In earlier classes, we have learnt about methods of representing data graphically and in tubular form. Here, we shall study various methods of finding a representative value of the given data. RD Sharma Solutions for Class 11 Maths Chapter 32 Statistics are provided here for students to practice and prepare for their board exams. Students are advised to refer to the pdf of RD Sharma Class 11 Maths Solutions in order to help them grasp the shortcut techniques and important formulas. For students to understand the concepts easily, the solutions are solved in a stepwise manner, as each step carries marks in the annual exam. Students who aim to improve their academic performance can use RD Sharma as a key source of reference material to achieve their goals. Students can download the readily available pdf of RD Sharma solutions, from the below mentioned links.
Chapter 32 – Statistics contains seven exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.
 Measures of dispersion.
 Range.
 Mean deviation.
 Mean deviation for ungrouped data or individual observations.
 Mean deviation of a discrete frequency distribution.
 Mean deviation of a grouped or continuous frequency distribution.
 Limitations of mean deviation.
 Variance and standard deviation.
 Variance of individual observations.
 Variance of a discrete frequency distribution.
 Variance of a grouped or continuous frequency distribution.
 Analysis of frequency distribution.
Download the pdf of RD Sharma Solutions for Class 11 Maths Chapter 32 – Statistics
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EXERCISE 32.1 PAGE NO: 32.6
1. Calculate the mean deviation about the median of the following observation :
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000
(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
Solution:
(i) 3011, 2780, 3020, 2354, 3541, 4150, 5000
To calculate the Median (M), let us arrange the numbers in ascending order.
Median is the middle number of all the observation.
2354, 2780, 3011, 3020, 3541, 4150, 5000
So, Median = 3020 and n = 7
By using the formula to calculate Mean Deviation,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)
x_{i} 
d_{i} = x_{i} â€“ 3020 
3011 
9 
2780 
240 
3020 
0 
2354 
666 
3541 
521 
4150 
1130 
5000 
1980 
Total 
4546 
= 1/7 Ã— 4546
= 649.42
âˆ´ The Mean Deviation is 649.42.
(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44
To calculate the Median (M), let us arrange the numbers in ascending order.
Median is the middle number of all the observation.
34, 38, 42, 44, 46, 48, 54, 55, 63, 70
Here the Number of observations are Even then Median = (46+48)/2 = 47
Median = 47 and n = 10
By using the formula to calculate Mean Deviation,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)
x_{i} 
d_{i} = x_{i} â€“ 47 
38 
9 
70 
23 
48 
1 
34 
13 
42 
5 
55 
8 
63 
16 
46 
1 
54 
7 
44 
3 
Total 
86 
= 1/10 Ã— 86
= 8.6
âˆ´ The Mean Deviation is 8.6.
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
To calculate the Median (M), let us arrange the numbers in ascending order.
Median is the middle number of all the observation.
30, 34, 38, 40, 42, 44, 50, 51, 60, 66
Here the Number of observations are Even then Median = (42+44)/2 = 43
Median = 43 and n = 10
By using the formula to calculate Mean Deviation,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)
x_{i} 
d_{i} = x_{i} â€“ 43 
30 
13 
34 
9 
38 
5 
40 
3 
42 
1 
44 
1 
50 
7 
51 
8 
60 
17 
66 
23 
Total 
87 
= 1/10 Ã— 87
= 8.7
âˆ´ The Mean Deviation is 8.7.
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
To calculate the Median (M), let us arrange the numbers in ascending order.
Median is the middle number of all the observation.
22, 24, 25, 27, 28, 29, 30, 31, 41, 42
Here the Number of observations are Even then Median = (28+29)/2 = 28.5
Median = 28.5 and n = 10
By using the formula to calculate Mean Deviation,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)
x_{i} 
d_{i} = x_{i} â€“ 28.5 
22 
6.5 
24 
4.5 
30 
1.5 
27 
1.5 
29 
0.5 
31 
2.5 
25 
3.5 
28 
0.5 
41 
12.5 
42 
13.5 
Total 
47 
= 1/10 Ã— 47
= 4.7
âˆ´ The Mean Deviation is 4.7.
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
To calculate the Median (M), let us arrange the numbers in ascending order.
Median is the middle number of all the observation.
34, 38, 43, 44, 47, 48, 53, 55, 63, 70
Here the Number of observations are Even then Median = (47+48)/2 = 47.5
Median = 47.5 and n = 10
By using the formula to calculate Mean Deviation,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)
x_{i} 
d_{i} = x_{i} â€“ 47.5 
38 
9.5 
70 
22.5 
48 
0.5 
34 
13.5 
63 
15.5 
42 
5.5 
55 
7.5 
44 
3.5 
53 
5.5 
47 
0.5 
Total 
84 
= 1/10 Ã— 84
= 8.4
âˆ´ The Mean Deviation is 8.4.
