The NCERT Solutions For Class 11 Maths Chapter 15 Statistics are available as a pdf on this page. The NCERT Solutions are authored by the most experienced educators in the teaching industry making the solution of every problem straightforward and justifiable. Every solution written in the pdf given underneath empowers the student to get ready for the test and accomplish it. These solutions assist a class 11 student with mastering the idea of Limits and Derivatives.
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Get detailed solution for all the questions listed under below exercises:
Exercise 15.1 Solutions : 12 Questions
Exercise 15.2 Solutions : 10 Questions
Exercise 15.3 Solutions : 5 Questions
Miscellaneous Exercise Solutions: 7 Questions
Access Solutions for Class 11 Maths Chapter 15 Exercise
Exercise 15.1 Page: 360
Find the mean deviation about the mean for the data in Exercises 1 and 2.
1. 4, 7, 8, 9, 10, 12, 13, 17
Solution:
First we have to find (xÌ…) of the given data
So, the respective values of the deviations from mean,
i.e., x_{i}Â â€“ xÌ… are, 10 â€“ 4 = 6, 10 â€“ 7 = 3, 10 â€“ 8 = 2, 10 â€“ 10 = 0,
10 â€“ 12 = â€“ 2, 10 â€“ 13 = â€“ 3, 10 â€“ 17 = â€“ 7
6, 3, 2, 1, 0, 2, 3, 7
Now absolute values of the deviations,
6, 3, 2, 1, 0, 2, 3, 7
MD = sum of deviations/ number of observations
= 24/8
= 3
So, the mean deviation for the given data is 3.
2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
First we have to find (xÌ…) of the given data
So, the respective values of the deviations from mean,
i.e., x_{i}Â â€“ xÌ… are, 50 â€“ 38 = 12, 50 70 = 20, 50 â€“ 48 = 2, 50 â€“ 40 = 10, 50 â€“ 42 = 8,
50 â€“ 55 = â€“ 5, 50 â€“ 63 = â€“ 13, 50 â€“ 46 = 4, 50 â€“ 54 = 4, 50 â€“ 44 = 6
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
Now absolute values of the deviations,
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
MD = sum of deviations/ number of observations
= 84/10
= 8.4
So, the mean deviation for the given data is 8.4.
Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Solution:
First we have to arrange the given observations into ascending order,
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
The number of observations is 12
Then,
Median = ((12/2)^{th} observation + ((12/2)+ 1)^{th} observation)/2
(12/2)^{th} observation = 6^{th} = 13
(12/2)+ 1)^{th} observation = 6 + 1
= 7^{th} = 14
Median = (13 + 14)/2
= 27/2
= 13.5
So, the absolute values of the respective deviations from the median, i.e., x_{i}Â â€“ 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
Mean Deviation,
= (1/12) Ã— 28
= 2.33
So, the mean deviation about the median for the given data is 2.33.
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Solution:
First we have to arrange the given observations into ascending order,
36, 42, 45, 46, 46, 49, 51, 53, 60, 72.
The number of observations is 10
Then,
Median = ((10/2)^{th} observation + ((10/2)+ 1)^{th} observation)/2
(10/2)^{th} observation = 5^{th} = 46
(10/2)+ 1)^{th} observation = 5 + 1
= 6^{th} = 49
Median = (46 + 49)/2
= 95
= 47.5
So, the absolute values of the respective deviations from the median, i.e., x_{i}Â â€“ M are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Mean Deviation,
= (1/10) Ã— 70
= 7
So, the mean deviation about the median for the given data is 7.
Find the mean deviation about the mean for the data in Exercises 5 and 6.
5.
x_{i} 
5 
10 
15 
20 
25 
f_{i} 
7 
4 
6 
3 
5 
Solution:
Let us make the table of the given data and append other columns after calculations.
X_{i} 
f_{i} 
f_{i}x_{i} 
x_{i}Â â€“ xÌ… 
f_{i} x_{i}Â â€“ xÌ… 
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 
The sum of calculated data,
The absolute values of the deviations from the mean, i.e., x_{i}Â â€“ xÌ…, as shown in the table.
6.
x_{i} 
10 
30 
50 
70 
90 
f_{i} 
4 
24 
28 
16 
8 
Solution:
Let us make the table of the given data and append other columns after calculations.
X_{i} 
f_{i} 
f_{i}x_{i} 
x_{i}Â â€“ xÌ… 
f_{i} x_{i}Â â€“ xÌ… 
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 
Find the mean deviation about the median for the data in Exercises 7 and 8.
