TU Solution

Mean (X̅) = ΣX = 48 = 4.8
N 10

Standard deviation (σX) = √ΣNX2 − (ΣNX)2

= √21902 − (4180)2

= √29.2 − 23.04

= √6.16

= 2.48

C. V. = σX 100% = 2.48 100% = 51.66%
X̅ 4.8

For MBS Program

Mean (Y̅) = ΣY = 64 = 6.4
N 10

And, Standard deviation MBS(σY) = √ΣNY2 − (ΣNY)2

= √41508 − (1640)2

= √43.8 − 40.96

= √2.84

= 1.69

C. V. = σY 100% = 1.69 100% = 26.41%
̅Y 6.4

Since CV of MBS is less than CV of BBS. Thus, MBS program is easier to teach.

QN. 16) (2074) The average weekly wages, standard deviation, and number of workers of
two factories are given below.

Average weekly wage Factory A Factory B
Standard deviation Rs. 4600 Rs. 4900
No of workers Rs. 50 Rs. 40
100 80

100

Calculate the combined standard deviation. Also, explain which factory has greater
variability in the distribution of weekly wages?

Solution: Let X and Y denotes the weekly wages of factory A and factory B respectively.
Then, ̅ = 4600, ̅ = 4900, = 50, = 40, 1 = 100, 2 = 80

Now, average weekly wages of all workers

̅ 12 = ̅ 1 + ̅ 2
1 + 2

̅ 12 = 4600 × 100 + 4900 × 80
100 + 80

̅ 12 = 460000 + 392000
180

̅ 12 = 852000
180

̅ 12 = 4733.33

Again,

d1 = ̅X − ̅X12 = 4600 − 4733.33 = −133.33
and d2 = ̅Y − ̅X12 = 4900 − 4733.33 = 166.67

variance of weekly wages,

122 = 100(502 + 133.332) + 80(402 + 166.672)
100 + 80

122 = 2027688.89 + 2350311.11
100 + 80

122 = 4378000
180

σ122 = 24322.22

Coefficient of Variance of factory A

× 100% = 50 × 100% = 1.09%
̅ 4600

Coefficient of Variance of factory B

× 100% = 40 × 100% = 0.82%
̅ 4900

Since, CV(A) = 1.09% > CV(B) = 0.82, thus factory A has greater variability in wages.

QN. 17) (2073)

101

A buyer obtained samples of electronic fan from two companies A and B. He got those
samples tasted in his laboratory for length of life in the number of hours. The following are
the result of these tests.

Length of the (hours) Numbers of electronic fans
Company A Company B

600 – 800 20 6

800 1000 32 20

1000 – 1200 52 84

1200 – 1400 20 24

1400 - 1600 16 6

What would you conclude as to which supplier's fan are more uniform for length of life?

Solution: Calculation of coefficient of variance

Length of the Mid- ′ = Company A Company B
(hours) value − 1100 f2 f2d' f2d'2
(X) f1 f1d' f1d'2 6 -12 24
600 – 800 700 200 20 -40 80 20 -20 20
800 -1000 -2 32 -32 32 84 0 0
1000 – 1200 900 52 0 0 24 24 24
1200 – 1400 -1 20 20 20 6 12 24
1400 - 1600 1100 16 32 64 140 4 92
0
1300
1
1500
2

140 -20 196

Here, n1 = 140, n2 = 140, Σf1d′ = −20, Σf1d′2 = 196, Σf2d′ = 4, Σf1d′2 = 92, h = 200

Company A

̅X = A + Σf1d′ ×h= 1100 + −20 × 200 = 1100 − 28.57 = 1071.43
n1 140

σA = √Σfn1d1′2 − (Σfn11d′)2 × h

= √119406 − (−14200)2 × 200

= √1.4 − 0.02 × 200
= √1.38 × 200
= 234.95

102

( ) = × 100% = 234.95 × 100% = 21.93%
̅ 1071.34

Company B

̅X = A + Σf2d′ × h = 1100 + 4 × 200 = 1100 + 5.71 = 1105.71
n2 140

σA = √Σfn1d1′2 − (Σfn11d′)2 × h

= √19420 − (1440)2 × 200

= √0.66 − 0.0.03 × 200

= √0.63 × 200

= 158.75

( ) = × 100% = 158.75 × 100% = 14.36%
̅ 1105.71

Since, CV(A) = 21.93% > CV(B) = 14.36%

Therefore, supplier's B fan are more uniform for length of life.

QN. 18) (2073, old)

The following is the wage distribution of the workers of a company working in two
different shifts.

Morning shift Day shift

No of workers 300 200
9000
Average wage 8000 121

Variance of wages distribution 100

Find;

a) Average wage of all workers of the company.
b) Standard deviation of wages of all workers of workers of the company.
c) Coefficient of variation of all workers of the company.
Solution:

Let X and Y denotes the weekly wages of factory A and factory B respectively.
Then, ̅ = 8000, ̅ = 9000, 2 = 100, = 121, 1 = 200, 2 = 200

Now, average weekly wages of all workers

103

̅ 12 = ̅ 1 + ̅ 2
1 + 2

̅ 12 = 8000 × 3000 + 9000 × 200
3000 + 200

̅ 12 = 24000000 + 1800000
3200

̅ 12 = 25800000
3200

̅ 12 = 8062.50

Average wage of all workers of the company is 8062.50

Again,

d1 = ̅X − ̅X12 = 8000 − 8062.50 = −62.50
and d2 = ̅Y − ̅X12 = 9000 − 8062.50 = 937.50

variance of weekly wages,

122 = 1( 12 + 12) + 1( 22 + 22)
1 + 2

122 = 3000(100 + (−62.50)2) + 200(121 + 937.502)
3000 + 200

122 = 12018750 + 157781250
3200

122 = 187800000
3200

122 = 58687.50

12 = 242.26

Therefore, standard deviation of wages of all workers of workers of the company is 242.26

Finally, Coefficient of Variance of all workers of the company,

12 × 100% = 242.26 × 100% = 3.0047%
̅ 12 8062.50

QN. 19) (2072)

The mean and variance of of the marks in statistics obtained by all the 50 students of a certain
college was computed as 60 and 100 respectively. Later on, it was discovered that the score 76
was wrongly taken as 67. Find the mean and standard deviation of scores when value is omitted.
Also, calculate the coefficient of variation of marks after ignoring wrong value.

Solution: Here given

Number of items (n) = 50
mean ( ̅ ) = 60

104

variance (σ2) = 100

Incorrect values = 67

Correct values = 76

Now, we have

̅ = ∑


⇒ 60 = ∑
50

∴ = 3000

variance σ2 = ΣX2 − (ΣnX)2
n

⇒ 100 = 2 − (350000)2
50

⇒ 100 = 2 − 3600
50

⇒ 100 + 3600 = 2
50

⇒ 3700 = 2
50

⇒ 2 = 185000

Omitting the incorrect value

= 3000 − 67 = 2933

Mean = ∑X = 2933 = 59.86
n−1 49

ΣX2 = 185000 − 672 = 180511

standard deciation σ = √nΣ−X21 − (nΣ−X1)2

= √18409511 − (294393)2

= √3683.9 − 3583.22

= √100.68

= 10.03

Again, CV = σ 100% = 10.03 100% = 16.76%
̅X 59.86

105

QN. 20) (2072) Two brands of tyres are tested for their life and following result were found:

Life (000km) 20 - 24 24 - 28 28 - 32 32 - 36 36 - 40

Brand X 8 15 12 8 7

Brand Y 6 20 14 5 5

Both brands are offering same price and adverting in favor of their brand saying that the brand has

consistence of life. If you are requested to decide to purchase type of tyre of one of these two

brands, which one do you prefer and why?

