Worked Out Examples

Example 1

If n = 50, x2 = 600, x = 150, Mo= 1.75, find the Pearsonian coefficient of skewness.

Solution:

Here, n = 50, x2 = 600, x = 150, Mo= 1.75

X=∑ x

n=

15050

=3

σ 2=∑ x2

n−(∑ x

n )2

=60050

−(15050 )

2

=12−9=3

σ =1. 732

Now, Sk(P) =

X−M o

σ=3−1 .75

1 .732=0.72

Example 2A frequency distribution gives the results: mean = 40, Mode = 39 and coefficient of skewness = 0.5. Find the standard deviation and coefficient of variation.Solution:

Here, Mean (X ) = 40, Mode (Mo) = 39 and Sk = 0.5

Sk(P)=

X−M o

σ

0.5 =

40−39σ

σ = 2

CV=

σX

×100 %= 240

×100 %=5 %

Example 3

From the following information of a factory relating to the wages, find as much information as you

can about the distribution of wages.

Arithmetic Mean 56.80Median 59.50S.D. 12.40Solution:We can obtain the following information from the above data:

1. Since Median = Rs. 59.50, we conclude that 50% of the wrokers in the factory obtain the wages above Rs. 59.50

2. Mode = 3Median-4Mean = 359.50 = Rs. 64.90.

3. C.V. =

σx×100 %=12 . 40×100

56 . 80=21 .83 %

4. Karl Pearson’s coefficient of skewness

Sk =

3( x−M d )σ

=3(56 . 80−59 . 80)12 . 40

=3(−2. 70 )12 . 40

=−0 .65

Example 4a. If the quartile coefficient of skewness is 0.5, quartile deviation is 8 and the first quartile is

16, find the median of the distribution.b. In a distribution, the difference of two quartiles is 15 and their sum is 35 and the median

is 20. Find the coefficient of skewness.

Solution:a. Sk(B) = 0.5, Q.D. = 8, Q1 = 16, Md = ?

Q.D. =

Q3−Q1

2

or

, 8 =

Q3−16

2

Q3 = 32

Now, Sk(B) =

Q3+Q1−2 M d

Q3−Q1

or, 0.5=

32+16−2M d

32−16or, 8 = 48-2Md

Md = 20

b. Q3-Q1 = 15……………….(i)Q3+Q1 = 35………………(ii)Solving equation (i) and (ii), we getQ3 = 25, Q1 = 10 and Md = 20

Coefficient of skewness Sk(B) =

Q3+Q1−2 M d

Q3−Q1

=

25+10−2×2025−10

= 0.33Example 5Calculate the Pearson’s measure of skewness on the basis of mean, mode and standard deviation.

Mid value (x) 14.5 15.5 16.5 17.5 18.5 19.5 20.5 21.5Frequency (f) 35 40 48 100 125 87 43 22

Here, the mid value x are the mid-values of class intervals having class width 1. So, the intervals are 14-15, 15-16...and so on.The continuous frequency distribution of the given data is as follows:

CALCULATION FOR MEAN, MODE AND S.D.Class Mid value (x) f d=x-18.5 fd fd2

14-1515-1616-1717-1818-1919-2020-2121-22

14.515.516.517.518.519.520.521.5

354048100125874322

-4-3-2-10123

-140-120-96-100

0878666

560360192100087172198

N = 500 fd = -217 fd2 = 1669Here, the maximum frequency is 125. Thus the corresponding class 18-19 is the modal class.Let, f1 = 125, f0 = 100, f2 = 87, l = 18, h = 1

Mode (Mo)= l+

h ( f 1−f 0 )2 f 1− f 0−f 2

=18+1(125−100 )

250−100−87=18+25

63=18. 397

Mean(x )=a+

∑ fd

N=18 .5+

(−217 )500

=18 .5+0 . 434=18. 066

S.D. ()=

1N √ N∑ fd 2−(∑ fd )2= 1

500√500×1669−(−217 )2

Now,

Sk =

x−M o

σ=18 . 066−18.397

1.775=−0 . 186

Hence, the distribution is slightly negatively skewed.

Example 6Calculate Karl Pearson’s coefficient of skewness from the data given below:

Daily Wages (Rs.) 40-50 50-60 60-70 70-80 80-90 90-10No. of workers 5 16 8 16 25 30

SolutionSince the maximum frequency viz. 30 occurs towards the end of the frequency distribution,

mode is ill-defined. Hence, we obtain the Karl Pearson’s coefficient of skewness using median.

COMPUTAION OF MEAN, MEDIAN AND S.D.Daily wage

(Rs.)Midvalue

(X)No. of workers

(f) d '=X−7510

fd ' fd '2 Less than c.f.

40-5050-6060-7070-8080-9090-100

455565758595

5168162530

-3-2-1012

-15-32-802530

45648025120

521294570100

N = f = 100 fd ' = 30 fd '2= 262

Mean = a+h

∑ fd '

N=75+10×

30100

=78

For Median, N

2=100

2=50 .

The c.f. just greater than 50 is 70. Hence, the corresponding class 80-90 is median class.