7.
x_{i} 
5 
7 
9 
10 
12 
15 
f_{i} 
8 
6 
2 
2 
2 
6 
Solution:
Let us make the table of the given data and append other columns after calculations.
X_{i} 
f_{i} 
c.f. 
x_{i}Â â€“ M 
f_{i} x_{i}Â â€“ M 
5 
8 
8 
2 
16 
7 
6 
14 
0 
0 
9 
2 
16 
2 
4 
10 
2 
18 
3 
6 
12 
2 
20 
5 
10 
15 
6 
26 
8 
48 
Now, N = 26, which is even.
Median is the mean of the 13^{th} and 14^{th} observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
Then,
Median = (13^{th} observation + 14^{th }observation)/2
= (7 + 7)/2
= 14/2
= 7
So, the absolute values of the respective deviations from the median, i.e., x_{i}Â â€“ M are shown in the table.
8.
x_{i} 
15 
21 
27 
30 
35 
f_{i} 
3 
5 
6 
7 
8 
Solution:
Let us make the table of the given data and append other columns after calculations.
X_{i} 
f_{i} 
c.f. 
x_{i}Â â€“ M 
f_{i} x_{i}Â â€“ M 
15 
3 
3 
13.5 
40.5 
21 
5 
8 
7.5 
37.5 
27 
6 
14 
1.5 
9 
30 
7 
21 
1.5 
10.5 
35 
8 
29 
6.5 
52 
Now, N = 30, which is even.
Median is the mean of the 15^{th} and 16^{th} observations. Both of these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.
Then,
Median = (15^{th} observation + 16^{th }observation)/2
= (30 + 30)/2
= 60/2
= 30
So, the absolute values of the respective deviations from the median, i.e., x_{i}Â â€“ M are shown in the table.
Find the mean deviation about the mean for the data in Exercises 9 and 10.
9.
Income per day in â‚¹ 
0 â€“ 100 
100 â€“ 200 
200 â€“ 300 
300 â€“ 400 
400 â€“ 500 
500 â€“ 600 
600 â€“ 700 
700 â€“ 800 
Number of persons 
4 
8 
9 
10 
7 
5 
4 
3 
Solution:
Let us make the table of the given data and append other columns after calculations.
Income per day in â‚¹ 
Number of persons f_{i} 
Mid â€“ points x_{i} 
f_{i}x_{i} 
x_{i}Â â€“ xÌ… 
f_{i}x_{i}Â â€“ xÌ… 
0 â€“ 100 
4 
50 
200 
308 
1232 
100 â€“ 200 
8 
150 
1200 
208 
1664 
200 â€“ 300 
9 
250 
2250 
108 
972 
300 â€“ 400 
10 
350 
3500 
8 
80 
400 â€“ 500 
7 
450 
3150 
92 
644 
500 â€“ 600 
5 
550 
2750 
192 
960 
600 â€“ 700 
4 
650 
2600 
292 
1160 
700 â€“ 800 
3 
750 
2250 
392 
1176 
50 
17900 
7896 
10.
Height in cms 
95 â€“ 105 
105 â€“ 115 
115 â€“ 125 
125 â€“ 135 
135 â€“ 145 
145 â€“ 155 
Number of boys 
9 
13 
26 
30 
12 
10 
Solution:
Let us make the table of the given data and append other columns after calculations.
Height in cms 
Number of boys f_{i} 
Mid â€“ points x_{i} 
f_{i}x_{i} 
x_{i}Â â€“ xÌ… 
f_{i}x_{i}Â â€“ xÌ… 
95 â€“ 105 
9 
100 
900 
25.3 
227.7 
105 â€“ 115 
13 
110 
1430 
15.3 
198.9 
115 â€“ 125 
26 
120 
3120 
5.3 
137.8 
125 â€“ 135 
30 
130 
3900 
4.7 
141 
135 â€“ 145 
12 
140 
1680 
14.7 
176.4 
145 â€“ 155 
10 
150 
1500 
24.7 
247 
100 
12530 
1128.8 
11. Find the mean deviation about median for the following data:
Marks 
0 10 
10 20 
20 â€“ 30 
30 â€“ 40 
40 â€“ 50 
50 â€“ 60 
Number of girls 
6 
8 
14 
16 
4 
2 
Solution:
Let us make the table of the given data and append other columns after calculations.
