NCERT Solutions for Class 11 Maths Chapter 13 Statistics

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 15
Chapter Name : Statistics

Exercise 15.1

Q1 Find the mean deviation about the mean for the data
4, 7, 8, 9, 10, 12, 13, 17

Answer. The given data is
4, 7, 8, 9, 10, 12, 13, 17
Mean of the data, x = ¯
¯¯ 4+7+8+9+10+12+13+17

8
=
80

8
= 10

The deviations of the respective observations from the mean x, i.e. x − x are ¯
¯¯
i
¯
¯¯

–6, – 3, –2, –1, 0, 2, 3, 7
The absolute values of the deviations, i.e. ∣∣x − x∣∣, are i
¯
¯¯

6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is
8
¯
¯¯
∑ ∣
∣xi −x ∣
∣ 6+3+2+1+0+2+3+7
¯
¯¯ i=1 24
M. D.(x) = = = = 3
8 8 8

Page : 360 , Block Name : Exercise 15.1

Q2 Find the mean deviation about the mean for the data
38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Answer. The given data is
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the given data,
38+70+48+40+42+55+63+46+54+44 500
¯
¯¯
x = = = 50
10 10

The deviations of the respective observations from the mean x, i.e. x − x are ¯
¯¯
i
¯
¯¯

–12, 20, –2, –10, –8, 5, 13, –4, 4, –6
The absolute values of the deviations, i.e. ∣∣x − x∣∣ , are i
¯
¯¯

12, 20, 2, 10, 8, 5, 13, 4, 4, 6
The required mean deviation about the mean is

10
∣ ¯
¯¯
∑
i=1
∣ xi − x ∣
∣
¯
¯¯
M. D.(x) =
10

12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6
=
10

84
=
10

= 8.4

Page : 360 , Block Name : Exercise 15.1

Q3 Find the mean deviation about the median for the data
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Answer. The given data is
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Here, the numbers of observations are 12, which is even.
Arranging the data in ascending order, we obtain
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
th th
12 12
( ) observation +( +1) observation
2 2

M =
2
th th
6 observation + 7 observation
=
2
13 + 14 27
= = = 13.5
2 2
The deviations of the respective observations from the median, i.e. xi − M ,

are

−3.5, −2.5, −2.5, −1.5, −1.5, −0.5, −0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

The absolute values of the deviations, |xi − M|
are

3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The required mean deviation about the median is
12
∑ |xi − M|
i=1
M. D.(M) =
12
3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5
=
12
28
= = 2.33
12

Page : 360 , Block Name : Exercise 15.1

Q4 Find the mean deviation about the median for the data
36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Answer. The given data is
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Here, the number of observations is 10, which is even.

Arranging the data in ascending order, we obtain
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
th th
10 10
( ) observation + ( + 1) observation
2 2

Median M = observation
2
th ′h
5 observation + 6 observation
=
2
46 + 49 95
= = = 47.5
2 2
The deviations of the respective observations from the median, i.e. xi − M ,

are

−11.5, −5.5, −2.5, −1.5, −1.5, −1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations, |xi − M|
are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Thus, the required mean deviation about the median is
10
∑ |xi − M| 11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5
i=1
MD ⋅ (M) = =
10 10

70
= = 7
10

Page : 360 , Block Name : Exercise 15.1

Q5 Find the mean deviation about the mean for the data
xi 5 10 15 20 25

fi 7 4 6 3 5

Answer.

5
N = ∑ fi = 25
i=1

5
∑ fi xi = 350
i=1

1 5 1
¯
¯¯
∴ x = ∑ f i xi = × 350 = 14
N i=1 25

1 5 1
¯
¯¯ ¯
¯¯
∴ MD(x) = ∑ f i ∣ xi − x ∣ = × 158 = 6.32
N i=1 25

Page : 360 , Block Name : Exercise 15.1

Q6 Find the mean deviation about the mean for the data
xi 10 30 50 70 90

fi 4 24 28 16 8

Answer.

