NCERT Solutions for Class 11 Science Maths Chapter 15 Statistics are provided here with simple stepbystep explanations. These solutions for Statistics are extremely popular among Class 11 Science students for Maths Statistics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of Class 11 Science Maths Chapter 15 are provided here for you for free. You will also love the adfree experience on Meritnation’s NCERT Solutions. All NCERT Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.
Page No 360:
Question 1:
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,
The deviations of the respective observations from the mean are
–6, – 3, –2, –1, 0, 2, 3, 7
The absolute values of the deviations, i.e., are
6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is
Page No 360:
Question 2:
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,
The deviations of the respective observations from the mean are
–12, 20, –2, –10, –8, 5, 13, –4, 4, –6
The absolute values of the deviations, i.e. , are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
The required mean deviation about the mean is
Page No 360:
Question 3:
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
The deviations of the respective observations from the median, i.e.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 absolute values of the deviations,, 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
Page No 360:
Question 4:
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
The deviations of the respective observations from the median, i.e.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,, 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
Page No 360:
Question 5:
Find the mean deviation about the mean for the data.
x_{i} 
5 
10 
15 
20 
25 
f_{i} 
7 
4 
6 
3 
5 
Answer:

x_{i}
f_{i}
f_{i} x_{i}
5
7
35
9
63
10
4
40
4
16
15
6
90
1
6
20
3
60
6
18
25
5
125
11
55
25
350
158
Page No 360:
Question 6:
Find the mean deviation about the mean for the data
x_{i} 
10 
30 
50 
70 
90 
f_{i} 
4 
24 
28 
16 
8 
Answer:

x_{i}
f_{i}
f_{i} x_{i}
10
4
40
40
160
30
24
720
20
480
50
28
1400
0
0
70
16
1120
20
320
90
8
720
40
320
80
4000
1280
Page No 360:
Question 7:
Find the mean deviation about the median for the data.
x_{i} 
5 
7 
9 
10 
12 
15 
f_{i} 
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.

x_{i}
f_{i}
c.f.
5
8
8
7
6
14
9
2
16
10
2
18
12
2
20
15
6
26
Here, N = 26, which is even.
Median is the mean of 13^{th} and 14^{th} observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
The absolute values of the deviations from median, i.e.are

x_{i} – M
2
0
2
3
5
8
f_{i}
8
6
2
2
2
6
f_{i} x_{i} – M
16
0
4
6
10
48
and
Page No 360:
Question 8:
Find the mean deviation about the median for the data
x_{i} 
15 
21 
27 
30 
35 
f_{i} 
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.

x_{i}
f_{i}
c.f.
15
3
3
21
5
8
27
6
14
30
7
21
35
8
29
Here, N = 29, which is odd.
observation = 15^{th} observation
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.are

x_{i} – M
15
9
3
0
5
f_{i}
3
5
6
7
8
f_{i} x_{i} – M
45
45
18
0
40
∴
Page No 361:
Question 9:
Find the mean deviation about the mean for the data.
Income per day 
Number of persons 
0100 
4 
100200 
8 
200300 
9 
300400 
10 
400500 
7 
500600 
5 
600700 
4 
700800 
3 
Answer:
The following table is formed.

Income per day
Number of persons f_{i}
Midpoint x_{i}
f_{i} x_{i}
0 – 100
4
50
200
308
1232
100 – 200
8
150
1200
208
1664
200 – 300
9
250
2250
108
972
300 – 400
10
350
3500
8
80
400 – 500
7
450
3150
92
644
500 – 600
5
550
2750
192
960
600 – 700
4
650
2600
292
1168
700 – 800
3
750
2250
392
1176
50
17900
7896
Here,
Page No 361:
Question 10:
Find the mean deviation about the mean for the data
Height in cms 
Number of boys 
95105 
9 
105115 
13 
115125 
26 
125135 
30 
135145 
12 
145155 
10 
Answer:
The following table is formed.