Marks 
Number of Girls f_{i} 
Cumulative frequency (c.f.) 
Mid â€“ points x_{i} 
x_{i}Â â€“ Med 
f_{i}x_{i}Â â€“ Med 
0 â€“ 10 
6 
6 
5 
22.85 
137.1 
10 â€“ 20 
8 
14 
15 
12.85 
102.8 
20 â€“ 30 
14 
28 
25 
2.85 
39.9 
30 â€“ 40 
16 
44 
35 
7.15 
114.4 
40 â€“ 50 
4 
48 
45 
17.15 
68.6 
50 â€“ 60 
2 
50 
55 
27.15 
54.3 
50 
517.1 
The class interval containingÂ N^{th}/2Â or 25^{th}Â item is 2030
So, 2030 is the median class.
Then,
Median = l + (((N/2) â€“ c)/f) Ã— h
Where, l = 20, c = 14, f = 14, h = 10 and n = 50
Median = 20 + (((25 â€“ 14))/14) Ã— 10
= 20 + 7.85
= 27.85
12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age (in years) 
16 â€“ 20 
21 â€“ 25 
26 â€“ 30 
31 â€“ 35 
36 â€“ 40 
41 â€“ 45 
46 â€“ 50 
51 â€“ 55 
Number 
5 
6 
12 
14 
26 
12 
16 
9 
[Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]
Solution:
The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.
Age 
Number f_{i} 
Cumulative frequency (c.f.) 
Mid â€“ points x_{i} 
x_{i}Â â€“ Med 
f_{i}x_{i}Â â€“ Med 
15.5 â€“ 20.5 
5 
5 
18 
20 
100 
20.5 â€“ 25.5 
6 
11 
23 
15 
90 
25.5 â€“ 30.5 
12 
23 
28 
10 
120 
30.5 â€“ 35.5 
14 
37 
33 
5 
70 
35.5 â€“ 40.5 
26 
63 
38 
0 
0 
40.5 â€“ 45.5 
12 
75 
43 
5 
60 
45.5 â€“ 50.5 
16 
91 
48 
10 
160 
50.5 â€“ 55.5 
9 
100 
53 
15 
135 
100 
735 
The class interval containingÂ N^{th}/2Â or 50^{th}Â item is 35.5 â€“ 40.5
So, 35.5 â€“ 40.5 is the median class.
Then,
Median = l + (((N/2) â€“ c)/f) Ã— h
Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100
Median = 35.5 + (((50 â€“ 37))/26) Ã— 5
= 35.5 + 2.5
= 38
Exercise 15.2 Page: 371
Find the mean and variance for each of the data in Exercise 1 to 5.
1. 6, 7, 10, 12, 13, 4, 8, 12
Solution:
So, xÌ… = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
Let us make the table of the given data and append other columns after calculations.
X_{i} 
Deviations from mean (x_{i}Â â€“ xÌ…) 
(x_{i}Â â€“ xÌ…)^{2} 
6 
6 â€“ 9 = 3 
9 
7 
7 â€“ 9 = 2 
4 
10 
10 â€“ 9 = 1 
1 
12 
12 â€“ 9 = 3 
9 
13 
13 â€“ 9 = 4 
16 
4 
4 â€“ 9 = â€“ 5 
25 
8 
8 â€“ 9 = â€“ 1 
1 
12 
12 â€“ 9 = 3 
9 
74 
We know that Variance,
Ïƒ^{2}Â = (1/8) Ã— 74
= 9.2
âˆ´Mean = 9 and Variance = 9.25
2. First n natural numbers
Solution:
We know that Mean = Sum of all observations/Number of observations
âˆ´Mean, xÌ… = ((n(n + 1))2)/n
= (n + 1)/2
and also WKT Variance,
By substitute that value of xÌ… we get,
WKT, (a + b)(a â€“ b) = a^{2} â€“ b^{2}
Ïƒ^{2}Â = (n^{2} â€“ 1)/12
âˆ´Mean = (n + 1)/2 and Variance = (n^{2} â€“ 1)/12
3. First 10 multiples of 3
Solution:
First we have to write the first 10 multiples of 3,
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
So, xÌ… = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10
= 165/10
= 16.5
Let us make the table of the data and append other columns after calculations.