5 3
N = ∑ fi = 80, ∑ fi xi = 4000
i=1 i=1

1 5 1
¯
¯¯
∴ x = ∑ f i xi = × 4000 = 50
N i=1 80

1 5 1
¯
¯¯ ¯
¯¯
MD(x) ∑ f i ∣ xi − x ∣ = × 1280 = 16
N i=1 80

Page : 360 , Block Name : Exercise 15.1

Q7 Find the mean deviation about the median for the data
xi 5 7 9 10 12 15

fi 8 6 2 2 2 6

Answer. The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the
following table.

Here, N = 26, which is even.
Median is the mean of 13 and 14 observations. th th

Both of these observations lie in the cumulative frequency 14, for which the corresponding
observation is 7.
th th
13 observation +14 observation 7+7
∴ Median = = = 7
2 2

The absolute values of the deviations from median, i.e. |xi − M| are

6 6
∑ fi = 26 ∑ ∑ fi |xi − M| = 84
i=1 and i=1

1 6 1
M. D.(M) = ∑ fi |xi − M| = × 84 = 3.23
N i=1 26

Page : 360 , Block Name : Exercise 15.1

Q8 Find the mean deviation about the median for the data
xi 15 21 27 30 35

fi 3 5 6 7 8

Answer. The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the
following table.

Here, N = 29, which is odd.
th

observation = 15 observation
29+1 th
∴ Median = ( )
2

This observation lies in the cumulative frequency 21, for which the corresponding observation is
30.
∴ Median = 30
The absolute values of the deviations from median, i.e. |x − M| are i

5 5
∑ fi = 29, ∑ fj |xi − M| = 148
i=1 i=1

1 5 1
∴ M. D.(M) = ∑ fi |xi − M| = × 148 = 5.1
N i=1 29

Page : 360 , Block Name : Exercise 15.1

Q9 Find the mean deviation about the mean for the data.

Answer. The following table is formed.

8 8
N = ∑ fi = 50, ∑ fi xi = 17900
i=1 i=1

1 8 1
¯
¯¯
∴ x = ∑ f i xi = × 17900 = 358
N i=1 50

1 8 1
¯
¯¯ ¯
¯¯
M. D.(x) = ∑ f i ∣ xi − x ∣ = × 7896 = 157.92
N i=1 50

Page : 361 , Block Name : Exercise 15.1

Q10 Find the mean deviation about the mean for the data

Answer. The following table is formed.

6 6
N = ∑ fi = 100, ∑ fi xi = 12530
i=1 i=1

1 6 1
¯
¯¯
∴ x = ∑ f i xi = × 12530 = 125.3
N i=1 100

1 6 1
¯
¯¯ ¯
¯¯
M. D.(x) = ∑ f i ∣ xi − x ∣ = × 1128.8 = 11.28
N i=1 100

Page : 361 , Block Name : Exercise 15.1

Q11 Find the mean deviation about median for the following data:

Answer. The following table is formed.

′h

The class interval containing the ( or 25th item is 20- 30.
N
)
2

Therefore, 20 - 30 is the median class.
It is known that,
N
−C
2
Median = l + × h
f

Here, l = 20, C = 14, f = 14, h = 10, and N = 50
Medium = 20 +
25−14 110
× 10 = 20 + = 20 + 7.85 = 27.85
14 14

Thus, mean deviation about the median is given by,
1 6 1
M. D.(M) = ∑ fi |xi − M| = × 517.1 = 10.34
N i=1 50

Page : 361 , Block Name : Exercise 15.1

Q12 Calculate the mean deviation about median age for the age distribution of 100 persons given
below:

Answer. The given data is not continuous. Therefore,
it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower
limit and adding 0.5 to the upper limit of each class interval.
The table is formed as follows.