Height in cms
Number of boys f_{i}
Midpoint x_{i}
f_{i} x_{i}
95105
9
100
900
25.3
227.7
105115
13
110
1430
15.3
198.9
115125
26
120
3120
5.3
137.8
125135
30
130
3900
4.7
141
135145
12
140
1680
14.7
176.4
145155
10
150
1500
24.7
247
Here,
Page No 361:
Question 11:
Find the mean deviation about median for the following data:
Marks 
Number of girls 
010 
6 
1020 
8 
2030 
14 
3040 
16 
4050 
4 
5060 
2 
Answer:
The following table is formed.

Marks
Number of girls f_{i}
Cumulative frequency (c.f.)
Midpoint x_{i}
x_{i} – Med.
f_{i} x_{i} – Med.
010
6
6
5
22.85
137.1
1020
8
14
15
12.85
102.8
2030
14
28
25
2.85
39.9
3040
16
44
35
7.15
114.4
4050
4
48
45
17.15
68.6
5060
2
50
55
27.15
54.3
50
517.1
The class interval containing theor 25^{th} item is 20 – 30.
Therefore, 20 – 30 is the median class.
It is known that,
Here, l = 20, C = 14, f = 14, h = 10, and N = 50
∴ Median =
Thus, mean deviation about the median is given by,
Page No 361:
Question 12:
Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age 
Number 
1620 
5 
2125 
6 
2630 
12 
3135 
14 
3640 
26 
4145 
12 
4650 
16 
5155 
9 
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.

Age
Number f_{i}
Cumulative frequency (c.f.)
Midpoint x_{i}
x_{i} – Med.
f_{i} x_{i} – Med.
15.520.5
5
5
18
20
100
20.525.5
6
11
23
15
90
25.530.5
12
23
28
10
120
30.535.5
14
37
33
5
70
35.540.5
26
63
38
0
0
40.545.5
12
75
43
5
60
45.550.5
16
91
48
10
160
50.555.5
9
100
53
15
135
100
735
The class interval containing theor 50^{th} item is 35.5 – 40.5.
Therefore, 35.5 – 40.5 is the median class.
It is known that,
Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100
Thus, mean deviation about the median is given by,
Page No 371:
Question 1:
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
Mean,
The following table is obtained.

x_{i}
6
–3
9
7
–2
4
10
–1
1
12
3
9
13
4
16
4
–5
25
8
–1
1
12
3
9
74
Page No 371:
Question 2:
Find the mean and variance for the first n natural numbers
Answer:
The mean of first n natural numbers is calculated as follows.
Page No 371:
Question 3:
Find the mean and variance for the first 10 multiples of 3
Answer:
The first 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10
The following table is obtained.

x_{i}
3
–13.5
182.25
6
–10.5
110.25
9
–7.5
56.25
12
–4.5
20.25
15
–1.5
2.25
18
1.5
2.25
21
4.5
20.25
24
7.5
56.25
27
10.5
110.25
30
13.5
182.25
742.5
Page No 371:
Question 4:
Find the mean and variance for the data
xi 
6 
10 
14 
18 
24 
28 
30 
f i 
2 
4 
7 
12 
8 
4 
3 
Answer:
The data is obtained in tabular form as follows.

x_{i}
f i
f_{i}x_{i}
6
2
12
–13
169
338
10
4
40
–9
81
324
14
7
98
–5
25
175
18
12
216
–1
1
12
24
8
192
5
25
200
28
4
112
9
81
324
30
3
90
11
121
363
40
760
1736
Here, N = 40,
Page No 371:
Question 5:
Find the mean and variance for the data
xi 
92 
93 
97 
98 
102 
104 
109 
f i 
3 
2 
3 
2 
6 
3 
3 
Answer:
The data is obtained in tabular form as follows.

x_{i}
f i
f_{i}x_{i}
92
3
276
–8
64
192
93
2
186
–7
49
98
97
3
291
–3
9
27
98
2
196
–2
4
8
102
6
612
2
4
24
104
3
312
4
16
48
109
3
327
9
81
243
22
2200
640
Here, N = 22,
Page No 371:
Question 6:
Find the mean and standard deviation using shortcut method.
x_{i} 
60 
61 
62 
63 
64 
65 
66 
67 
68 
f_{i} 
2 
1 
12 
29 
25 
12 
10 
4 
5 
Answer:
The data is obtained in tabular form as follows.