Now, Median = l+

( N2

−cf )×h

f=80+(50−45 )×10

25=82

S.D.() =h×√∑ fd '2

N−(∑ fd '

N )2

=10×√262100

−(30100 )

2

=15 .91

Sk=

3 (Mean−Median )σ

=3 (78−82 )15 .91

=−0.754

Example 7

Calculate the coefficient of skewness from the following data by using quartiles (i.e. Bowley’s coefficient).Marks No. of students Marks No. of studentsAbove 0 180 Above 60 65Above 15 160 Above 75 20Above 30 130 Above 90 5Above 45 100

Solution:Here ‘more than’ cumulative frequency distribution is given. To compute quartiles, we first express it as a continuous frequency distribution as:

COMPUTAION OF QUARTILESMARKS No. of students (f) Less than c.f.0-1515-3030-45

180-60 = 20160-130 = 30130-100 = 30

205080

45-6060-7575-90Above 90

100-60 = 3565-20 = 4520-5 = 155

115160175180

Total N = f = 180

Here, N

2=180

2=90 ,

N4=180

4=45 and

3 N4=135

The c.f. just greater than N

2=90

is 115. Hence, the corresponding class 45-60 is the median class

M d=l+ h

f (N2

−cf )=45+1535

(90−80 )=49 .29

The c.f. just greater than N

4=45

is 50. Hence, the corresponding class 15 – 30 contains Q1

∴Q1=L+ hf (N

4−c)=45+15

35(90−80 )=49 .29

The c.f just greater than

3N4

=135is 160. Hence the corresponding class 60 – 75 contains Q3.

∴Q3=L+ hf ( 3N

4−c)=60+10

45(135−115 )=66 . 67

Coefficient of skewness Sk(B) =

Q3+Q1−2 M d

Q3−Q1

=66 .67+27 .50−2×49. 2966 .67−27 . 50

=−0 . 126

Exercise

1.a. Explain the concept of skewness. Draw the sketch of a skewed frequency

distribution and show the position of mean, median and mode when the distribution is asymmetric.

b. Define Pearson’s measure of skewness. What are the difference between relative measure and the absolute measure of skewness.

2.a. In a asymmetric distribution, the mean and mode are 32.1 and 35.4 respectively.

Calculate the median.b. In n = 2, x = 240, x2 = 4860 and Mo= 18.6, find the coefficient of skewness

based on mean and standard deviation.c. The sum of 50 observations is 500 and the sum of their squares is 6000 and

median is 12. Compute the coefficient of variation and coefficient of skewness.3.

a. In a distribution the Pearson’s coefficient of skewness is 0.4 and its coefficient of variation is 30%. Its mode is 88. Find the mean and median.

b. Pearson’s coefficient of skewness of a distribution is 0.5. its median and mode are respectively 42 and 36. Find the coefficient of variation.

4.

a. In a symmetric distribution, the first quartile is 142 and the semi-interquartile range is 18. Find the median.

b. Find the coefficient of skewness from the information:Sum of two quartiles = 22 Mode = 11Difference of two quartiles = 8 Mean = 8

c. In a frequency distribution, coefficient of skewness based on quartiles is 0.6. if the sum of upper and lower quartiles is 100 and median is 38, find the values of upper and lower quartiles.

5. a. Consider the following wages distribution of two factories.

Factory A Factory BArithmetic Mean 50 45Mode 45 50Variance 100 100

Is the distribution of factory A same as the distribution B regarding the degree of variation and skewness?

6. a. Calculate the coefficient of skewness based on mean and median from the

following distribution:Wage (Rs.) 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80No. of workers 6 12 22 48 56 32 18 6

b. From the following data find out Karl Pearson’s coefficient of skewnessMarks: 10 11 12 13 14 15No. of Candidates: 2 4 10 8 5 1

c. Calculate Karl Pearson’s coefficient of skewness from the following dataMarks FrequencyAbove 30Above 40Above 50Above 60Above 70

605126250

d. From the following data of age of employees, calculate mean, mode and coefficient of skewness.Age below (yrs.) 25 30 35 40 45 50 55No. of employees 8 20 40 65 80 92 100

7.a. From the following frequency distribution, calculate the Bowley’s coefficient of

skewness.Monthly Income (in Rs.) No. of workers

Below 100 10100-150 25150-200 145200-250 220

250-300 70300 and below 30

b. Compute Bowley’s coefficient of skewness from the following data:Marks 10-15 15-20 20-25 25-30 30-35No. of workers 8 12 20 18 2

Answers2 a. 34.3 b. 0.625c. 44.7%, -1.343 a. 100.96 b. 40%4 a. 160 b. 0.5 c. 70.305 C.V. (A) = 20 C.V. (B) = 22.2 Sk (A) = +0.5 C.V. (B) =-0.5, different with regard to both e.v. and s.k.

6 a. Sk= -0.085 b. 0.3478 c. 0.93 d. X = 37.25 yrs, Mo= 36.67yrs, Sk = 0.077 a. -0.1022 b. -0.08