The class interval containing theor 50th item is 35.5 – 40.5.
Therefore, 35.5 – 40.5 is the median class.
It is known that,
N
−C
2
Median = l + × h
f

Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100
50−37 13×5
∴ Median = 35.5 + × 5 = 35.5 + = 35.5 + 2.5 = 38
26 26

Thus, mean deviation about the median is given by,/

1 g 1
M. D.(M) = ∑ ft |xi − M| = × 735 = 7.35
N i=1 100

Page : 361 , Block Name : Exercise 15.1

Exercise 15.2

Q1 Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12

Answer. 6, 7, 10, 12, 13, 4, 8, 12
8
∑i=1 xi
Mean ,x =
6+7+10+12+13+4+8+12 72
¯
¯¯
= = = 9
n 8 8

The following table is obtained.

2
Varience (σ ) = 2 1 8 1
¯
¯¯
∑ (xi − x) = × 74 = 9.25
n i=1 8

Page : 371 , Block Name : Exercise 15.2

Q2 Find the mean and variance for the data rst n natural numbers

Answer. The mean of rst n natural numbers is calculated as follows.
Sum of all observations
Mean =
Number of observations

n(n+1)

2 n+1
∴ Mean = =
n 2

n 2
2 1 ¯
¯¯
Variance (σ ) = ∑ (xi − x)
n i=1

2
1 n n+1
= ∑ [xi − ( )]
n i=1 2

2
1 n 2 1 n n+1 1 n n+1
= ∑ x − ∑ 2( ) xi + ∑ ( )
n i=1 i n i=1 2 n i=1 2

2
1 n(n+1)(2n+1) n+1 n(n+1) (n+1)
= − ( )[ ] + × n
n 6 n 2 4n

2 2
(n+1)(2n+1) (n+1) (n+1)
= − +
6 2 4
2
(n+1)(2n+1) (n+1)
= −
6 4

4n+2−3n−3
= (n + 1) [ ]
12

(n+1)(n−1)
=
12

2
n −1
=
12

Page : 371 , Block Name : Exercise 15.2

Q3 Find the mean and variance for the data rst 10 multiples of 3

Answer. The rst 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10
10
∑ xi
Mean x = ¯
¯¯ i=1

10
=
165

10
= 16.5

The following table is obtained.

2
Varience (σ ) = 2 1 10 1
¯
¯¯
∑ (xi − x) = × 742.5 = 74.25
n i=1 10

Page : 371 , Block Name : Exercise 15.2

Q4 Find the mean and variance for the data

Answer. The data is obtained in tabular form as follows.

7
∑ fi xi = 760
i=1

7

Here , N = 40 ∴ x =
∑ f i xi 760
i=1
¯
¯¯
= = 19
N 40

1 7 2 1
2 ¯
¯¯
Variance = (σ ) = ∑ fi (xi − x) = × 1736 = 43.4
N i=1 40

Page : 371 , Block Name : Exercise 15.2

Q5 Find the mean and variance for the data

Answer. The data is obtained in tabular form as follows.

7
∑ fi xi = 2200
1=1

7
Here , N = 22 ∴ x =
¯
¯¯ 1 1
∑ f i xi = × 2200 = 100
N i=1 22

1 7 2 1
2 ¯
¯¯
Variance (σ ) = ∑ fi (xi − x) = × 640 = 29.09
N i=1 22

Page : 371 , Block Name : Exercise 15.2

Q6 Find the mean and standard deviation using short-cut method.

Answer. The data is obtained in tabular form as follows.

9
∑ fi y
Mean , x = A
¯
¯¯ i=1

N
i
× h = 64 +
0

100
× 1 = 64 + 0 = 64

2
2 9 9
h ⎡ ⎤
2 2
Variance , σ = N ∑ fi y − ( ∑ fi yi )
2 i
N ⎣ ⎦
i=1 i=1

1
= [100 × 286 − 0]
2
100

= 2.86

∴ S tan dard deviation (σ) = √2.86 = 1.69

Page : 371 , Block Name : Exercise 15.2

Q7 Find the mean and variance for the following frequency distribution.

Answer.