Solution:

Solution: Calculation of coefficient of variance

Life (000km) Mid- ′ = Brand X Brand Y
value − 30 f1 f1d' f1d'2 f2 f2d' f2d'2
20 – 24 (X) 8 -16 32 6 -12 24
24 - 28 4 15 -15 15 20 -20 20
28 - 32 22 -2 12 0 0 14 0 0
32 - 36 888 555
36 - 40 26 -1 7 14 28 5 10 20
50 -17 69
30 0

34 1

38 2

50 -9 83

Here, n = 50, Σf1d′ = −9, Σf1d′2 = 83, Σf2d′ = −17, Σf1d′2 = 69, h = 4

For company A

X̅ = A + Σf1d′ × h = 30 + −9 × 4 = 30 − 0.72 = 29.28
n 50

σX = √Σfn1d1′2 − (Σfn11d′)2 × h

= √8530 − (−509)2 × 4

= √1.66 − 0.0324 × 4

= √1.6276 × 4

= 5.10

CV(A) = σX × 100% = 5.10 × 100% = 17.42%
X̅ 29.28

For company B

106

̅ = + 2 ′ ×ℎ = 30 + −17 × 4 = 30 − 1.36 = 28.64
2 50

σY = √Σf1d′2 − (Σfn1d′)2 × h
n

= √6509 − (−5107)2 × 4

= √1.38 − 0.12 × 4

= √1.26 × 4

= 4.49

CV(A) = σY × 100% = 4.49 × 100% = 15.68%
X̅ 28.64

Since, CV(A) = 17.42% > CV(B) = 15.68%

Therefore, I prefer to purchase tyre of brand Y, because the brand Y has consistence of life.

QN. 21) (2072 old) For the group of 200 candidates, the mean and the standard deviation were
found to be 40 and 15 respectively. Later on, it was discovered that the score 43 and 53 was
misread as 34 and 35 respectively. Find the correct mean and standard deviation.

Solution: Here given

Number of items (n) = 200
mean ( ̅ ) = 40
standard deviation ( ) = 15

Incorrect values = 34 and 53

Correct values = 43 and 35

Now, we have

̅ = ∑


40 = ∑
200

∴ = 8000

Now, Correct ΣX = 8000 − 34 − 53 + 43 + 35 = 7991

Correct mean = Correct ∑X = 7991 = 39.955
n 200

Again

107

standard deciation σ = √ΣnX2 − (ΣnX)2

r 15 = √2Σ0X02 − (8200000)2

or 225 + 1600 = ΣX2
200

or ΣX2 = 365000

correct ΣX2 = 365000 − 342 − 532 + 432 + 352

correct ΣX2 = 365000 − 342 − 532 + 432 + 352 = 364109

Correct standard deciation σ = √ΣnX2 − (ΣnX)2

= √362401009 − (7290901)2

= √1820.545 − 1596.40

= √224.145
= 14.97

QN. 22) (2071)

The combined mean and variances of salaries of 250 workers of city A and city B are 560 and
5497 respectively. The mean and variances of the salaries of 150 workers of city B are 500 and 81
respectively. Find the variance of salaries of workers of city A.

Solution:

Let X and Y denotes the salaries of city A and city B respectively.
Then ̅ 12 = 560, ( ) = 5497, ̅ = 500, ( ) = 81, = 250, 2 = 150, ( ) =?

Now, we have

̅ 1 + ̅ 2 1 = − 2 = 250 − 150 = 100,
1 + 2
̅ 12 =

560 = ̅ × 100 + 500 × 150
100 + 150

560 = 100 ̅ + 75000
250

108

140000 = 100 ̅ + 75000

100 ̅ = 140000 − 75000

̅ = 65000 = 650
100

Average salary of city A is Rs. 650.

Again,

d1 = ̅X − ̅X12 = 650 − 560 = 90

and d2 = ̅Y − ̅X12 = 500 − 560 = −60

122 = 1( 12 + 12) + 2( 22 + 22)
1 + 2

⇒ 5497 = 100( 12 + (90)2) + 150(81 + (−60)2)
100 + 150

⇒ 5497 = 100( 12 + 8100) + 552150
250

⇒ 1374250 = 100 12 + 810000 + 552150

⇒ 100 12 = 12100

⇒ 12 = 121

Therefore, variance of city A is 121.

QN. 23) (2071 old) The mode of the following distribution is Rs. 24, find the mean deviation from
mean.

Expenditure in Rs. 0 - 10 10 – 20 20 - 30 30 – 40 40 - 50
No of families 14 23 27 - 15
Solution: Let the missing frequency be a. Then

Expenditure in Rs. No of families (f)

0 – 10 14
10 – 20 23
20 – 30 27
30 – 40 a
40 – 50 15

N = 79 + a

Since, the mode of the class is 24, so the model class is 20 – 30.
Here, L = 20, f1 = 27, fo = 23, f2 = a, h = 20

109

Now, mode (Mo) = L + f1 − fo f2 × h
2f1 − fo −

⇒ 24 = 20 + 2 × 27 − 23 − × 10
27 − 23

⇒ 24 − 20 = 4 × 10
31 −

⇒ 4 = 40
31 −

⇒ 124 − 4 = 40

⇒ 124 − 40 = 4

⇒ = 21

Again, calculation of mean deviation

Expenditure No of Mid-value fX | − ̅ | | − ̅ |
families (f) (X)
in Rs. 5 70 20 280
14 15 345 10 230
0 – 10 23 25 675 0 0
10 – 20 27 35 735 10 210
20 – 30 21 45 675 20 300
30 – 40 15 =2500 | −
40 – 50 N = 100 ̅ |=1020

Now, mean(̅X) = ΣfX = 2500 = 25
n 100

Mean deviation (MD) = Σf|X − ̅X| = 1020 = 10.20
n 100

QN. 24) (2071 old) The mean weight of 150 students is 60kg. The mean weight of boys is 70 kg
with a standard deviation of 10kg. For the girls, the mean weight is 55kg with the standard
deviation 15kg. Find the number of boys and the number of girls in the class and calculate
combined standard deviation.

Solution: Let the number of boys be n1 and the number of girls be n2 respectively.

Then n1 + n2 = 150 or n1 = 150 – n2 ……….. (i)

Let the weight of boys be X1 and the weight of girls be X1.

Here, ̅X1 = 70, X̅2 = 55, σ1 = 10, σ2 = 15, ̅X12 = 60, σ12 =?

̅X12 = ̅Xn1 + ̅Yn2
n1 + n2

110

60 = 70 × (150 − 2) + 55 × 2
150

9000 = 10500 − 70 2 + 55 2

15 2 = 1500

2 = 1500 = 50
15

From equation (i)

n1 = 150 – 50 = 100

Therefore, number of boys is 100 and number of girls is 50.

Again,

1 = ̅ − ̅ 12 = 70 − 60 = 10
2 = ̅ − ̅ 12 = 55 − 60 = −5

σ122 = n1(σ12 + d12) + n2(σ22 + d22)
n1 + n2

⇒ 122 = 100(102 + (10)2) + 50(152 + (−5)2)
50 + 100

⇒ σ122 = 20000 + 12500
150

⇒ 122 = 32500
150

⇒ 122 =216.667

⇒ 12 = 14.72

Therefore, combined standard deviation is 14.72.

Analytical Question Answer

QN. 25) (2077 old)

A factory produces two types electric motors A and B. In an experiment relating to their life, the
following result were obtained:

are the result of these tests.

Life (in years) Numbers of motors

0–2 Model A Model B
2–4
4–6 15
6–8
97

12 15

11 19

111

8 - 10 89

a) Find which model of motor has greater uniformity of life? Give reason.

b) Calculate the combined mean.

c) Calculate the combined standard deviation.

Solution:

Calculation of Coefficient of variation:

Length of the Mid- ′ = − 5 Company A Company B
life (hours) value 2 f2 f2d' f2d'2
(X) f1 f1d' f1d'2 5 -10 20
0–2 -2 1 -2 4 7 -7 7
2–4 1 -1 9 -9 9 15 0 0
4–6 0 12 0 0 19 19 19
6–8 3 1 11 11 11 9 18 36
8 - 10 2 8 16 32 55 20 82
5 41 16 56

7

9

Here, n1 = 41, n2 = 55, Σf1d′ = 16, Σf1d′2 = 56, Σf2d′ = 20, Σf1d′2 = 82, h = 2

Company A

X̅ = A + Σf1d′ × h = 5 + 16 ×2 = 5 + 0.78 = 5.78
n1 41

σA = √Σfn1d1 ′2 − (Σfn11d′ 2 × h

)

= √5416 − (4116)2 × 2

= √1.3659 − 0.1521 × 2

= √1.2138 × 2

= 2.20

( ) = × 100% = 2.20 × 100% = 38.06%
̅ 5.78

Company B

112

X̅ = A + Σf2d′ ×h =5+ 20 × 2 = 5 + 0.73 = 5.73
n2 55

σB = √Σfn1d1′2 − (Σfn11d′)2 × h

= √5825 − (5250)2 × 2

= √1.50 − 0.13 × 2

= √1.37 × 2

= 2.34

( ) = × 100% = 2.34 × 100% = 40.83%
̅ 5.73

Motor of model A has greater uniformity of life, as CV(A) = 38.06% < CV(B) = 40.83%.

b) Combined mean

̅ 12 = ̅ 1 + ̅ 2
1 + 2

= 5.78 × 41 + 5.73 × 55
41 + 55

= 236.98 + 315.15
96

= 552.13
96

= 5.75

c) Combined standard deviation

1 = ̅ − ̅ 12 = 5.78 − 5.75 = 0.03
2 = ̅ − ̅ 12 = 5.73 − 5.75 = −0.02

12 = √ 1( 12 + 12) + 2( 22 + 22)
1 + 2

= √41(2.202 + 0.032 ) + 55(2.342 + (−0.02)2)
41 + 55

113

= √198.476969+ 301.18

= √4996.69569

= 2.69

QN. 26) (2077) A factory produces two types electric lamps A and B. In an experiment relating to
their life, the following result were obtained:

Length of Life (in No. of lamps A No. of lamps B
hours)
5 6
500 - 700 11 10
700 - 900 15 18
900 - 1100 10 9
1100 - 1300 8 7
1300 - 1500 6 6
1500 - 1700 5 4
1700 - 1900

Compare the variability of the length of life of two varieties of electric lamps using coefficient of
variation and state which lamp is more uniform regarding to the length of life. Also, calculate the
combined standard deviation.