X_{i} 
Deviations from mean (x_{i}Â â€“ xÌ…) 
(x_{i}Â â€“ xÌ…)^{2} 
3 
3 â€“ 16.5 = 13.5 
182.25 
6 
6 â€“ 16.5 = 10.5 
110.25 
9 
9 â€“ 16.5 = 7.5 
56.25 
12 
12 â€“ 16.5 = 4.5 
20.25 
15 
15 â€“ 16.5 = 1.5 
2.25 
18 
18 â€“ 16.5 = 1.5 
2.25 
21 
21 â€“ 16.5 = â€“ 4.5 
20.25 
24 
24 â€“ 16.5 = 7.5 
56.25 
27 
27 â€“ 16.5 = 10.5 
110.25 
30 
30 â€“ 16.5 = 13.5 
182.25 
742.5 
Then, Variance
= (1/10) Ã— 742.5
= 74.25
âˆ´Mean = 16.5 and Variance = 74.25
4.
x_{i} 
6 
10 
14 
18 
24 
28 
30 
f_{i} 
2 
4 
7 
12 
8 
4 
3 
Solution:
Let us make the table of the given data and append other columns after calculations.
X_{i} 
f_{i} 
f_{i}x_{i} 
Deviations from mean (x_{i}Â â€“ xÌ…) 
(x_{i}Â â€“ xÌ…)^{2} 
f_{i}(x_{i}Â â€“ xÌ…)^{2} 
6 
2 
12 
6 â€“ 19 = 13 
169 
338 
10 
4 
40 
10 â€“ 19 = 9 
81 
324 
14 
7 
98 
14 â€“ 19 = 5 
25 
175 
18 
12 
216 
18 â€“ 19 = 1 
1 
12 
24 
8 
192 
24 â€“ 19 = 5 
25 
200 
28 
4 
112 
28 â€“ 19 = 9 
81 
324 
30 
3 
90 
30 â€“ 19 = 11 
121 
363 
N = 40 
760 
1736 
5.
x_{i} 
92 
93 
97 
98 
102 
104 
109 
f_{i} 
3 
2 
3 
2 
6 
3 
3 
Solution:
Let us make the table of the given data and append other columns after calculations.
X_{i} 
f_{i} 
f_{i}x_{i} 
Deviations from mean (x_{i}Â â€“ xÌ…) 
(x_{i}Â â€“ xÌ…)^{2} 
f_{i}(x_{i}Â â€“ xÌ…)^{2} 
92 
3 
276 
92 â€“ 100 = 8 
64 
192 
93 
2 
186 
93 â€“ 100 = 7 
49 
98 
97 
3 
291 
97 â€“ 100 = 3 
9 
27 
98 
2 
196 
98 â€“ 100 = 2 
4 
8 
102 
6 
612 
102 â€“ 100 = 2 
4 
24 
104 
3 
312 
104 â€“ 100 = 4 
16 
48 
109 
3 
327 
109 â€“ 100 = 9 
81 
243 
N = 22 
2200 
640 
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 
Solution:
Let the assumed mean A = 64. Here h = 1
We obtain the following table from the given data.
X_{i} 
Frequency f_{i} 
Y_{i} = (x_{i} â€“ A)/h 
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 
0 
286 
Mean,
Where A = 64, h = 1
So, xÌ… = 64 + ((0/100) Ã— 1)
= 64 + 0
= 64
Then, variance,
Ïƒ^{2} = (1^{2}/100^{2}) [100(286) â€“ 0^{2}]
= (1/10000) [28600 â€“ 0]
= 28600/10000
= 2.86
Hence, standard deviation = Ïƒ = âˆš2.886
= 1.691
âˆ´Â Mean = 64 and Standard Deviation = 1.691
Find the mean and variance for the following frequency distributions in Exercises 7 and 8.
7.
Classes 
0 â€“ 30 
30 â€“ 60 
60 â€“ 90 
90 â€“ 120 
120 â€“ 150 
150 â€“ 180 
180 â€“ 210 
Frequencies 
2 
3 
5 
10 
3 
5 
2 
Solution:
Let us make the table of the given data and append other columns after calculations.