Mean,

7
∑ f i yi 2
i=1
¯
¯¯
x = A + × h = 105 + × 30 = 105 + 2 = 107
N 30
2
2 7 7
h ⎡ ⎤
2 2
Variance (σ ) = N ∑ fi y − (∑ f i y )
2 i i
N ⎣ ⎦
i=1 i=1

2
(30)
2
= [30 × 76 − (2) ]
2
(30)

= 2280 − 4

= 2276

Page : 371 , Block Name : Exercise 15.2

Q8 Find the mean and variance for the following frequency distribution.

Answer.

5
∑ f 1 yi
Mean , x = A +
¯
¯¯ i=1

N
× h = 25 +
10

50
× 10 = 25 + 2 = 27

2
2 5 5
h ⎡ ⎤
2 2
Variance (σ ) = N ∑ fi y − ( ∑ fi yi )
2 i
N ⎣ ⎦
i=1 i=1

2
(10)
2
= [50 × 68 − (10) ]
2
(50)

1 3300
= [3400 − 100] =
25 25

= 132

Page : 372 , Block Name : Exercise 15.2

Q9 Find the mean, variance and standard deviation using short-cut method

Answer.

q
∑ fi yi
Mean , x = A +
¯
¯¯ i=1

N
× h = 92.5 +
6

60
× 5 = 92.5 + 0.5 = 93

2
2 9 9
h ⎡ ⎤
2 2
Variance (σ ) = N ∑ fi y − ( ∑ fi yi )
2 i
N ⎣ ⎦
i=1 i=1

2
(5)
2
= [60 × 254 − (6) ]
2
(60)
25
= (15204) = 105.58
3600

∴ Stan dard deviation (σ) = √105.58 = 10.27

Page : 372 , Block Name : Exercise 15.2

Q10 The diameters of circles (in mm) drawn in a design are given below:

Answer.

Here, N = 100, h = 4

Let the assumed mean, A, be 42.5.
5
∑ fi yi
Mean , x = A +
¯
¯¯ i=1

N
× h = 42.5 +
25

100
× 4 = 43.5

2
2 s s
h ⎡ ⎤
2 2
Variance (σ ) = N ∑ fi y − ( ∑ fi yi )
2 i
N ⎣ ⎦
i=1 i=1

16
2
= [100 × 199 − (25) ]
10000

16
= [19900 − 625]
10000

16
= × 19275
10000

= 30.84

∴ Stan dard deviation (σ) = 5.55

Page : 372 , Block Name : Exercise 15.2

Exercise 15.3

Q1 From the data given below state which group is more variable, A or B?

Answer. Firstly, the standard deviation of group A is calculated as follows.

Here, h = 10, N = 150, A = 45
7
∑ xi (−6)×10
i=1
Mean = A + × h = 45 + = 45 − 0.4 = 44.6
N 150

2
2
2 h 7 2 7
σ = (N ∑ fi y − (∑ fi y ) )
1 2 i=1 i i=1 i
N

100
2
= (150 × 342 − (−6) )
22500

1
= (51264)
225

= 227.84

∴ Stan dard deviation (σ1 ) = √227.84 = 15.09

The standard deviation of group B is calculated as follows.

7
∑ fi y (−6)×10
i=1 i
Mean = A + × h = 45 + = 45 − 0.4 = 44.6
N 150

2
2
2 h 7 2 7
σ = [N ∑ fi y − (∑ fi y ) ]
2 2 i=1 i i=1 i
N

100
2
= [150 × 366 − (−6) ]
22500

1
= [54864] = 243.84
225

∴ Stan dard deviation (σ2 ) = √243.84 = 15.61

Since the mean of both the groups is same, the group with greater standard deviation will be more
variable.
Thus, group B has more variability in the marks.

Page : 375 , Block Name : Exercise 15.3

Q2 From the prices of shares X and Y below, nd out which is more stable in value:
x 35 54 52 53 58 52 50 51 49

Y 108 107 105 105 106 104 104 104 101

Answer. The prices of the shares X are
35, 54, 52, 53, 56, 58, 52, 50, 51, 49
Here, the number of observations, N = 10
1 10 1
¯
¯¯
∴ Mean, x = ∑ xi = × 510 = 51
N i=1 10

The following table is obtained corresponding to shares X.