Solution:

Calculation of Coefficient of variation:

Length of Mid- ′ − 1200 Company A Company B
Life (in value 200
hours) (X) = f1 f1d' f1d'2 f2 f2d' f2d'2
500 - 700 600 5 -15 45 6 - 18 54
-3 11 - 22 44 10 - 20 40
700 - 900 800 15 - 15 15 18 - 18 18
-2 10 0 0 900
900 - 1100 1000 888 777
-1 6 12 24 6 12 24
1100 - 1300 1200 5 15 45 4 12 36
0 60 -17 181
1300 - 1500 1400 60 -25 179
1500 - 1700 1600 1
1700 - 1900
1800 2

3

Here, n1 = 60, n2 = 60, Σf1d′ = −17, Σf1d′2 = 181, Σf2d′ = −25, Σf1d′2 = 179, h = 200

114

Company A

̅ = + 1 ′ × ℎ = 1200 + −17 × 200 = 1200 − 56.67 = 1143.33
1 60

σA = √Σfn1d1 ′2 − (Σfn11d′ 2 × h

)

= √16801 − (−6107)2 × 200

= √3.0167 − 0.0784 × 200

= √2.9383 × 200

= 342.83

( ) = × 100% = 342.83 × 100% = 29.99%
̅ 1143.33

Company B

̅ = + 2 ′ ×ℎ = 1200 + −25 × 200 = 1200 − 83.33 = 1116.67
2 60

= √ 1 1 ′2 − ( 11 ′)2 × ℎ

= √16709 − (−6205)2 × 200

= √2.98 − 0.18 × 200

= √2.8 × 200

= 1.67 × 200

= 334

( ) = × 100% = 334 × 100% = 29.91%
̅ 1116.67

Electric lamps B has greater uniformity of life, as CV(A) = 29.99%% < CV(B) = 29.91%.

Combined mean

̅ 12 = ̅ 1 + ̅ 2
1 + 2

115

= 1143.33 × 60 + 1116.67 × 60
60 + 60

= 68599.8 + 67000.2
120

= 135600
120

= 1130

Combined standard deviation

1 = ̅ − ̅ 12 = 1143.33 − 1130 = 13.33
2 = ̅ − ̅ 12 = 1116.67 − 1130 = −13.33

12 = √ 1( 12 + 12) + 2( 22 + 22)
1 + 2

= √60(342.832 + 13.332 ) + 60(3342 + (−13.33)2)
60 + 60

= √60(117710.09182+0 111733.689)

= √229414230.7869

= 338.71

QN. 27) (2067) Monthly income (in 000 Rs.) distribution of 400 families of a certain town
are given below:

Income, below (000 Rs.) 80 120 160 200 240 280
35 165 310 380 400
No of families 10

a) Calculate the appropriate measure of central tendency from the above distribution
and justify for your choice of measure.

b) Calculate the most suitable measures of dispersion and support for your choice.
c) Find the lowest income of richest 20% of the families.
d) Find the highest income of the poorest 20% of the families.
e) Find the limits of income of middle 50% families.

116

Solution: The given distribution is open ended class interval. The median is appropriate
measure of central tendency and quartile deviation is most appropriate measure of
dispersion.

Calculation of median and quartile deviation

Income (in 000 Rs.) No of families (f) Cumulative frequency (c.f.)
below 80 10 10
80 - 120 25 35
120 -160 130 165
160 - 200 145 310
200 – 240 70 380
240 – 280 20 400

N = 400

a) N 400
2 2
For median = = 200

cf just greater or equal to 200 is 320, so the median class is 160 – 200.

Here, L = 160, f = 145, cf = 165, h = 40

Now, median(Md) = L + N − cf × h
2 f

= 160 + 200 − 165 × 40
145

= 160 + 35 × 40
145

= 160 + 9.65517

= 169.65517

Therefore, median income is Rs169655.20

b) Calculation of quartile deviation

For lower quartile, N = 400 = 100
4 4

cf just greater or equal to 100 is 165, so the first quartile class is 120 – 160.

Here, L = 120, f = 130, cf = 35, h = 40

Now, Q1 = L + N − cf × h
4 f

117

= 120 + 100 − 35 × 40
130

= 120 + 65 × 40
130

= 120 + 20

= 140

For upper quartile, 3N = 3 × 400 = 300
4 4

cf just greater or equal to 300 is 310, so the upper quartile class is 160 – 200.

Here, L = 160, f = 145, cf = 165, h = 40

Now, upper quartile (Q3) = L + N − cf × h
4 f

= 160 + 300 − 165 × 40
145

= 160 + 135 × 40
145

= 160 + 37.2414

= 197.2414

Quartile Deviation (Q. D. ) = 3 − 1 = 197.2414 − 140 = 28.6207
2 2

Therefore, quartile deviation is Rs. 28620.70

c) Calculation of 80th percentile

80 = 80 × 400 = 320
100 100

cf just greater or equal to 320 is 380, so the median class is 200 – 240.

Here, L = 200, f = 70, cf = 310, h = 40

80 = + 80 − × ℎ
100


= 200 + 320 − 310 × 40
70

= 200 + 10 × 40
70

118

= 200 + 5.71430
= 205.71430

Therefore, lowest income of richest 20% of family is Rs. 205714.30

c) Calculation of 20th percentile

20 = 20 × 400 = 80
100 100

cf just greater or equal to 80 is 380, so the median class is 120 – 160.

Here, L = 120, f = 130, cf = 35, h = 40

20 = + 80 − × ℎ
100


= 120 + 80 − 35 × 40
130

= 120 + 45 × 40
130

= 120 + 13.8461538

= 133.84620

Therefore, highest income of poorest 20% of family is Rs. 133846.20

e) Calculation of P25 and P75

25 = 25 × 400 = 100
100 100

cf just greater or equal to 100 is 165, so the median class is 120 – 160.

Here, L = 120, f = 130, cf = 35, h = 40

20 = + 25 − × ℎ
100


= 120 + 100 − 35 × 40
130

= 120 + 65 × 40
130

= 120 + 20

= 140

119

75 = 75 × 400 = 300
100 100

cf just greater or equal to 300 is 310, so the median class is 160 – 200.

Here, L = 160, f = 145, cf = 165, h = 40

20 = + 75 − × ℎ
100


= 160 + 300 − 165 × 40
145

= 160 + 135 × 40
145

= 160 + 37.24140

= 197.24140

limits of income of middle 50% families is Rs. 140000 and Rs. 197241.40

QN. 28) (2075) What do you understand by measure of dispersion and distinguish between
absolute and relative measure of dispersion. A sample of 50 cars, each of two makes X and
Y, is taken and their average running life in years in recorded.

Life (no of years) 0 - 5 5 - 10 10 - 15 15 - 20 20 – 25

Make X 6 10 20 12 2

Make Y 8 12 17 10 3

Which of these two makes car would you prefer to buy if the costs of both makes are same

regarding consisting.