Classes 
Frequency f_{i} 
Mid â€“ points x_{i} 
f_{i}x_{i} 
(x_{i}Â â€“ xÌ…) 
(x_{i}Â â€“ xÌ…)^{2} 
f_{i}(x_{i}Â â€“ xÌ…)^{2} 
0 â€“ 30 
2 
15 
30 
92 
8464 
16928 
30 â€“ 60 
3 
45 
135 
62 
3844 
11532 
60 â€“ 90 
5 
75 
375 
32 
1024 
5120 
90 â€“ 120 
10 
105 
1050 
2 
4 
40 
120 â€“ 150 
3 
135 
405 
28 
784 
2352 
150 â€“ 180 
5 
165 
825 
58 
3364 
16820 
180 â€“ 210 
2 
195 
390 
88 
7744 
15488 
N = 30 
3210 
68280 
8.
Classes 
0 â€“ 10 
10 â€“ 20 
20 â€“ 30 
30 â€“ 40 
40 â€“50 
Frequencies 
5 
8 
15 
16 
6 
Solution:
Let us make the table of the given data and append other columns after calculations.
Classes 
Frequency f_{i} 
Mid â€“ points x_{i} 
f_{i}x_{i} 
(x_{i}Â â€“ xÌ…) 
(x_{i}Â â€“ xÌ…)^{2} 
f_{i}(x_{i}Â â€“ xÌ…)^{2} 
0 â€“ 10 
5 
5 
25 
22 
484 
2420 
10 â€“ 20 
8 
15 
120 
12 
144 
1152 
20 â€“ 30 
15 
25 
375 
2 
4 
60 
30 â€“ 40 
16 
35 
560 
8 
64 
1024 
40 â€“50 
6 
45 
270 
18 
324 
1944 
N = 50 
1350 
6600 
9. Find the mean, variance and standard deviation using shortcut method
Height in cms 
70 â€“ 75 
75 â€“ 80 
80 â€“ 85 
85 â€“ 90 
90 â€“ 95 
95 â€“ 100 
100 â€“ 105 
105 â€“ 110 
110 â€“ 115 
Frequencies 
3 
4 
7 
7 
15 
9 
6 
6 
3 
Solution:
Let the assumed mean, A = 92.5 and h = 5
Let us make the table of the given data and append other columns after calculations.
Height (class) 
Number of children Frequency f_{i} 
Midpoint X_{i} 
Y_{i} = (x_{i} â€“ A)/h 
Y_{i}^{2} 
f_{i}y_{i} 
f_{i}y_{i}^{2} 
70 â€“ 75 
2 
72.5 
4 
16 
12 
48 
75 â€“ 80 
1 
77.5 
3 
9 
12 
36 
80 â€“ 85 
12 
82.5 
2 
4 
14 
28 
85 â€“ 90 
29 
87.5 
1 
1 
7 
7 
90 â€“ 95 
25 
92.5 
0 
0 
0 
0 
95 â€“ 100 
12 
97.5 
1 
1 
9 
9 
100 â€“ 105 
10 
102.5 
2 
4 
12 
24 
105 â€“ 110 
4 
107.5 
3 
9 
18 
54 
110 â€“ 115 
5 
112.5 
4 
16 
12 
48 
N = 60 
6 
254 
Mean,
Where, A = 92.5, h = 5
So, xÌ… = 92.5 + ((6/60) Ã— 5)
= 92.5 + Â½
= 92.5 + 0.5
= 93
Then, Variance,
Ïƒ^{2} = (5^{2}/60^{2}) [60(254) â€“ 6^{2}]
= (1/144) [15240 â€“ 36]
= 15204/144
= 1267/12
= 105.583
Hence, standard deviation = Ïƒ = âˆš105.583
= 10.275
âˆ´Â Mean = 93, variance = 105.583 and Standard Deviation = 10.275
10. The diameters of circles (in mm) drawn in a design are given below:
Diameters 
33 â€“ 36 
37 â€“ 40 
41 â€“ 44 
45 â€“ 48 
49 â€“ 52 
No. of circles 
15 
17 
21 
22 
25 
Calculate the standard deviation and mean diameter of the circles.
[Hint first make the data continuous by making the classes as 32.536.5, 36.540.5, 40.544.5, 44.5 â€“ 48.5, 48.5 â€“ 52.5 and then proceed.]
Solution:
Let the assumed mean, A = 42.5 and h = 4
Let us make the table of the given data and append other columns after calculations.