2 1 10 1
¯¯ 2
¯
Variance (σ ) = ∑ (xi − x) = × 350 = 35
1 N i=1 10

∴ Stan dard deviation (σ1 ) = √35 = 5.91

σ1 5.91
C.V. (Shares X) = × 100 = × 100 = 11.58
x 51

The prices of share Y are
108, 107, 105, 105, 106, 107, 104, 103, 104, 101
1 10 1
¯
¯¯
∴ Mean, y = ∑ yi = × 1050 = 105
N i=1 10

The following table is obtained corresponding to shares Y.

1 10 2 1
2 ¯
¯¯
Variance (σ ) = ∑ (yi − y ) = × 40 = 4
2 N i=1 10

∴ Stan dard deviation (σ2 ) = √4 = 2

σ2 2
∴ C.V. ( Shares Y ) = × 100 = × 100 = 1.9
y 105

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Thus, the prices of shares Y are more stable than the prices of shares X.

Page : 375 , Block Name : Exercise 15.3

Q3 An analysis of monthly wages paid to workers in two rms A and B, belonging to the same
industry, gives the following results:

(i) Which rm A or B pays larger amount as monthly wages?
(ii) Which rm, A or B, shows greater variability in individual wages?

Answer. (i) Monthly wages of rm A = Rs 5253
Number of wage earners in rm A = 586
∴Total amount paid = Rs 5253 × 586
Monthly wages of rm B = Rs 5253
Number of wage earners in rm B = 648
∴Total amount paid = Rs 5253 × 648
Thus, rm B pays the larger amount as monthly wages as the number of wage earners in rm B are
more than the number of wage earners in rm A.

2

(ii) Variance of the distribution of wages in rm A = 100
(σ )
1

∴ Standard deviation of the distribution of wages in rm
A ((σ1 ) = √100 = 10

2
variance of the distribution of wages in firm B(σ )
2 =121

2
∴ Standard deviation of the distribution of wages in firm B (σ ) = √121 = 11
2

The mean of monthly wages of both the rms is same i.e., 5253. Therefore, the rm with greater
standard deviation will have more variability.
Thus, rm B has greater variability in the individual wages.

Page : 375 , Block Name : Exercise 15.3

Q4 The following is the record of goals scored by team A in a football session:

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25
goals.
Find which team may be considered more consistent?

Answer. The mean and the standard deviation of goals scored by team A are calculated as follows.

5
∑ f 1 xi
i=1 50
= 5
= = 2
25
∑i=1 fi

Thus, the mean of both the teams is same.
2
1
σ = √N ∑ f x2 − (∑ f x )
i i i i
N

1
2
= √25 × 130 − (50)
25

1
= √750
25

1
= × 27.38
25

= 1.09

The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same i.e., 2.
Therefore, the team with lower standard deviation will be more consistent.
Thus, team A is more consistent than team B.

Page : 376 , Block Name : Exercise 15.3

Q5 The sum and sum of squares corresponding to length x (in cm) and weight y
(in gm) of 50 plant products are given below:
50 50 2 50 50 2
∑ xi = 212, ∑ x = 902.8, ∑ yi = 261, ∑ y = 1457.6
i=1 i=1 i i=1 i=1 i

Which is more varying, the length or weight?