Solution:

Calculation of Coefficient of variation:

Life (in Mid- ′ = − 12.5 Make X Make Y
years) value 5
(X) f1 f1d' f1d'2 f2 f2d' f2d'2
0–5 2.5 -2 6 -12 24 8 -16 32
5 – 10 -1 10 -10 10 12 -12 12
10 - 15 7.5 0 20 0 0 17 0 0
15 - 20 1 12 12 12 10 10 10
20 - 25 12.5 2 248 3 6 12
50 -6 54 50 -12 66
17.5

22.5

Here, n1 = 50, n2 = 50, Σf1d′ = −6, Σf1d′2 = 54, Σf2d′ = −12, Σf1d′2 = 66, h = 5

120

Company A

X̅ = A + Σf1d′ ×h= 12.5 + −6 × 5 = 12.5 − 0.6 = 11.90
n1 50

σX = √Σfn1d1′2 − (Σfn11d′)2 × h

= √5540 − (−506)2 × 5

= √1.08 − 0.014 × 5

= √1.066 × 5

= 5.16

( ) = × 100% = 5.16 × 100% = 43.36%
̅ 11.9

Company B

̅ = + 2 ′ × ℎ = 12.5 + −12 × 5 = 12.5 − 1.2 = 11.30
2 50

= √ 1 1 ′2 − ( 11 ′)2 × ℎ

= √5660 − (−5102)2 × 5

= √1.32 − 0.06 × 5

= √1.26 × 5

= 1.12 × 5

= 5.62

( ) = × 100% = 5.62 × 100% = 49.73%
̅ 11.30

Electric lamps B has greater uniformity of life, as CV(X) = 43.36%% < CV(Y) = 49.73%.

QN. 29) (2074) Two companies ONIDA and BALTRA of home applications profit distribution
are given as follows:

Profit (million Rs.) 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12

121

ONIDA 2 7 11 20 12 2
BALTRA 5 4 9 13 15 4

a) An inverter wants to invest according to the profit of the company; in which
company it is better to invest?

b) Also find the average profit and variances of both companies together.

Solution: Let the profit of ONIDA company and BALTRA company be denoted by X and Y
Respectively. Then

Calculation of Coefficient of variation:

Length of the Mid- ′ = − 5 ONIDA BALTRA
life (hours) value 2 f2 f2d' f2d'2
(X) f1 f1d' f1d'2 5 -10 20
0–2 -2 2 -4 8 4 -4 4
2–4 1 -1 7 -7 7 900
4–6 0 11 0 0 13 13 13
6–8 3 1 20 20 20 15 30 60
8 - 10 2 7 14 28 4 12 36
10 - 12 5 3 3 9 27 50 41 133
50 32 90
7

9

11

Here, n1 = 50, n2 = 50, Σf1d′ = 32, Σf1d′2 = 90, Σf2d′ = 41, Σf1d′2 = 133, h = 2

̅X = A + Σf1d′ × h = 5 + 32 ×2 = 5 + 1.28 = 6.28
n1 50

σX = √Σfn1d1′2 − (Σfn11d′)2 × h

= √9500 − (5320)2 × 2

= √1.8 − 0.41 × 2

= √1.39 × 2

= 2.36

( ) = × 100% = 2.36 × 100% = 37.58%
̅ 6.28

Again,

122

̅Y = A + Σf2d′ ×h =5+ 41 × 2 = 5 + 1.64 = 6.64
n2 50

σY = √Σfn1d1′2 − (Σfn11d′)2 × h

= √15303 − (4501)2 × 2

= √2.66 − 0.67 × 2

= √2 × 2

= 2.82

( ) = × 100% = 2.82 × 100% = 40.47%
̅ 6.64

ONIDA company is better to invest, as CV(A) = 38.06% < CV(B) = 40.83%.

b) Combined mean

̅ 12 = ̅ 1 + ̅ 2
1 + 2

= 6.28 × 50 + 6.64 × 50
50 + 50

= 314 + 332
100

= 646
100

= 6.46

Average profit of both companies is Rs. 6.64 million.

Combined variance of both companies

1 = ̅ − ̅ 12 = 6.28 − 6.46 = −0.18
2 = ̅ − ̅ 12 = 6.64 − 6.46 = 0.18

122 = 1( 12 + 12) + 2( 22 + 22)
1 + 2

= 50(2.362 + (−0.18)2 ) + 50(2.822 + 0.182)
50 + 50

123

= 280.1 + 399.24
100

= 679.34
100

= 6.7934

QN. 30) (2072 old) The following two samples describes the age of students in Model
Campus and Galaxy Campus BBS program:

Model Campus 25 31 29 24 26 23 27 28 29 26
Galaxy Campus 29 28 35 34 30 28 29 30 34 27

a. Calculate the mean and standard deviation of age of students.

b. If homogeneity in age of the students is a positive factor for learning, which of the
two campuses will be easier to teach?

c. Calculate the mean, variance, and coefficient of variation of age of students of
both campuses taken together.

d. Which campus's students are more intelligent and why?

Solution: Let the age of students' of Model college be X and that of Galaxy college be Y.
Then

Model Galaxy X2 Y2 XY
Campus (X) Campus (Y)
625 841 725
25 29 961 784 868
31 28 841 1225 1015
29 35 576 1156 816
24 34 676 900 780
26 30 529 784 644
729 841 783
23 28 784 900 840
27 29 841 1156 986
676 729 702
28 30 2 = 7238 2 =9316 =8159
29 34
26 27
= 268 =304

Model Campus

a) Mean and standard deviation of each campus

̅ = = 268 = 26.80
10

124

= √ 2 − ( )2 = √712038 − (21608)2 = √723.8 − 718.24 = 2.36

̅ = = 304 = 30.40
10

= √ 2 − ( )2 = √931106 − (31004)2 = √931.6 − 924.16 = 2.73

b) variance of each campus

( ) = 100% = 2.36 100% = 8.81%
̅ 26.80

( ) = 100% = 2.73 100% = 8.98%
̅ 30.40

Model campus will be easier to teach as, CV(X) = 8.81% < CV(Y) = 8.98%

c) Mean, variance, and coefficient of variance of both campus

̅ 12 = ̅ 1 + ̅ 2
1 + 2

= 26.80 × 10 + 30.40 × 10
10 + 10

= 268 + 304
20

= 572
20

= 28.60

Combined variance of both companies

1 = ̅ − ̅ 12 = 26.80 − 28.60 = −1.8
2 = ̅ − ̅ 12 = 30.40 − 28.60 = 1.8

122 = 1( 12 + 12) + 2( 22 + 22)
1 + 2

= 10(2.362 + (−1.8)2 ) + 10(2.732 + 1.82)
10 + 10

= 88.096 + 106.929
20

125

= 195.025
20

= 9.75125

CV of both companies

12 100% = 3.12 100% = 10.91%
̅ 12 28.60

d) Intelligence can not be determined on the basis of age of students.

QN. 31) (2072, old) The following two samples describes the age of students in morning
MBS and evening MBS program of public campus.

Morning MBS 24 30 28 23 25 22 26 27 28 25
Evening MBS 29 28 35 34 30 28 29 30 34 27
a. Calculate the mean and standard deviation of age of students of each program.

b. If homogeneity in age of the students is a positive factor for learning, which of the
two campuses will be easier to teach?

c. Calculate the mean, variance, and coefficient of variation of age of students of
both programs taken together.

d. Which programs' students are more intelligent and why?

Solution: Same as Question no 28.

126

Chapter – 5

Skewness, Kurtosis and Moments

Brief Answer Question

QN. 1) (2078) Find the Karl Pearson’s Coefficient of Skewness, when mean = 45, mode = 48 and
standard deviation = 15.

Solution: Here, mean (X̅ ) = 45, mode (Mo) = 48, and standard deviation (σ) = 15

Karl Person′s coefficient of skewness(Skp)

= 3(Mean − Median) = 3(45 − 50) = −1.5
σ 10

Since, = - 1.5 < 0, distribution is negatively skewed.

QN. 2) (2078) The quartile deviation of the distribution is 2 and the difference between P90 and P10
is 8. Calculate the coefficient of Kurtosis.

Solution: Here given Quartile deviation (Q.D.) = 2, P90 – P10 = 8

coefficient of kurtosis

K = 1 (Q3 − Q1) = 2 = 0.25
2 8
P90 − P10

QN. 2 (2077 back) If mean = 45, median = 50, and standard deviation = 10. Find the Karl Person's
coefficient of skewness and interpret the result.

Solution: Here given Mean(X̅) = 45, Median(Md) = 50 and s. d. (σ) = 10

Karl Person′s coefficient of skewness(Skp)

= 3(Mean − Median) = 3(45 − 50) = −1.5
σ 10

Since, = - 1.5 < 0, distribution is negatively skewed.