x_{i}
f_{i}
y_{i}^{2}
f_{i}y_{i}
f_{i}y_{i}^{2}
60
2
–4
16
–8
32
61
1
–3
9
–3
9
62
12
–2
4
–24
48
63
29
–1
1
–29
29
64
25
0
0
0
0
65
12
1
1
12
12
66
10
2
4
20
40
67
4
3
9
12
36
68
5
4
16
20
80
100
220
0
286
Mean,
Page No 371:
Question 7:
Find the mean and variance for the following frequency distribution.
Classes 
030 
3060 
6090 
90120 
120150 
150180 
180210 
Frequencies 
2 
3 
5 
10 
3 
5 
2 
Answer:

Class
Frequency f_{i}
Midpoint x_{i}
y_{i}^{2}
f_{i}y_{i}
f_{i}y_{i}^{2}
030
2
15
–3
9
–6
18
3060
3
45
–2
4
–6
12
6090
5
75
–1
1
–5
5
90120
10
105
0
0
0
0
120150
3
135
1
1
3
3
150180
5
165
2
4
10
20
180210
2
195
3
9
6
18
30
2
76
Mean,
Page No 372:
Question 8:
Find the mean and variance for the following frequency distribution.
Classes 
010 
1020 
2030 
3040 
4050 
Frequencies 
5 
8 
15 
16 
6 
Answer:
Class 
Frequency f_{i} 
Midpoint x_{i} 
y_{i}^{2} 
f_{i}y_{i} 
f_{i}y_{i}^{2} 

010 
5 
5 
–2 
4 
–10 
20 
1020 
8 
15 
–1 
1 
–8 
8 
2030 
15 
25 
0 
0 
0 
0 
3040 
16 
35 
1 
1 
16 
16 
4050 
6 
45 
2 
4 
12 
24 
50 
10 
68 
Mean,
Page No 372:
Question 9:
Find the mean, variance and standard deviation using shortcut method
Height in cms 
No. of children 
7075 
3 
7580 
4 
8085 
7 
8590 
7 
9095 
15 
95100 
9 
100105 
6 
105110 
6 
110115 
3 
Answer:
Class Interval 
Frequency f_{i} 
Midpoint x_{i} 
y_{i}^{2} 
f_{i}y_{i} 
f_{i}y_{i}^{2} 

7075 
3 
72.5 
–4 
16 
–12 
48 
7580 
4 
77.5 
–3 
9 
–12 
36 
8085 
7 
82.5 
–2 
4 
–14 
28 
8590 
7 
87.5 
–1 
1 
–7 
7 
9095 
15 
92.5 
0 
0 
0 
0 
95100 
9 
97.5 
1 
1 
9 
9 
100105 
6 
102.5 
2 
4 
12 
24 
105110 
6 
107.5 
3 
9 
18 
54 
110115 
3 
112.5 
4 
16 
12 
48 
60 
6 
254 
Mean,
Page No 372:
Question 10:
The diameters of circles (in mm) drawn in a design are given below:
Diameters 
No. of children 
3336 
15 
3740 
17 
4144 
21 
4548 
22 
4952 
25 
Answer:
Class Interval 
Frequency f_{i} 
Midpoint x_{i} 
f_{i}^{2} 
f_{i}y_{i} 
f_{i}y_{i}^{2} 

32.536.5 
15 
34.5 
–2 
4 
–30 
60 
36.540.5 
17 
38.5 
–1 
1 
–17 
17 
40.544.5 
21 
42.5 
0 
0 
0 
0 
44.548.5 
22 
46.5 
1 
1 
22 
22 
48.552.5 
25 
50.5 
2 
4 
50 
100 
100 
25 
199 
Here, N = 100, h = 4
Let the assumed mean, A, be 42.5.
Mean,
Page No 375:
Question 1:
From the data given below state which group is more variable, A or B?
Marks 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
Group A 
9 
17 
32 
33 
40 
10 
9 
Group B 
10 
20 
30 
25 
43 
15 
7 
Answer:
Firstly, the standard deviation of group A is calculated as follows.
Marks 
Group A f_{i} 
Midpoint x_{i} 
y_{i}^{2} 
f_{i}y_{i} 
f_{i}y_{i}^{2} 