Answer. ∑
50 50 2
xi = 212, ∑ x = 902.8
i=1 i=1 i

Here , N = 50
50
∑ yi
Mean , x = ¯
¯¯ i=1

N
=
212

50
= 4.24

50
1 2
2 ¯
¯¯
Variance (σ ) = ∑ (xi − x)
1
N
i=1

50
1 2
= ∑ (xi − 4.24)
50
i=1

50
1
2
= ∑ [x − 8.48xi + 17.97]
i
50
i=1

50 50
1
2
= [∑ x − 8.48 ∑ xi + 17.97 × 50]
i
50
i=1 i=1

1
= [902.8 − 8.48 × (212) + 898.5]
50

1
= [1801.3 − 1797.76]
50

1
= × 3.54
50

= 0.07

∴ Stan dard deviation, σ1 ( Length ) = √0.07 = 0.26

Standard deviation 0.26
∴ C.V. ( Length ) = × 100 = × 100 = 6.13
Mean 4.24

50 50 2
∑ y = 261, ∑ y = 1457.6
i=1 i i=1 1

Mean , y =
1 50 1
¯
¯¯
∑ yi = × 261 = 5.22
N i=1 50

30
1 2
2 ¯
¯¯
Variance (σ ) = ∑ (yi − y )
2
N
i=1

50
1 2
= ∑ (yi − 5.22)
50
i=1

50
1
2
= ∑ [y − 10.44yi + 27.24]
i
50
i=1

50 50
1
2
= [∑ y − 10.44 ∑ yi + 27.24 × 50]
i
50
i=1 i=1

1
= [1457.6 − 10.44 × (261) + 1362]
50

1
= [2819.6 − 2724.84]
50

1
= × 94.76
50

= 1.89

∴ Stan dard deviation, σ2 ( Weight ) = 1.37

Stan dard deviation 1.37
∴ C.V. ( Weight ) = × 100 = × 100 = 26.24
Mean 5.22

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the
lengths.

Page : 376 , Block Name : Exercise 15.3

Miscellaneous Exercise

Q1 The mean and variance of eight observations are 9 and 9.25, respectively.
If six of the observations are 6, 7, 10, 12, 12 and 13, nd the remaining two observations.

Answer. Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.
6+7+10+12+12+13+x+y
¯
¯¯
Mean, x = = 9
8

⇒ 60 + x + y = 72

⇒ x + y = 12 . . . (1)

1 8 2
¯
¯¯
Variance = 9.25 = ∑ (xi − x)
n i=1

1 2 2 2 2 2 2 2 2 2
9.25 = [(−3) + (−2) + (1) + (3) + (3) + (4) + x + y − 2 × 9(x + y) + 2 × (9) ]
8

1 2 2
9.25 = [9 + 4 + 1 + 9 + 9 + 16 + x + y − 18(12) + 162] [using (1)]
8

1 2 2
9.25 = [48 + x + y − 216 + 162]
8

1 2 2
9.25 = [x + y − 6]
8

2 2
⇒ x + y = 80 . . . (2)

From (1), we obtain

2 2
x + y + 2xy = 144 … (3)

From (2) and (3), we obtain

2xy = 64 … (4)

Subtracting ( 4 ) from (2), we obtain

2 2
x + y − 2xy = 80 − 64 = 16

⇒ x – y = ± 4 … (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 4, when x – y = 4
x = 4 and y = 8, when x – y = –4
Thus, the remaining observations are 4 and 8.

Page : 380 , Block Name : Miscellaneous Exercise

Q2 The mean and variance of 7 observations are 8 and 16, respectively.
If ve of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

Answer. Let the remaining two observations be x and y.
The observations are 2, 4, 10, 12, 14, x, y.

2+4+10+12+14+x+y
¯
¯¯
Mean, x = = 8
7

⇒ 56 = 42 + x + y

⇒ x + y = 14 . . . (1)

1 7 2
¯
¯¯
Variance = 16 = ∑ (xi − x)
n i=1

1 2
2 2 ¯
¯¯ 2 2 2 2
16 = [(−6) + (−4) + (xi − x) + (6) + x + y − 2 × 8(x + y) + 2 × (8) ]
7

1 2 2 2
16 = [(−6) + 4 + 16 + 36 + x + y − 16(14) + 2(64)]
7

1 2 2
16 = [36 + 16 + 4 + 16 + 36 + x + y − 16(14) + 2(64)] [using (1)]
7
1 2 2
16 = [108 + x + y − 224 + 128]
7

1 2 2
16 = [12 + x + y ]
7

2 2
⇒ x + y = 112 − 12 = 100

2 2
x + y = 100 … (2)

From ( 1), we obtain

2 2
x + y + 2xy = 196 … (3)

From (2) and (3), we obtain

2xy = 196 − 100

⇒ 2xy = 96 … (4)

Subtracting (4) from (2), we obtain

2 2
x + y − 2xy = 100 − 96

⇒ x – y = ± 2 … (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 6 when x – y = 2
x = 6 and y = 8 when x – y = – 2
Thus, the remaining observations are 6 and 8.