QN. 3 (2076) In a moderately skewed frequency distribution, the mean, median, and standard
deviation are 10, 8.50 and 2 respectively. Find the Karl Person's coefficient of skewness.

Solution: Here given Mean(̅X) = 10, Median(Md) = 8.50 and s. d. (σ) = 2

Karl Person′s coefficient of skewness(Skp)

= 3(Mean − Median) = 3(10 − 8.50) = 2.25
σ 2

AS, = 2.25 > 0, distribution is positively skewed.

QN. 4 (2076) First four moments about mean of a certain distribution are 0, 16, -30, and 40
respectively. Calculate the coefficient of kurtosis and interpret the result.

127

Solution: Here given μ1 = 0, μ2 = 16, μ3 = −30, μ4 = 40

Now, Coefficient of kurtosis β2 = μ4 = 40 = 40 = 0.15625
μ22 162 256

As, 2 < 3, the distribution is platykurtic.

QN. 5 (2075) For the group of 10 items ΣX = 452, ΣX2 = 24,270, and mode = 43.7. Find the

Pearson's coefficient of skewness.

Solution: Now,

Mean (X̅) = ΣX = 452 = 45.2
N 10

Standard deviation (σ) = √ΣNX2 − (ΣNX)2

= √2412070 − (41501)2 = √2427 − 2043.042 = 19.60

Pearson′s coefficient of skewness Skp = Mean − Mode = 45.2 − 43.7 = 1.5 = 0.08
SD 19.60 19.60

, = 0.08 > 0, the distribution is positively skewed.

QN. 6 (2074) First four moments about the arbitrary number 5 are – 0.1, 2.82, 0.2, and 20.58. Find
the second and third moment.

Solution: Here given μ1′ = −0.1, μ′2 = 2.82, μ3 = 0.2 and μ4 = 20.58
Now, μ2 = μ2′ − (μ1′ )2 = 2.82 – (- 0.1)2 = 2.81
μ3 = μ3′ − 3μ2′ μ1′ + 2(μ1′ )3 = 0.2 − 3 × 2.82 × −0.1 + 2(−0.1)3

= 0.2 + 0.846 + 0.002 = 1.048

QN. 7 (2073) First four moments about mean are 0, 3, 9 and 21. Test the symmetry.

Solution: Here given μ1 = 0, μ2 = 3, μ3 = 9 and μ4 = 21

Coefficient of skewness β1 = μ23 = 93 = 81 = 3
μ23 33 27

As, 1 = 3 > 0, the distribution is positively skewness.

QN. 8 (2072) List the five number summary from the following daily sales data (in Rs. 000) of
seven different shops:

Sales (in 000) 50 20 80 10 60 30 70

Shops ABCDE F G

128

Solution: Now arranging the numbers in ascending order

10, 20, 30, 50, 60, 70, 80

Numbers of shops (N) = 7

Lower quartile (Q1) = (N + 1) th item = 7 + 1 th = 2nd item = 20
4 4

Median (Md) = (N + 1) th item = 7 + 1 th = 4th item = 50
2 2

Upper quartile (Q3) = 3(N + 1) th item = 3(7 + 1) th = 6th item = 70
4 4

Five number summaries

Smallest number = 10, lower quartile = 20, median = 50,

upper quartile = 70 & largest number = 80

QN. 9 (2072) Test for the normality of the distribution on the basis of the following information

Lower quartile (Q1) = 41.5, Upper quartile (Q3) = 58.25,
10th percentile (P10) = 31.428 90th percentile (P90) = 70

Solution: Now, percentile coefficient of kurtosis

= 1 ( 3 − 1) = 1 (58.25 − 41.5) = 8.375 = 0.27
2 2 70 − 31.428 30.572
90 − 10

k = 0.27 > 0.263, the distribution is leptokurtic.

QN. 10 (2072/071)

The standard deviation of a symmetrical distribution is 7. what must be the value of the 4th
moment about the mean in order that the distribution is mesokurtic?

Solution: Here standard deviation (σ) = 7 ∴ μ2 = σ2 = 72 = 49

For the distribution to be mesokurtic

2 = 3

⇒ 4 = 3
22

⇒ 4 = 3
492

∴ 4 = 3 × 2401 = 7203

Descriptive Question Answer

129

QN. 11) (2078) Calculate the Karl Pearson’s coefficient of Skewness and interpret the result.

Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 – 70 70 - 80

Frequency 4 5 14 17 25 18 10 7

Solution:

Calculation of Pearsonian measure of skewness d′ = X − 35 fd' fd'2
Class Frequency c.f. Mid-value(X) 10

0 – 10 4 4 5 -3 -12 144

10 – 20 5 9 15 - 2 -10 100

20 – 30 14 23 25 -1 -14 196

30 – 40 17 40 35 0 00

40 – 50 25 65 45 1 25 625

50 – 60 18 83 55 2 36 1296

60 - 70 10 93 65 3 30 900

70 - 80 7 100 75 4 28 784

100 83 4045

Here, N = 100, A = 30, h = 10, Σfd' = 83, Σfd'2 = 4045

Now, Mean(̅X) = A + Σfd′ × h = 30 + 83 × 10 = 30 + 8.3 = 38.30
N 100

For, median N = 100 = 50
2 2

c.f. just greater or equal to 50 is 65, the median class is 40 – 50

Here, L = 40, f = 25, cf = 40, h = 10

Median (Md) = L + N − cf × h
2 f

= 40 + 50 − 40 × 10
25

= 40 + 10 × 10
25

= 40 + 4

130

= 44

Standard deviation (σ) = √Σfd′2 − (ΣNfd′ 2 × h
N
)

= √4100405 − (18030)2 × 10

= √40.45 − 0.6889 × 10

= √39.76 × 10

= 63.10

Pearsonian coefficient of skewness(Skp) = 3(X̅ − Md) = 3(38.30 − 44) = −0.27
σ 63.10

Since Skp < 0, the distribution is negatively skewed.

QN. 10 (2077) Calculate the Pearsonian measure of skewness for the following distribution:

Wage per hour (Rs.) 300-330 330-360 360-390 390-420 420-450 540-480

Number of workers 2 4 26 47 15 6

Solution:

Calculation of Pearsonian measure of skewness

Wages per No. of c.f. Mid- ′ = − 405 fd' fd'2
hour (Rs.) workers value(X) 30

300 - 330 2 2 315 -3 - 6 18

330 – 360 4 6 345 -2 - 8 16

360 – 390 26 32 375 -1 - 26 26

390 – 420 47 79 405 0 00

420 – 450 15 94 435 1 15 15

450 - 480 6 100 465 2 12 24

100 -13 99

Here, N = 100, A = 405, h = 30, Σfd' = - 13, Σfd'2 = 99

Now, Mean(X̅) = A + Σfd′ × h = 405 + −13 × 30 = 405 − 3.9 = 401.1
N 100

131

For, median N = 100 = 50
2 2

c.f. just greater or equal to 50 is 79, the median class is 390 – 420

Here, L = 390, f = 47, cf = 32, h = 30

Median (Md) = L + N − cf × h
2 f

= 390 + 50 − 32 × 30
47

= 390 + 18 × 30
47

= 390 + 11.49

= 401.49

Standard deviation (σ) = √Σfd′2 − (ΣNfd′ 2 × h
N
)

= √19090 − (−10103)2 × 30

= √0.99 − 0.0169 × 30

= √0.9731 × 30

= 29.59

Pearsonian coefficient of skewness(Skp) = 3(̅X − Md) = 3(401.1 − 401.49) = −0.04
σ 29.59

Since Skp < 0, the distribution is negatively skewed.

QN. 10 (2077, old) Calculate the Karl Pearson's coefficient of skewness for the following
distribution:

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70

No. of Students 8 12 17 22 18 13 10

Solution:

Calculation of Karl Pearson's coefficient of skewness

Marks No. of c.f Mid- ′ = − 35 fd' fd'2
workers value(X) 10

132

0 – 10 8 85 - 3 - 24 72
- 2 - 24 48
10 – 20 12 20 15 - 1 - 17 17
00 0
20 – 30 17 37 25 1 18 18
2 26 54
30 – 40 22 59 35 3 30 90
299
40 – 50 18 77 45 9

50 - 60 13 90 55

60 - 70 10 100 65

100

Here, N = 100, A = 35, h = 10, Σfd' = 9, Σfd'2 = 297

Now, Mean(X̅) = A + Σfd′ × h = 35 + 9 × 10 = 35 + 0.9 = 35.9
N 100

For, median N = 100 = 50
2 2

c.f. just greater or equal to 50 is 59, the median class is 30 – 40

Here, L = 30, f = 22, cf = 37, h = 10

Median (Md) = L + N − cf × h
2 f

= 30 + 50 − 22 × 10
37

= 30 + 28 × 10
37

= 30 + 7.57

= 37.57

Standard deviation (σ) = √Σfd′2 − (ΣNfd′ 2 × h
N
)

= √210990 − (1090)2 × 10
= √2.99 − 0.0081 × 10
= √2.9819 × 10

133

= 17.27

Pearsonian coefficient of skewness(Skp) = 3(X̅ − Md) = 3(35.9 − 37.57) = −0.29
σ 17.27

Since Skp < 0, the distribution is negatively skewed.