Height (class) 
Number of children (Frequency f_{i}) 
Midpoint X_{i} 
Y_{i} = (x_{i} â€“ A)/h 
Y_{i}^{2} 
f_{i}y_{i} 
f_{i}y_{i}^{2} 
32.5 â€“ 36.5 
15 
34.5 
2 
4 
30 
60 
36.5 â€“ 40.5 
17 
38.5 
1 
1 
17 
17 
40.5 â€“ 44.5 
21 
42.5 
0 
0 
0 
0 
44.5 â€“ 48.5 
22 
46.5 
1 
1 
22 
22 
48.5 â€“ 52.5 
25 
50.5 
2 
4 
50 
100 
N = 100 
25 
199 
Mean,
Where, A = 42.5, h = 4
So, xÌ… = 42.5 + (25/100) Ã— 4
= 42.5 + 1
= 43.5
Then, Variance,
Ïƒ^{2} = (4^{2}/100^{2})[100(199) â€“ 25^{2}]
= (1/625) [19900 â€“ 625]
= 19275/625
= 771/25
= 30.84
Hence, standard deviation = Ïƒ = âˆš30.84
= 5.553
âˆ´Â Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.
Exercise 15.3 Page: 375
1. From the data given below state which group is more variable, A or B?
Marks 
10 â€“ 20 
20 â€“ 30 
30 â€“ 40 
40 â€“ 50 
50 â€“ 60 
60 â€“ 70 
70 â€“ 80 
Group A 
9 
17 
32 
33 
40 
10 
9 
Group B 
10 
20 
30 
25 
43 
15 
7 
Solution:
For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.
Coefficient of variation (C.V.) = (Ïƒ/ xÌ…) Ã— 100
Where, Ïƒ = standard deviation, xÌ… = mean
For Group A
Marks 
Group A f_{i} 
Midpoint X_{i} 
Y_{i} = (x_{i} â€“ A)/h 
(Y_{i})^{2} 
f_{i}y_{i} 
f_{i}(y_{i})^{2} 
10 â€“ 20 
9 
15 
((15 â€“ 45)/10) = 3 
(3)^{2} = 9 
â€“ 27 
81 
20 â€“ 30 
17 
25 
((25 â€“ 45)/10) = 2 
(2)^{2} = 4 
â€“ 34 
68 
30 â€“ 40 
32 
35 
((35 â€“ 45)/10) = â€“ 1 
(1)^{2} = 1 
â€“ 32 
32 
40 â€“ 50 
33 
45 
((45 â€“ 45)/10) = 0 
0^{2} 
0 
0 
50 â€“ 60 
40 
55 
((55 â€“ 45)/10) = 1 
1^{2} = 1 
40 
40 
60 â€“ 70 
10 
65 
((65 â€“ 45)/10) = 2 
2^{2} = 4 
20 
40 
70 â€“ 80 
9 
75 
((75 â€“ 45)/10) = 3 
3^{2} = 9 
27 
81 
Total 
150 
6 
342 
Where A = 45,
and y_{i} = (x_{i} â€“ A)/h
Here h = class size = 20 â€“ 10
h = 10
So, xÌ… = 45 + ((6/150) Ã— 10)
= 45 â€“ 0.4
= 44.6
Ïƒ^{2} = (10^{2}/150^{2}) [150(342) â€“ (6)^{2}]
= (100/22500) [51,300 â€“ 36]
= (100/22500) Ã— 51264
= 227.84
Hence, standard deviation = Ïƒ = âˆš227.84
= 15.09
âˆ´Â C.V for group A = (Ïƒ/ xÌ…) Ã— 100
= (15.09/44.6) Ã— 100
= 33.83
Now, for group B.
Marks 
Group B f_{i} 
Midpoint X_{i} 
Y_{i} = (x_{i} â€“ A)/h 
(Y_{i})^{2} 
f_{i}y_{i} 
f_{i}(y_{i})^{2} 
10 â€“ 20 
10 
15 
((15 â€“ 45)/10) = 3 
(3)^{2} = 9 
â€“ 30 
90 
20 â€“ 30 
20 
25 
((25 â€“ 45)/10) = 2 
(2)^{2} = 4 
â€“ 40 
80 
30 â€“ 40 
30 
35 
((35 â€“ 45)/10) = â€“ 1 
(1)^{2} = 1 
â€“ 30 
30 
40 â€“ 50 
25 
45 
((45 â€“ 45)/10) = 0 
0^{2} 
0 
0 
50 â€“ 60 
43 
55 
((55 â€“ 45)/10) = 1 
1^{2} = 1 
43 
43 
60 â€“ 70 
15 
65 
((65 â€“ 45)/10) = 2 
2^{2} = 4 
30 
160 
70 â€“ 80 
7 
75 
((75 â€“ 45)/10) = 3 
3^{2} = 9 
21 
189 
Total 
150 
6 
592 
Where A = 45,
h = 10
So, xÌ… = 45 + ((6/150) Ã— 10)
= 45 â€“ 0.4
= 44.6
Ïƒ^{2} = (10^{2}/150^{2}) [150(592) â€“ (6)^{2}]
= (100/22500) [88,800 â€“ 36]
= (100/22500) Ã— 88,764
= 394.50
Hence, standard deviation = Ïƒ = âˆš394.50
= 19.86
âˆ´Â C.V for group B = (Ïƒ/ xÌ…) Ã— 100
= (19.86/44.6) Ã— 100
= 44.53
By comparing C.V. of group A and group B.