Page : 380 , Block Name : Miscellaneous Exercise

Q3 The mean and standard deviation of six observations are 8 and 4, respectively.
If each observation is multiplied by 3, nd the new mean and new standard deviation of the
resulting observations.

Answer.
Let the observations be x1 , x2 , x3 , x4 , x5 , and x6 .

It is given that mean is 8 and standard deviation is 4 .

x1 +x2 +x3 +x4 +x5 +x6
¯
¯¯
Mean, x = = 8
6

If each observation is multiplied by 3 and the resulting observations are yi, then

1
yi = 3xi i.e.. xi = yi , for i = 1 to 6
3
y1 + y2 + y3 + y4 + y5 + y6
¯
¯¯
∴ New mean, y =
6

3 (x1 + x2 + x3 + x4 + x5 + x6 )
=
6

= 3 × 8 [using (1)]

= 24

1 6 2
Standard deviation, σ = √
¯
¯¯
∑ (xi − x)
n i=1

1 6 2
2 ¯
¯¯
∴ (4) = ∑ (xj − x)
6 i=1

6 2
¯
¯¯
∑ (xi − x) = 96 . . . (2)
i=1

From (1) and (2), it can be observed that,

¯
¯¯ ¯
¯¯
y = 3x

¯
¯¯ 1 ¯
¯¯
x = y
3

¯
¯¯
Substituting the values of x i and x in (2), we obtain

2
6 1 1
∑ ( yi − y) = 96
i=1 3 3

6 2
¯
¯¯
⇒ ∑ (yi − y ) = 864
i=1

Therefore, variance of new observations = ( 1

6
× 864) = 144

Hence, the standard deviation of new observations is √144 = 12

Page : 380 , Block Name : Miscellaneous Exercise

Q4
¯
¯¯ 2
Given that x is the mean and σ is the variance of n observations x1 , x2 , … . xn .

Prove that the mean and variance of the observations ax1 , ax2 , ax3 , … … , axn are

¯
¯¯ 2 2
ax and a σ , respectively, (a ≠ 0)

Answer.
The given n observations are x1 , x2 … xn

¯
¯¯
Mean = x

2
Variance = σ

1 n 2
2 ¯
¯¯
∴ σ = ∑ yi (xi − x) . . . (1)
n i=1

If each observation is multiplied by a and the new observations are yi, then
1
yi = axi i.e., xi = yi
a

1 n 1 n a n 1 n
¯
¯¯ ¯
¯¯ ¯
¯¯
∴ y = ∑ yi = ∑ axi = ∑ xi = a x (x = ∑ xi )
n i=1 n i=1 n i=1 n i=1

Therefore, mean of the observations, ax , ax … ax 1 2 n, is

¯
¯¯
Substituting the values of x i and x in (1), we obtain

2
2 1 n 1 1
σ = ∑ ( yi − y)
n i=1 a a

1 n 2
2 2 ¯
¯¯
⇒ a σ = ∑ (yi − y )
n i=1

Thus, the variance of the observations, ax , ax … ax , is a σ 1 2 n
2 2

Page : 380 , Block Name : Miscellaneous Exercise

Q5 The mean and standard deviation of 20 observations are found to be 10 and 2,respectively.
On rechecking, it was found that an observation 8 was incorrect.
Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.