2. Calculate the mean deviation from the mean for the following data :
(i) 4, 7, 8, 9, 10, 12, 13, 17
(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
Solution:
(i) 4, 7, 8, 9, 10, 12, 13, 17
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [4 + 7 + 8 + 9 + 10 + 12 + 13 + 17]/8
= 80/8
= 10
Number of observations, â€˜nâ€™ = 8
x_{i} 
d_{i} = x_{i} â€“ 10 
4 
6 
7 
3 
8 
2 
9 
1 
10 
0 
12 
2 
13 
3 
17 
7 
Total 
24 
= 1/8 Ã— 24
= 3
âˆ´ The Mean Deviation is 3.
(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [13 + 17 + 16 + 14 + 11 + 13 + 10 + 16 + 11 + 18 + 12 + 17]/12
= 168/12
= 14
Number of observations, â€˜nâ€™ = 12
x_{i} 
d_{i} = x_{i} â€“ 14 
13 
1 
17 
3 
16 
2 
14 
0 
11 
3 
13 
1 
10 
4 
16 
2 
11 
3 
18 
4 
12 
2 
17 
3 
Total 
28 
= 1/12 Ã— 28
= 2.33
âˆ´ The Mean Deviation is 2.33.
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44]/10
= 500/10
= 50
Number of observations, â€˜nâ€™ = 10
x_{i} 
d_{i} = x_{i} â€“ 50 
38 
12 
70 
20 
48 
2 
40 
10 
42 
8 
55 
5 
63 
13 
46 
4 
54 
4 
44 
6 
Total 
84 
= 1/10 Ã— 84
= 8.4
âˆ´ The Mean Deviation is 8.4.
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [36 + 72 + 46 + 42 + 60 + 45 + 53 + 46 + 51 + 49]/10
= 500/10
= 50
Number of observations, â€˜nâ€™ = 10
x_{i} 
d_{i} = x_{i} â€“ 50 
36 
14 
72 
22 
46 
4 
42 
8 
60 
10 
45 
5 
53 
3 
46 
4 
51 
1 
49 
1 
Total 
72 
= 1/10 Ã— 72
= 7.2
âˆ´ The Mean Deviation is 7.2.
(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [57 + 64 + 43 + 67 + 49 + 59 + 44 + 47 + 61 + 59]/10
= 550/10
= 55
Number of observations, â€˜nâ€™ = 10
x_{i} 
d_{i} = x_{i} â€“ 55 
57 
2 
64 
9 
43 
12 
67 
12 
49 
6 
59 
4 
44 
11 
47 
8 
61 
6 
59 
4 
Total 
74 
= 1/10 Ã— 74
= 7.4
âˆ´ The Mean Deviation is 7.4.
3. Calculate the mean deviation of the following income groups of five and seven members from their medians:
I Income in â‚¹ 
II Income in â‚¹ 
4000 
3800 
4200 
4000 
4400 
4200 
4600 
4400 
4800 
4600 
4800 

5800 
Solution:
Let us calculate the mean deviation for the first data set.
Since the data is arranged in ascending order,
4000, 4200, 4400, 4600, 4800
Median = 4400
Total observations = 5
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ M
x_{i} 
d_{i} = x_{i} â€“ 4400 
4000 
400 
4200 
200 
4400 
0 
4600 
200 
4800 
400 
Total 
1200 
= 1/5 Ã— 1200
= 240
Let us calculate the mean deviation for the second data set.
Since the data is arranged in ascending order,
3800, 4000, 4200, 4400, 4600, 4800, 5800
Median = 4400
Total observations = 7
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ M
x_{i} 
d_{i} = x_{i} â€“ 4400 
3800 
600 
4000 
400 
4200 
200 
4400 
0 
4600 
200 
4800 
400 
5800 
1400 
Total 
3200 
= 1/7 Ã— 3200
= 457.14
âˆ´ The Mean Deviation of set 1 is 240 and set 2 is 457.14
4. The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i)Â Find the mean deviation from the median.
(ii)Â Find the mean deviation from the mean also.
Solution:
(i)Â Find the mean deviation from the median
Let us arrange the data in ascending order,
15.2, 27.9, 30.2, 32.5, 40.0, 52.3, 52.8, 55.2, 72.9, 79.0
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ M
The number of observations are Even then MedianÂ = (40+52.3)/2 = 46.15
Median = 46.15
Number of observations, â€˜nâ€™ = 10
x_{i} 
d_{i} = x_{i} â€“ 46.15 
40.0 
6.15 
52.3 
6.15 
55.2 
9.05 
72.9 
26.75 
52.8 
6.65 
79.0 
32.85 
32.5 
13.65 
15.2 
30.95 
27.9 
19.25 
30.2 
15.95 
Total 
167.4 
= 1/10 Ã— 167.4
= 16.74
âˆ´ The Mean Deviation is 16.74.