C.V of Group B > C.V. of Group A
So, Group B is more variable.
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 
Solution:
From the given data,
Let us make the table of the given data and append other columns after calculations.
X (x_{i}) 
Y (y_{i}) 
X_{i}^{2} 
Y_{i}^{2} 
35 
108 
1225 
11664 
54 
107 
2916 
11449 
52 
105 
2704 
11025 
53 
105 
2809 
11025 
56 
106 
8136 
11236 
58 
107 
3364 
11449 
52 
104 
2704 
10816 
50 
103 
2500 
10609 
51 
104 
2601 
10816 
49 
101 
2401 
10201 
Total = 510 
1050 
26360 
110290 
We have to calculate Mean for x,
Mean xÌ… = âˆ‘x_{i}/n
Where, n = number of terms
= 510/10
= 51
Then, Variance for x =
= (1/10^{2})[(10 Ã— 26360) â€“ 510^{2}]
= (1/100) (263600 â€“ 260100)
= 3500/100
= 35
WKT Standard deviation = âˆšvariance
= âˆš35
= 5.91
So, coefficient of variation = (Ïƒ/ xÌ…) Ã— 100
= (5.91/51) Ã— 100
= 11.58
Now, we have to calculate Mean for y,
Mean È³ = âˆ‘y_{i}/n
Where, n = number of terms
= 1050/10
= 105
Then, Variance for y =
= (1/10^{2})[(10 Ã— 110290) â€“ 1050^{2}]
= (1/100) (1102900 â€“ 1102500)
= 400/100
= 4
WKT Standard deviation = âˆšvariance
= âˆš4
= 2
So, coefficient of variation = (Ïƒ/ xÌ…) Ã— 100
= (2/105) Ã— 100
= 1.904
By comparing C.V. of X and Y.
C.V of X > C.V. of Y
So, Y is more stable than X.
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 wages 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?
Solution:
From the given table,
Mean monthly wages of firm A = Rs 5253
and Number of wage earners = 586
Then,
Total amount paid = 586 Ã— 5253
= Rs 3078258
Mean monthly wages of firm B = Rs 5253
Number of wage earners = 648
Then,
Total amount paid = 648 Ã— 5253
= Rs 34,03,944
(i) So, firm B pays larger amount as monthly wages.
(ii) Variance of firm A = 100
We know that, standard deviation (Ïƒ)= âˆš100
=10
Variance of firm B = 121
Then,
Standard deviation (Ïƒ)=âˆš(121 )
=11
Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.
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?
Solution:
From the given data,
Let us make the table of the given data and append other columns after calculations.
Number of goals scored x_{i} 
Number of matches f_{i} 
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 
Total 
25 
50 
130 
Since C.V. of firm B is greater
âˆ´Â Team A is more consistent.
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?
Solution:
First we have to calculate Mean for Length x,
Miscellaneous Exercise Page: 380
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.
Solution:
2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.
Solution:
Hence, from equation (i) and (v) we have:
When x â€“ y = 2 then x = 8 and y = 6
And, when x â€“ y = â€“ 2 then x = 6 and y = 8
âˆ´Â the remaining observations are 6 and 8
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.
Solution:
We know that,
4. Given that xÌ… is the mean and Ïƒ^{2} is the variance of n observations x_{1}, x_{2}, â€¦,x_{n} . Prove that the mean and variance of the observations ax_{1}, ax_{2}, ax_{3}, â€¦., ax_{n} are axÌ… and a^{2}Ïƒ^{2}, respectively, (a â‰ 0).