1020 
9 
15 
–3 
9 
–27 
81 
2030 
17 
25 
–2 
4 
–34 
68 
3040 
32 
35 
–1 
1 
–32 
32 
4050 
33 
45 
0 
0 
0 
0 
5060 
40 
55 
1 
1 
40 
40 
6070 
10 
65 
2 
4 
20 
40 
7080 
9 
75 
3 
9 
27 
81 
150 
–6 
342 
Here, h = 10, N = 150, A = 45
The standard deviation of group B is calculated as follows.
Marks 
Group B f_{i} 
Midpoint x_{i} 
y_{i}^{2} 
f_{i}y_{i} 
f_{i}y_{i}^{2} 

1020 
10 
15 
–3 
9 
–30 
90 
2030 
20 
25 
–2 
4 
–40 
80 
3040 
30 
35 
–1 
1 
–30 
30 
4050 
25 
45 
0 
0 
0 
0 
5060 
43 
55 
1 
1 
43 
43 
6070 
15 
65 
2 
4 
30 
60 
7080 
7 
75 
3 
9 
21 
63 
150 
–6 
366 
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 No 375:
Question 2:
From the prices of shares X and Y below, find out which is more stable in value:
X 
35 
54 
52 
53 
56 
58 
52 
50 
51 
49 
Y 
108 
107 
105 
105 
106 
107 
104 
103 
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
The following table is obtained corresponding to shares X.
x_{i} 

35 
–16 
256 
54 
3 
9 
52 
1 
1 
53 
2 
4 
56 
5 
25 
58 
7 
49 
52 
1 
1 
50 
–1 
1 
51 
0 
0 
49 
–2 
4 
350 
The prices of share Y are
108, 107, 105, 105, 106, 107, 104, 103, 104, 101
The following table is obtained corresponding to shares Y.
y_{i} 

108 
3 
9 
107 
2 
4 
105 
0 
0 
105 
0 
0 
106 
1 
1 
107 
2 
4 
104 
–1 
1 
103 
–2 
4 
104 
–1 
1 
101 
–4 
16 
40 
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 No 375:
Question 3:
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
Firm A 
Firm B 