(ii)Â Find the mean deviation from the mean also.
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [40.0 + 52.3 + 55.2 + 72.9 + 52.8 + 79.0 + 32.5 + 15.2 + 27.9 + 30.2]/10
= 458/10
= 45.8
Number of observations, â€˜nâ€™ = 10
x_{i} 
d_{i} = x_{i} â€“ 45.8 
40.0 
5.8 
52.3 
6.5 
55.2 
9.4 
72.9 
27.1 
52.8 
7 
79.0 
33.2 
32.5 
13.3 
15.2 
30.6 
27.9 
17.9 
30.2 
15.6 
Total 
166.4 
= 1/10 Ã— 166.4
= 16.64
âˆ´ TheÂ Mean Deviation is 16.64
5. In question 1(iii), (iv), (v) find the number of observations lying betweenÂ \(\overline{X} – M.D\) andÂ \(\overline{X} + M.D\), where M.D. is the mean deviation from the mean.
Solution:
(iii)Â 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51]/10
= 455/10
= 45.5
Number of observations, â€˜nâ€™ = 10
x_{i} 
d_{i} = x_{i} â€“ 45.5 
34 
11.5 
66 
20.5 
30 
15.5 
38 
7.5 
44 
1.5 
50 
4.5 
40 
5.5 
60 
14.5 
42 
3.5 
51 
5.5 
Total 
90 
= 1/10 Ã— 90
= 9
Now
\(\overline{X} – M.D\) = 45.5 – 9 = 36.5 \(\overline{X} + M.D\) = 45.5 + 9 = 54.5So, There are total 6 observation betweenÂ \(\overline{X} – M.D\) andÂ \(\overline{X} + M.D\)
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42]/10
= 299/10
= 29.9
Number of observations, â€˜nâ€™ = 10
x_{i} 
d_{i} = x_{i} â€“ 29.9 
22 
7.9 
24 
5.9 
30 
0.1 
27 
2.9 
29 
0.9 
31 
1.1 
25 
4.9 
28 
1.9 
41 
11.1 
42 
12.1 
Total 
48.8 
= 1/10 Ã— 48.8
= 4.88
Now
So, There are total 5 observation betweenÂ
andÂ
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
We know that,
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)Where, d_{i} = x_{i} â€“ x
So, let â€˜xâ€™ be the mean of the given observation.
x = [38 + 70 + 48 + 34 + 63 + 42 + 55 + 44 + 53 + 47]/10
= 494/10
= 49.4
Number of observations, â€˜nâ€™ = 10
x_{i} 
d_{i} = x_{i} â€“ 49.4 
38 
11.4 
70 
20.6 
48 
1.4 
34 
15.4 
63 
13.6 
42 
7.4 
55 
5.6 
44 
5.4 
53 
3.6 
47 
2.4 
Total 
86.8 
= 1/10 Ã— 86.8
= 8.68
Now
EXERCISE 32.2 PAGE NO: 32.11
1. Calculate the mean deviation from the median of the following frequency distribution:
Heights in inches 
58 
59 
60 
61 
62 
63 
64 
65 
66 
No. of students 
15 
20 
32 
35 
35 
22 
20 
10 
8 
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
We know, Median is the Middle term,
So, Median = 61
Let x_{i}Â =Heights in inches
And, f_{i}Â = Number of students
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M = x_{i} â€“ 61 
f_{i} d_{i} 
58 
15 
15 
3 
45 
59 
20 
35 
2 
40 
60 
32 
67 
1 
32 
61 
35 
102 
0 
0 
62 
35 
137 
1 
35 
63 
22 
159 
2 
44 
64 
20 
179 
3 
60 
65 
10 
189 
4 
40 
66 
8 
197 
5 
40 
N = 197 
Total = 336 
N=197
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/197 Ã— 336
= 1.70
âˆ´ The mean deviation is 1.70.
2. The number of telephone calls received at an exchange in 245 successive on2minute intervals is shown in the following frequency distribution:
Number of calls 
0 
1 
2 
3 
4 
5 
6 
7 
Frequency 
14 
21 
25 
43 
51 
40 
39 
12 
Compute the mean deviation about the median.
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
We know, Median is the even term, (3+5)/2 = 4
So, Median = 8
Let x_{i}Â =Number of calls
And, f_{i}Â = Frequency
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M = x_{i} â€“ 61 
f_{i} d_{i} 
0 
14 
14 
4 
56 
1 
21 
35 
3 
63 
2 
25 
60 
2 
50 
3 
43 
103 
1 
43 
4 
51 
154 
0 
0 
5 
40 
194 
1 
40 
6 
39 
233 
2 
78 
7 
12 
245 
3 
36 
Total = 366 

Total = 245 
N = 245
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/245 Ã— 336
= 1.49
âˆ´ The mean deviation is 1.49.