Answer. (i) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2
1 20
¯
¯¯
x = ∑ xi
n i=1

1 20
10 = ∑ xi
20 i=1

20
⇒ ∑ xi = 200
i=1

That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
∴ Correct mean =
Correct sum 192
= = 10.1
19 19

Standard deviation σ = √ 1 n 1 n n 2 1 1
2 2 2
∑ x − (∑ (∑ ) = √ x − (x)
n i=1 1 n
2 i=1 i=1 n 4

1 n 2 2
⇒ 2 = √ Incorrect ∑ x − (10)
20 l=1 1

1 n 2
⇒ 4 = Incorrect ∑ x − 100
20 k=1

n 2
⇒ incorrect ∑ x = 2080
H =1
n

2 2 2
∴ Correct ∑ x = Incorrect ∑ x − (8)
1 i

k=1

= 2080 − 64

= 2016
2
Correct ∑ x
= √
2
∴ Correct standard deviation − ( Correct mean )
n

2016
2
= √ − (10.1)
19

= √1061.1 − 102.1

= √4.09

= 2.02

(ii) When 8 is replaced by 12,
Incorrect sum of observations = 200

∴ Correct sum of observations = 200 – 8 + 12 = 204
Correct sum 204
∴ Correct mean = = = 10.2
20 20

1 n 1 n 2 1 n
2 2 ¯
¯¯ 2
Standard deviation σ = √ ∑ x − (∑ xi ) = √ ∑ x − (x)
n i=1 i n
2 i=1 n i=1 i

1 n 2 2
⇒ 2 = √ Incorrect ∑ x − (10)
20 i=1 i

n
1
2
⇒4 = Incorrect ∑ x − 100
i
20
i=1

n

2
⇒ Incorrect ∑ x = 2080
i

i=1

n n

2 2 2 2
∴ Correct ∑ x = Incorrect ∑ x − (8) + (12)
i i

i=1 i=1

= 2080 − 64 + 144

= 2160

2
Correct ∑ x
i 2
∴ Correct standard deviation = √ − ( Correct mean )
n

2160
2
= √ − (10.2)
20

= √108 − 104.04

= 13.96

= 1.98

Page : 380 , Block Name : Miscellaneous Exercise

Q6 The mean and standard deviation of marks obtained by 50 students of a class in three subjects,
Mathematics, Physics and Chemistry are given below:

Which of the three subjects shows the highest variability in marks and which shows the lowest?

Answer. Standard deviation of Mathematics = 12
Standard deviation of Physics = 15
Standard deviation of Chemistry = 20
The coef cient of variation (C.V.) is given by .
Standard deviation
× 100
Mean

12
C.V. (in Mathematics )= × 100 = 28.57
42

15
C.V. (in Physics) = × 100 = 46.87
32

20
C.V. (in Chemistry )= × 100 = 48.89
40.9

The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in
Mathematics.

Page : 380 , Block Name : Miscellaneous Exercise

Q7 The mean and standard deviation of a group of 100 observations were found to be 20 and 3,
respectively.
Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18.

Find the mean and standard deviation if the incorrect observations are omitted.

Answer. Number of observations (n) = 100
¯
¯¯
Incorrect mean (x) = 20

Incorrect standard deviation (σ) = 3

1 100
⇒ 20 = ∑ xi
100 i=1

100
⇒ ∑ xi = 20 × 100 = 2000
i=1

∴ Incorrect sum of observations = 2000
⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940
Correct sum 1940
∴ Correct mean = = = 20
100−3 97

1 n 1 n 2 1 n 2 ¯
¯¯ 2
Standard deviation (σ) = √ ∑ xi − (∑ xi ) = √ ∑ x − (x)
n i=1 n
2 i=1 n i=1 i

1
2 2
⇒3 = √ × Incorrect ∑ x − (20)
i
100

2
⇒ Incorrect ∑ x = 100(9 + 400) = 40900
i

n n

2 2 2 2 2
Correct ∑ x = Incorrect ∑ x − (21) − (21) − (18)
t i

i=1 i=1

= 40900 − 441 − 441 − 324

= 39694

2
Correct ∑ x
i 2
∴ Correct standard deviation = √ − ( Correct mean )
n

39694
2
= √ − (20)
97

= √9.216

= 3.036

Page : 380 , Block Name : Miscellaneous Exercise