No. of wage earners 
586 
648 
Mean of monthly wages 
Rs 5253 
Rs 5253 
Variance of the distribution of wages 
100 
121 
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
Answer:
(i) Monthly wages of firm A = Rs 5253
Number of wage earners in firm A = 586
∴Total amount paid = Rs 5253 × 586
Monthly wages of firm B = Rs 5253
Number of wage earners in firm B = 648
∴Total amount paid = Rs 5253 × 648
Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.
(ii) Variance of the distribution of wages in firm A = 100
∴ Standard deviation of the distribution of wages in firm
A ((σ_{1}) =
Variance of the distribution of wages in firm = 121
∴ Standard deviation of the distribution of wages in firm
The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.
Thus, firm B has greater variability in the individual wages.
Page No 376:
Question 4:
The following is the record of goals scored by team A in a football session:
No. of goals scored 
0 
1 
2 
3 
4 
No. of matches 
1 
9 
7 
5 
3 
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.
No. of goals scored 
No. of matches 
f_{i}x_{i} 
x_{i}^{2} 
f_{i}x_{i}^{2} 
0 
1 
0 
0 
0 
1 
9 
9 
1 
9 
2 
7 
14 
4 
28 
3 
5 
15 
9 
45 
4 
3 
12 
16 
48 
25 
50 
130 
Thus, the mean of both the teams is same.
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 No 376:
Question 5:
The sum and sum of squares corresponding to length x (in cm) and weight y
(in gm) of 50 plant products are given below:
Which is more varying, the length or weight?
Answer:
Here, N = 50
∴ Mean,
Mean,
Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.
Page No 380:
Question 1:
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, find 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.
From (1), we obtain
x^{2} + y^{2} + 2xy = 144 …(3)
From (2) and (3), we obtain
2xy = 64 … (4)
Subtracting (4) from (2), we obtain
x^{2} + y^{2 }– 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 No 380:
Question 2:
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 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.
From (1), we obtain
x^{2} + y^{2} + 2xy = 196 … (3)
From (2) and (3), we obtain
2xy = 196 – 100
⇒ 2xy = 96 … (4)
Subtracting (4) from (2), we obtain
x^{2} + y^{2 }– 2xy = 100 – 96
⇒ (x – y)^{2} = 4
⇒ 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 No 380:
Question 3:
The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Answer:
Let the observations be x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, and x_{6}.
It is given that mean is 8 and standard deviation is 4.
If each observation is multiplied by 3 and the resulting observations are y_{i}, then
From (1) and (2), it can be observed that,
Substituting the values of x_{i} and in (2), we obtain
Therefore, variance of new observations =
Hence, the standard deviation of new observations is
Page No 380:
Question 4:
Given that is the mean and σ^{2} is the variance of n observations x_{1}, x_{2} … x_{n}. Prove that the mean and variance of the observations ax_{1}, ax_{2}, ax_{3} …ax_{n }are and a^{2} σ^{2}, respectively (a ≠ 0).
Answer:
The given n observations are x_{1}, x_{2} … x_{n}.
Mean =
Variance = σ^{2}
If each observation is multiplied by a and the new observations are y_{i}, then
Therefore, mean of the observations, ax_{1}, ax_{2} … ax_{n}, is .
Substituting the values of x_{i}and in (1), we obtain
Thus, the variance of the observations, ax_{1}, ax_{2} … ax_{n}, is a^{2} σ^{2}.
Page No 380:
Question 5:
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
That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
∴ Correct mean
$\mathrm{Standard}\mathrm{deviation},\mathrm{\sigma}=\sqrt{\frac{1}{n}\sum _{i=1}^{n}{{x}_{i}}^{2}\frac{1}{{n}^{2}}{\left(\sum _{i=1}^{n}{x}_{i}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2=\sqrt{\frac{1}{n}\sum _{i=1}^{n}{{x}_{i}}^{2}{\left(\frac{1}{n}\sum _{i=1}^{n}{x}_{i}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2=\sqrt{\frac{1}{n}\sum _{i=1}^{n}{{x}_{i}}^{2}{\left(\overline{x}\right)}^{2}}\left[\mathrm{as},\frac{1}{n}\sum _{i=1}^{n}x=\overline{)x}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 2=\sqrt{\frac{1}{20}\times \mathrm{Incorrect}\sum _{i=1}^{n}{{x}_{i}}^{2}{\left(10\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow 4=\frac{1}{20}\times \mathrm{Incorrect}\sum _{i=1}^{n}{{x}_{i}}^{2}100\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{20}\times \mathrm{Incorrect}\sum _{i=1}^{n}{{x}_{i}}^{2}=104\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Incorrect}\sum _{i=1}^{n}{{x}_{i}}^{2}=2080\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{correct}\sum _{i=1}^{n}{{x}_{i}}^{2}=\mathrm{Incorrect}\sum _{i=1}^{n}{{x}_{i}}^{2}{\left(8\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{correct}\sum _{i=1}^{n}{{x}_{i}}^{2}=208064=2016\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{correct}\mathrm{Standard}\mathrm{Deviation}=\sqrt{\frac{1}{n}\mathrm{correct}\sum _{i=1}^{n}{{x}_{i}}^{2}{\left(\mathrm{correct}\mathrm{mean}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{correct}\mathrm{Standard}\mathrm{Deviation}=\sqrt{\frac{1}{19}\times 2016{\left(\frac{192}{19}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{correct}\mathrm{Standard}\mathrm{Deviation}=\sqrt{\frac{2016}{19}{\left(\frac{192}{19}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{correct}\mathrm{Standard}\mathrm{Deviation}=\frac{\sqrt{1440}}{19}=\frac{12\sqrt{10}}{19}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{correct}\mathrm{Standard}\mathrm{Deviation}=\frac{12\times 3.162}{19}=1.997$
(ii) When 8 is replaced by 12,
Incorrect sum of observations = 200
∴ Correct sum of observations = 200 – 8 + 12 = 204
Page No 380:
Question 6:
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Subject 
Mathematics 
Physics 
Chemistry 
Mean 
42 
32 
40.9 
Standard deviation 
12 
15 
20 
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 coefficient of variation (C.V.) is given by .
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 No 380:
Question 7:
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
Incorrect standard deviation
∴ Incorrect sum of observations = 2000
⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940
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