Solution:
From the question it is given that, n observations are x_{1}, x_{2},â€¦..x_{n}
Mean of the n observation = xÌ…
Variance of the n observation = Ïƒ^{2}
As we know that,
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
Solution:
(i) If wrong item is omitted,
From the question it is given that,
The number of observations i.e. n = 20
The incorrect mean = 20
The incorrect standard deviation = 2
(ii) If it is replaced by 12,
From the question it is given that,
The number of incorrect sum observations i.e. n = 200
The correct sum of observations n = 200 â€“ 8 + 12
n = 204
Then, correct mean = correct sum/20
= 204/20
= 10.2
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?
Solution:
From the question it is given that,
Mean of Mathematics = 42
Standard deviation of Mathematics = 12
Mean of Physics = 32
Standard deviation of physics = 15
Mean of Chemistry = 40.9
Standard deviation of chemistry = 20
As we know that,
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.
Solution:
From the question it is given that,
The total number of observations (n) = 100
Incorrect mean, (xÌ…) = 20
And, Incorrect standard deviation (Ïƒ) = 3
NCERT Solutions for Class 11 Maths Chapter 15 â€“ Statistics
The topics of Class 11 Chapter 15 â€“ Statistics of NCERT Solution are as follows:
15.1 Introduction
This section talks about the concept of central tendency, mean and median [during even and the odd number of observations] with examples. It introduces the concept of measure of dispersion.
The values which cluster around the middle or centre of the distribution are measures of central tendency. They are mean, median and mode.
In a class, mean can be used to find the average marks scored by the students.
When calculating the height of students, the median can be used to find the middlemost value.
15.2 Measures of Dispersion
This section defines measures of dispersion, different measures of dispersion [range, quartile deviation, mean deviation, standard deviation]Â
Measures of dispersion explain the relationship with measures of central tendency. For example, the spread of data tells how well the mean represents the data. If the spread is large, then mean is not representative of data.
15.3 Range
This section defines the range, its formula and an example.
The range gives the variability of scores using the maximum and minimum values in the set.
In a cricket match, Batsman A range = 121 â€“ 0 = 121 and Batsman B range = 52 â€“ 22 = 30
Range A > Range B. So, the scores are spread in case of A whereas for B they are close to each other.
15.4 Mean Deviation
This section defines mean deviation, the formula for mean deviation.
The concept of mean deviation can be used by biologists in the comparison of different animal weights and decide what would be a healthy weight.
15.4.1Mean deviation for ungrouped data
The process of obtaining the mean deviation for ungrouped data is elaborated in this section.
Find the mean, deviations from the mean, absolute deviations and substitute the values in the mean deviation formula and arrive at the answer.
15.4.2 Mean deviation for grouped data
The process of obtaining mean deviation for discrete and continuous frequency distributions are elaborated in this section with solved examples.
15.4.3 Limitations of mean deviation
 If in a series the degree of variability is very high, then median will not be a representative of data. Hence the mean deviation calculated about such median cannot be relied on.
 If the sum of deviations from the mean is greater than the sum of deviations from the median, then the mean deviation about mean is not very specific.
 The absolute mean deviation calculated canâ€™t be subjected to further algebraic treatment. It canâ€™t be used as an appropriate measure of dispersion.
15.5 Variance and standard deviation
15.5.1 Standard Deviation
This section involves the variance and standard deviation definition, formula and solved examples on the discrete and continuous frequency distribution.Â
A science test was taken by a class of students. The mean score of the test was 85% on calculation. The teacher found the standard deviation of other scores and noticed that a very small standard deviation exists which suggests that most of the students scored very close to 85%.
15.5.2 Shortcut method to find Variance and standard deviation
This section deals with the simpler way of calculating the standard deviation with a few illustrations.Â
15.6 Analysis of Frequency Distributions
This section deals with the process of comparing the variability of two series having the same mean, coefficient of variation with few solved problems.
A few points on Chapter 15 Statistics
 Range is defined as the difference between the maximum and minimum value of the given data.
 If there exists series with equal means, then the series with lesser standard deviation is more consistent or less scattered.Â
 The addition or subtraction of a positive number to each data point of the data set will not affect the variance.
The solutions give substitute strategies and clarifications to take care of problems which makes the student feel sure while taking the test. Additionally, taking care of many muddled issues upgrades the information on mathematical ability in the students. The solutions cater to all the vital inquiries which a student should and ought to have aced to show up for the test. The BYJUâ€™S subject specialists who have written these solutions have complete knowledge about the question paper setting and the marks distributed across the chapters.