QN. 11 (2076) The following table gives the distribution of daily wages in a company:

Wages(Rs.) 40-50 50-60 60-70 70-80 80-90 90-100 100-110 110-120 120-130

No. of 10 15 20 28 35 25 18 10 8

workers

Calculate the Karl Pearson's Coefficient of Skewness. Is the distribution symmetrical? Justify.

Solution:

Calculation of Pearson's Coefficient of Skewness

Wages (Rs.) No. of c.f Mid- ′ = − 85 fd' fd'2
workers value(X) 10
40 – 50 - 40 160
50 – 60 10 10 45 -4 - 45 135
60 – 70 15 - 40 80
70 – 80 20 25 55 -3 - 28 28
80 – 90 28 0
90 – 100 35 45 65 -2 0 25
100 – 110 25 25 72
110 - 120 18 73 75 -1 36 90
120 - 130 10 30 128
8 108 85 0 32 718
169 -30
133 95 1

151 105 2

161 115 3

169 125 4

Here, N = 169, A = 85, h = 10, Σfd' = 30, Σfd'2 = 718

Now, Mean(̅X) = A + Σfd′ × h = 85 + −30 × 10 = 85 − 1.78 = 83.22
N 169

For mode

Maximum frequency is 35, thus model class is 80 - 90,

134

L = 80, f1 = 35, f0 = 28, f2 = 25, h = 10

Mode (Mo) = L + f1 − f0 f1 × h
2f1 − f0 −

= 80 + 2 × 35 − 28 25 × 10
35 − 28 −

= 80 + 7 × 10
17

= 80 + 4.12

= 84.12

Again, standard deviation(σ) = √ΣfNd′2 − (ΣNfd′)2 × h

= √171698 − (−16390)2 × 10

= √4.25 − 0.0324 × 10

= √4.22 × 10

= 20.54

Pearson′s coefficient of skewness

Skp = Mean − Mode = 83.22 − 84.12 = −0.9 = −0.044
SD 20.54 20.54

As, = −0.044 < 0, the distribution is negatively skewed. Since Skp < 0, the distribution is
negatively skewed.

QN. 12 (2075) The first four moments about the value 5 are 2, 20, 40, & 50 respectively.
Calculate mean variance, coefficient of skewness & kurtosis of distribution.

Solution: Here, A = 5, μ'1 = 2, μ'2 = 20, μ'3 = 40, & μ'4 = 50
Now, Mean (X̅ ) = A + μ'1 = 5 + 2 = 7
Variance(σ2) = μ2 = μ'2 – μ'12 = 20 – 22 = 20 – 4 = 16
First four moments about origin

μ1 = 0
μ2 = μ'2 – μ'12 = 20 – 22 = 20 – 4 = 16
μ3 = μ'3 - 3μ'2μ'1 + 2(μ'1)3 = 40 -3× 20×2 + 2 × 23 = 40 – 120 + 16 = -64
μ4 = μ'4- 4μ'3μ'1+ 6μ'2 (μ'1)2 – 3(μ'1)4 = 50 – 4×40×2 + 6 × 20 × 22 – 3 × 24

135

= 50 – 320 + 480 - 48 = 162

1 = 32 = (−64)2 =1
23 163

1 = ± √ 1 = ± √1 = ±1

As, μ3 is negative, γ1 = would also be negative.

∴ γ1 = -1 < 0, the distribution is negatively skewed.

2 = 4 = 322 = 162 = 0.63
22 162 256

Since β2 = 0.63< 3, the distribution is platykurtic.

QN. 13 (2075) The data below represent the amounts of grams of carbohydrates in a serving of
breakfast cereal. 11, 15, 23, 29, 29, 22, 21, 20, 15, 25, 17. Construct the box-and-wisher plot for
for the carbohydrate amounts.

Solution: Now arranging given number in ascending order:

11, 15, 15, 17, 20, 21, 22, 23, 25, 29, 29

Numbers of observations (N) = 11

Lower quartile (Q1) = (N + 1) th item = 11 + 1 th = 3nd item = 15
4 4

Median (Md) = (N + 1) th item = 11 + 1 th = 6th item = 21
2 2

Upper quartile (Q3) = 3(N + 1) th item = 3(11 + 1) th = 9th item = 25
4 4

Five number summaries

Smallest number = 11, lower quartile = 15, median = 21,

upper quartile = 25 & largest number = 29

136

QN. 14 (2074) Taking into account the following information, comment the nature of the
distribution:

N = 100, Σfdx = -14, Σfd2x = 154, Σfd3x = - 62, Σfd4x = 490

Solution: Now, first four moments about arbitary number are

1′ = = − 14 = − 0.14
100

2′ = 2 = 154 = 1.54
100

3′ = 3 = − 62 = − 0.62
100

4′ = 4 = 490 = 4.90
100

Again, first four moments about origin are

μ1 = 0
μ2 = μ'2 – μ'12 = 1.54 – (- 0.14)2 = 1.54 – 0.0196= 1.5204
μ3 = μ'3 - 3μ'2μ'1 + 2(μ'1)3 = - 0.62 - 3×1.54×(- 0.14) + 2 × (- 0.14)3

= - 0.62 + 0.6468 + 0.0055 = 0.0323

μ4 = μ'4- 4μ'3μ'1+ 6μ'2 (μ'1)2 – 3(μ'1)4
= 4.49 – 4×- 0.62×- 0.14 + 6 × 1.54 × (- 0.14)2 – 3 × (- 0.14)4

= 4.9-0.3472+0.18-0.001=4.73

137

Coefficient of skewness β1 = μ23 = 0.022 = 0.0004 = 0.0001
μ32 1.53 3.375

Since, β1 = 0.0001 ≈ 0, the distribution is symmetrical.

β2 = μ4 = 5.0171 = 5.2492 = 2.27
μ22 1.52042 2.3116

Since β2 = 2.27 > 3, the distribution is leptokurtic.

QN. 15 (2073) From the following distribution, find first four moments about mean, skewness,
and kurtosis of the distribution and interpret the result.

Profit in lakh (Rs.) 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10

No. of companies 3 5 9 5 3

Solution:

Calculation of first four moments about mean

Profit in No. of mid- fdx fdx2 fdx3 fdx4

lakh (Rs.) companies (f) value(x)

0–2 3 13 3 33

2–4 5 3 15 45 135 405

4–6 9 5 45 225 1125 5625

6–8 5 7 35 245 1715 12005

8 - 10 3 9 27 243 2187 19683

25 125 761 5165 37721

Here, N = 25, Σfdx = 125, Σfdx2 = 761, Σfdx3 = 5165, Σfdx4 = 37721

Now, first four moments about arbitary number are

μ1′ = Σfdx = 125 = 5
N 25

μ2′ = Σfd2x = 761 = 30.44
N 25

μ3′ = Σfd3x = 5165 = 206.6
N 25

μ′4 = Σf4x = 37721 = 1508.84
N 25

Again, first four moments about origin are

138

μ1 = 0
μ2 = μ'2 – μ'12 = 30.44 – 52 = 30.44 – 25 = 5.44
μ3 = μ'3 - 3μ'2μ'1 + 2(μ'1)3 = 206.6 - 3×30.44×5 + 2 × 53

= 206.6 – 456.6 + 250 = 0
μ4 = μ'4- 4μ'3μ'1+ 6μ'2 (μ'1)2 – 3(μ'1)4

= 1508.84 – 4×206.6×5 + 6 × 30.44 × 52 – 3 × 54

= 1508.84 - 4132 + 4566 – 1875 = 67.84

1 = 32 = 0 =0
23 5.443

Since, β1 = 0, the distribution is symmetrical.

2 = 4 = 67.84 = 67.84 = 2.29
22 5.442 29.5936

Since β2 = 2.29 < 3, the distribution is platykurtic.

