0568 ≈1.57

Page � of �9 10

=16

4

8

8

8 x 84 x 16

=91

3

3

3 x 3 x 31 x 3 x 9 3

3

The formula for the Interquartile Range (IQR) is:IQR = Q3 - Q1 ; i = (q/100) x n Q1 = the upper quartile ; i1 = n/4Q3 = the lower quartile ; i3 = [(n/4) x 3]

Calculating Interquartile Range (IQR):22, 24, 36, 21, 25, 30, 20, 25

Step 1: 20, 21, 22, 24, 25, 25, 30, 36

�

The Lower limit is = Q1 – 1.5 (Q3 – Q1) = 21 - 1.5 (25 – 21) = 15

The Upper limit is = Q3 + 1.5 (Q3 – Q1) = 25 + 1.5 (25 – 21) = 31

�

InterquartileRange(IQR)

W.P(20) : 3-2 Example Crude Data

iQ1 =Q1100(n)=

25100(8)=2⇒ Q1 ≈21

iQ2 =Q2100(n)=

50100(8)= 4⇒ Q1 ≈24

iQ3 =Q3100(n)=

75100(8)=6⇒ Q1 ≈25

⇒ IQR=25-21=4ConstructingaBoxandWhiskerPlot

W.P(21) : 3-2 Example : (Crude Data)

*

Interquartile range = 25 – 21 = 4

Median XmaximumXminimum Q1 Q3Q225% 25%25% 25%

21 2524 3120 36Outliers

Page � of �10 10

Mr. Osama A. Alwusaidi | Yanbu University College | GS Dept. | Saudi Arabia | instructor(B.Ed.), Mathematics. (Ed.M.), Testing, Measurement and Statistics (Psychology)

Mobile: +966-544115001 | Phone: +966-43932961 | Cisco:1574 | Fax: +966-43925394Email: alwusaidio@rcyci.edu.sa | Site: www.rcyci.edu.sa | Address: P.O.Box 31387

Yanbu Industrial City 41912 Saudi Arabia