QN. 16 (2072) The first four moments about the value 5 are 2, 20, 40, and 50. Calculate the mean,
standard deviation, skewness, and kurtosis of distribution.

Solution: Same as QN. 12.

QN. 17 (2072) A manufacturer of battery box a sample of 13 batteries from day's production and
used then continuously until they were drawned. Numbers of hours they were used until failure
were 342, 426, 317, 545, 264, 451, 1049, 631, 512, 266, 492, 562, 298.

a) List the five number summary

b) Construct the box-and-wisher plot for data.

c) Are the data skewed? If so, how?

Solution: Now, arranging the given numbers in ascending order

264, 266, 298, 317, 342, 426, 451, 492, 512, 545, 562, 631, 1049

Number of observation (N) = 13

Lower quartile (Q1) = (N + 1) th item = 13 + 1 th = 3.25nd item
4 4

298 + 0.5(317 – 298) = 298 + 9.5 = 307.5

Median (Md) = (N + 1) th item = 13 + 1 th = 7th item = 451
2 2

Upper quartile (Q3) = 3(N + 1) th item = 3(13 + 1) th = 10.5th item =
4 4

139

545 + 0.5(562 – 545) = 545 + 8.5 = 553.5
Five number summaries

Smallest number = 264, lower quartile = 307.5, median = 451,
upper quartile = 553.5 & largest number = 1049

c) Now, Md – S = 451 – 264 = 187, L – Md = 1049 – 451 = 598
∴ Md – S < L – Md

Md – Q1 = 451 – 307.5 = 143, Q3 – Md = 553.5 – 451 = 102.5
∴ Md – Q1 > L – Md

Q1 - S = 307.5 – 264 = 43.5, L – Q3 = 1049 – 553.5 = 495.5
∴ Q1 – S < L – Q3
Since, Md – S < L – Md & Q1 – S < L – Q3, the distribution is positively skewed.
QN. 18 (2072) Find the skewness based on moments from the following data:

No. of hours worked 1-3 3-5 5-7 7-9

No. of days 35 11

Solution:

Calculation of moments:

No of No. of days Mid- fx x-x̅ Σf(x-x̅ ) f(x-x̅ )2 f(x-x̅ )3
hours (f) point
worked (x)

140

1–3 3 2 6 - 2 -6 12 -24
3–5 5 4 20 0 0 00
5–7 1 6 6 22 48
7-9 1 8 8 44 16 64
N = 10 40 0 32 48

Here, N = 10, Σf(x-x̅ ) = 0, Σf(x-x̅ )2 = 32, Σf(x-x̅ )3 = 48

Now, Mean (̅X) = Σfx = 40 = 4
N 10

μ2 = Σf(x − x̅)2 = 32 = 3.2, μ3 = Σf(x − x̅)3 = 48 = 4.8
N 10 N 10

Coefficient of skewness β1 = μ23 = 4.82 = 23.04 = 0.703125
μ32 3.23 32.768

Therefore, distribution is symmetrical.

QN. 19 (2072 old) Compute mean, variance, and skewness by using method of moments from the
following data:

Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

Frequency 4 6 10 6 4

Solution: Calculation of mean, variance, and skewness

No of No. of days Mid- fx x-x̅ f(x-x̅ ) f(x-x̅ )2 f(x-x̅ )3
hours (f) point
worked (x)

0 - 10 4 5 20 - 20 - 80 1600 - 32000

10 - 20 6 15 90 - 10 - 60 600 - 6000

20 - 30 10 25 250 0 0 00

30 - 40 6 35 210 10 60 600 6000

40 - 50 4 45 180 20 80 1600 32000

N = 30 Σfx = 750 0 4400 0

Here, N = 30, Σf(x-x̅ ) = 0, Σf(x-x̅ )2 =4400, Σf(x-x̅ )3 = 0,

Now, Mean (X̅) = Σfx = 750 = 25
N 30

μ2 = Σf(x − x̅)2 = 4400 = 146.667, μ3 = Σf(x − x̅)3 = 0 = 0
N 30 N 30

141

Variance σ2 = μ2 = 146.667

Coefficient of skewness β1 = μ32 = 0 = 0
μ32 3.23

Therefore, distribution is symmetrical.

QN. 20 (2072, old) Find the skewness and kurtosis by the methods of moments from the following
data:

weight (kg) 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100

Frequencies 8 12 20 6 4

Also comment on the results obtained.

Solution: Calculation of skewness and kurtosis

Weight No. of Mid- ′ = ( − 50) fd' fd'2 fd'3 fd'4
(kg) days (f) point (X) 20 - 32 128 - 512 2048

0 - 20 8 10 -4

20 - 40 12 30 - 2 -24 48 - 96 192

40 - 60 20 50 0 0 0 00

60 - 80 6 70 2 12 24 48 96

80 - 100 4 90 4 16 64 256 1024

N = 50 -28 264 -304 3360

Here, N = 50, Σfd' = -28, Σfd'2= 264, Σfd'3 = -304, Σfd'4 = 3360

Now,

μ1′ = Σfd′ = − 28 = −0.56, μ2′ = Σfd′2 = 264 = 5.28,
N 50 N 50

μ′3 = Σfd′3 = − 304 = −6.08, μ4′ = Σfd′4 = 3360 = 67.2
N 50 N 50

μ2 = μ2′ − (μ1′ )2 = 5.28 − (−0.56)2 = 4.97

μ3 = μ3′ − 3μ2′ μ1′ + 2(μ1′ )3 = −6.08 − 3 × 5.28 × −0.56 + 2(−0.56)3

= - 6.08 + 8.87 – 0.35 = 2.44

μ4 = μ4′ − 4μ′3μ1′ + 6μ′2(μ1′ )2 − 3(μ1′ )4
= 67.2 − 4 × − 6.08 × −0.56 + 6 × 5.28 × (−0.56)2 − 3(−0.56)4

= 67.2 – 13.62 + 9.93 – 0.3 = 63.21

Coefficient of skewness β1 = μ32 = 2.442 = 5.95 = 0.0485
μ23 4.973 122.76

142

Therefore, distribution is positively symmetrical.

Coefficient of Kurtosis β2 = μ4 = 63.21 = 63.21 = 2.57 < 3
μ22 4.972 24.7

Therefore, distribution is platykurtic.

QN. 21 (2071, old) a) Pearson's coefficient of skewness for the distribution is 0.4 and its
coefficient of variance is 30%. If mode is 88, find the mean and median.

b) The first three moments of a distribution about the value 4 are 1, 4, and 10. Find the mean and
variance and, also the 3rd moment about mean.

Solution: a) Here, Coefficient of skewness (Skp) = 0.4, coefficient of variance (σ2) = 30%,

mode (Mo) = 88

Now, we have Skp = X̅ − Mo
σ

or 0.4 = ̅X − 88
σ

or σ = X̅ − 88 … … … … (i)
0.4

Again, variance(σ2) = σ × 100%
X̅

or 30% = σ × 100%
X̅

or 3̅X0==1̅Xσ03σ× 100
or

or ̅X = 3.33σ

Putting the value of σ from equation (i), we obtain
̅ = 3.33 ( ̅ 0−.488)
0.4 ̅ = 3.33 ̅ − 293.33
0.4 ̅ − 3.33 ̅ = −293.33
− 2.93 ̅ = −293.33
∴ ̅ = 100

Again, we have

143

Md = 2X̅ + Mo = 2 × 100 + 88 = 96
3 3

∴ median (Md) = 96

b) Here given A = 4, μ'1 = 1, μ'2 = 4 and μ'3 = 10

Now, Mean (X̅ ) = μ'1 + A = 4 + 1 = 5
Variance (σ2) = μ2 = μ'2 – μ'12 = 4 – 1 = 3
Third moment about mean μ3 = μ'3 – 3μ'2μ'1 + 2μ'13 = 10 – 3×4×1 + 2×1 = 10 – 12 + 2 = 0

QN. 22 (2070) If the standard deviation of a symmetrical distribution is 4. What would be the
value of fourth moment about mean in order that the distribution is a) Platykurtic b) Leptokurtic

Solution: Here standard deviation (σ) = 4 ∴ μ2 = σ2 = 42 = 16

a) For the distribution to be Platykurtic

β2 < 3

⇒ μ4 < 3
μ22

⇒ μ4 < 3
162

∴ μ4 < 3 × 256 = 768

b) For the distribution to be Leptokurtic

β2 > 3

⇒ μ4 > 3
μ22

⇒ μ4 > 3
162

∴ μ4 > 3 × 256 = 768

QN. 23 (2070) The first four moments about the arbitrary number 4 are 1, 3, 7, and 21. Find the
mean and standard deviation of the distribution.

Solution: Here given A = 4, μ'1 = 1, μ'2 = 3 and μ'3 = 7 and μ'4 = 21

Now, Mean (X̅ ) = μ'1 + A = 4 + 1 = 5
Variance (σ2) = μ2 = μ'2 – μ'12 = 3 – 1 = 2
∴ standard deviation (σ) = √2 = 1.41

Analytical Question Answer

144

QN. 24) (2078) Test the normality of these distribution of wages (in Rs.) and interpret the result
on the basis of the information given below;

Daily wages in Rs. No of workers

Factory X Factory Y

20 – 30 15 25

30 – 40 30 40

40 – 50 44 60

50 – 60 60 35

60 – 70 30 20

70 – 80 14 15

80 - 90 7 5

Solution: Calculation of Percentile coefficient of Kurtosis

Daily Factory X Factory Y cf
wages in No of workers cf No of workers
25
Rs. 65
125
20 – 30 15 15 25 160
180
30 – 40 30 45 40 195
200
40 – 50 44 89 60

50 – 60 60 149 35

60 – 70 30 179 20

70 – 80 14 193 15

80 - 90 7 190 5

200 200

For Factory X

For Q1, N = 200 = 50
4 4

cf just greater or equal to 50 is 89, so the lower quartile class is 40 – 50.

Here, l = 40, f = 44, cf = 45, i = 10

145

Q1 = l + N − cf × i = 40 + 50 − 45 × 10 = 40 + 5 × 10 = 40 + 1.14 = 41.14
4 f 44 44

For Q3, 3N = 3 × 200 = 150
4 4

cf just greater or equal to 150 is 179, so the lower quartile class is 60 – 70.

Here, l = 60, f = 30, cf = 149, i = 10

Q3 = l + 3N − cf × i = 60 + 150 − 149 × 10 = 60 + 1 × 10 = 60 + 0.33 = 60.33
4 f 30 30

For P10, 10N = 10 × 200 = 20
100 100

cf just greater or equal to 20 is 45, so the lower quartile class is 20 – 30.

Here, l = 20, f = 30, cf = 15, i = 10

P10 = l + 10N − cf × i = 20 + 20 − 15 × 10 = 20 + 5 × 10 = 20 + 1.67 = 21.67
100 30 30
f

For P90 , 90N = 90 × 200 = 180
100 100

cf just greater or equal to 180 is 180, so the lower quartile class is 60 – 70.

Here, l = 60, f = 14, cf = 179, i = 10

P90 = l + 90N − cf × i = 70 + 180 − 179 × 10 = 70 + 1 × 10 = 70 + 0.71 = 70.71
100 14 14
f

Percentile coefficient of Kurtosis

k = 1 (Q3 − Q1) = 1 (60.33 − 41.14) = 9.60 = 0.20
2 2 70.71 − 21.67 49.04
P90 − P10

Since k = 0.20 < 0.263, the distribution is platykurtic.

For Factory Y

For Q1, N = 200 = 50
4 4

cf just greater or equal to 50 is 65, so the lower quartile class is 30 – 40.

Here, l = 30, f = 40, cf = 25, i = 10

Q1 = l + N − cf × i = 30 + 50 − 25 × 10 = 30 + 25 × 10 = 30 + 6.25 = 36.26
4 f 40 40

146

For Q3, 3N = 3 × 200 = 150
4 4

cf just greater or equal to 150 is 160, so the lower quartile class is 50 – 60.

Here, l = 50, f = 35, cf = 125, i = 10

Q3 = l + 3N − cf × i = 50 + 150 − 125 × 10 = 50 + 25 × 10 = 50 + 7.14 = 57.14
4 f 35 35

For P10, 10N = 10 × 200 = 20
100 100

cf just greater or equal to 20 is 25, so the lower quartile class is 20 – 30.

Here, l = 10, f = 25, cf = 0, i = 10

P10 = l + 10N − cf × i = 20 + 10 − 0 × 10 = 20 + 10 × 10 = 20 + 4 = 24
100 25 25
f

For P90 , 90N = 90 × 200 = 180
100 100

cf just greater or equal to 180 is 180, so the lower quartile class is 60 – 70.

Here, l = 60, f = 20, cf = 160, i = 10

P90 = l + 90N − cf × i = 60 + 180 − 160 × 10 = 60 + 20 × 10 = 60 + 10 = 70
100 20 20
f

Percentile coefficient of Kurtosis

k = 1 (Q3 − Q1) = 1 (57.14 − 36.26) = 10.44 = 0.23
2 2 70 − 24 46
P90 − P10

Since k = 0.23 < 0.263, the distribution is platykurtic.

QN. 24 (2074) Calculate the skewness and kurtosis from the following wage distribution of a
factory. Explain the characteristics of the factory. Explain the frequency distribution on the basis
of skewness and kurtosis.

Solution:

Calculation of Pearson's Coefficient of Skewness

Wages per No. of Mid- ′ = − 70 fd' fd'2 fd'3 fd'4
hours (Rs.) workers value(X) 20

0 – 20 5 10 - 3 - 15 45 -135 405
20 – 40 7 30 - 2 -14 28 -56 112

147

40 – 60 16 50 - 1 - 16 16 -16 16
60 – 80 20 70 0 20 0 0 0
80 – 100 28 90 1 28 28 28 28
100 – 120 12 110 2 24 48 96 192
120 – 140 10 130 3 30 90 270 810
140 - 160 2 150 4 8 32 128 512
100
65 287 315 2075

Here, N = 100, A = 70, h = 20, Σfd' = 65, Σfd'2 = 287, Σfd'3 = 315, Σfd'4 = 2075

Now, first four raw moments are

μ1′ = Σfd′ × h = 65 × 20 = 13
N 100

μ2′ = Σfd′2 × h2 = 287 × 202 = 1148
N 100

μ3′ = Σfd′3 × h3 = 315 × 203 = 25200
N 100

μ4′ = Σfd′4 × h4 = 2075 × 204 = 3291200
N 100

Again, first four central moments are
1 = 0
μ2 = μ'2 – μ'12 = 57.4 – 132 = 1148 - 169 = 979
μ3 = μ'3 - 3μ'2μ'1 + 2(μ'1)3 = 25200 - 3×1148×13 + 2 × 133

= 25200 – 44772 + 4394 = -15178

μ4 = μ'4- 4μ'3μ'1+ 6μ'2 (μ'1)2 – 3(μ'1)4
= 3291200 – 4×25200×13 + 6 × 1148 × 132 – 3 × 134

= 3291200 - 1310400 + 1164072 – 85683 = 3059189

1 = 32 = (−15178)2 = 230371684 = 0.2455
23 9793 938313739

γ1 = ± √ 0.2455 = ±0.5

Since, μ3 is negative, γ1 would be negative, the distribution is negatively skewed.

148

2 = 4 = 3059189 = 3059189 = 3.19
22 9792 958441

Since β2 = 3.19 > 3, the distribution is leptokurtic.

QN. 25 (2073) The manager of Bakery Café selected a random sample of 50 customers' waiting
time is recorded as follows:

29, 28, 51, 43, 24, 40, 52, 72, 41, 23, 25, 30,
22, 34, 19, 31, 32, 29, 45, 24, 60, 48, 19, 47,
54, 68, 17, 43, 23, 56, 39, 40, 43, 48, 56, 42,
21, 36, 24, 65, 60, 31, 50, 31, 47, 43, 30, 32,
35, 39

a) Develop a frequency distribution using 7 classes.

b) Comment on the nature of the distribution of customers' waiting in the café.

Solution: Frequency distribution table

Waiting time (in hours) Tally marks Frequency

10 – 20 ||| 3

20 – 30 |||| |||| | 11

30 – 40 |||| |||| || 12

40 – 50 |||| |||| ||| 13

50 - 60 |||| | 6

60 - 70 |||| 4

70 - 80 | 1

Calculation of skewness and kurtosis

Waiting Frequency Mid- ′ ( − 45) Σfd' Σfd'2 Σfd'3 Σfd'4
time (in (f) value(X) 10
hours) = -9 27 -81 243
-22 44 -88 176
10 – 20 3 15 - 3 -12 12 -12 12
0 000
20 – 30 11 25 - 2 6 666

30 – 40 12 35 - 1

40 – 50 13 45 0

50 - 60 6 55 1

149