3. Calculate the mean deviation about the median of the following frequency distribution:
x_{i} 
5 
7 
9 
11 
13 
15 
17 
f_{i} 
2 
4 
6 
8 
10 
12 
8 
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
We know, N = 50
Median = (50)/2 = 25
So, the median Corresponding to 25 is 13
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M = x_{i} â€“ 61 
f_{i} d_{i} 
5 
2 
2 
8 
16 
7 
4 
6 
6 
24 
9 
6 
12 
4 
24 
11 
8 
20 
2 
16 
13 
10 
30 
0 
0 
15 
12 
42 
2 
24 
17 
8 
50 
4 
32 
Total = 50 
Total = 136 
N = 50
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/50 Ã— 136
= 2.72
âˆ´ The mean deviation is 2.72.
4. Find the mean deviation from the mean for the following data:
(i)
x_{i} 
5 
7 
9 
10 
12 
15 
f_{i} 
8 
6 
2 
2 
2 
6 
Solution:
To find the mean deviation from the mean, firstly let us calculate the mean.
By using the formula,
x_{i} 
f_{i} 
Cumulative Frequency (x_{i}f_{i}) 
d_{i} = x_{i} â€“ Mean 
f_{i} d_{i} 
5 
8 
40 
4 
32 
7 
6 
42 
2 
12 
9 
2 
18 
0 
0 
10 
2 
20 
1 
2 
12 
2 
24 
3 
6 
15 
6 
90 
6 
36 
Total = 26 
Total = 234 
Total = 88 
= 234/26
= 9
= 88/26
= 3.3
âˆ´ The mean deviation is 3.3
(ii)
x_{i} 
5 
10 
15 
20 
25 
f_{i} 
7 
4 
6 
3 
5 
Solution:
To find the mean deviation from the mean, firstly let us calculate the mean.
By using the formula,
x_{i} 
f_{i} 
Cumulative Frequency (x_{i}f_{i}) 
d_{i} = x_{i} â€“ Mean 
f_{i} d_{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 
Total = 25 
Total = 350 
Total = 158 
= 350/25
= 14
= 158/25
= 6.32
âˆ´ The mean deviation is 6.32
(iii)
x_{i} 
10 
30 
50 
70 
90 
f_{i} 
4 
24 
28 
16 
8 
Solution:
To find the mean deviation from the mean, firstly let us calculate the mean.
By using the formula,
x_{i} 
f_{i} 
Cumulative Frequency (x_{i}f_{i}) 
d_{i} = x_{i} â€“ Mean 
f_{i} d_{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 
Total = 80 
Total = 4000 
Total = 1280 
= 4000/80
= 50
= 1280/80
= 16
âˆ´ The mean deviation is 16
5. Find the mean deviation from the median for the following data :
(i)
x_{i} 
15 
21 
27 
30 
f_{i} 
3 
5 
6 
7 
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
We know, N = 21
Median = (21)/2 = 10.5
So, the median Corresponding to 10.5 is 27
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M 
f_{i} d_{i} 
15 
3 
3 
15 
45 
21 
5 
8 
9 
45 
27 
6 
14 
3 
18 
30 
7 
21 
0 
0 
Total = 21 
Total = 46 
Total = 108 
N = 21
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/21 Ã— 108
= 5.14
âˆ´ The mean deviation is 5.14
(ii)
x_{i} 
74 
89 
42 
54 
91 
94 
35 
f_{i} 
20 
12 
2 
4 
5 
3 
4 
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
We know, N = 50
Median = (50)/2 = 25
So, the median Corresponding to 25 is 74
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M 
f_{i} d_{i} 
74 
20 
4 
39 
156 
89 
12 
6 
32 
64 
42 
2 
10 
20 
80 
54 
4 
30 
0 
0 
91 
5 
42 
15 
180 
94 
3 
47 
17 
85 
35 
4 
50 
20 
60 
Total = 50 
Total = 189 
Total = 625 
N = 50
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/50 Ã— 625
= 12.5
âˆ´ The mean deviation is 12.5
(iii)
Marks obtained 
10 
11 
12 
14 
15 
No. of students 
2 
3 
8 
3 
4 
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
We know, N = 20
Median = (20)/2 = 10
So, the median Corresponding to 10 is 12
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M 
f_{i} d_{i} 
10 
2 
2 
2 
4 
11 
3 
5 
1 
3 
12 
8 
13 
0 
0 
14 
3 
16 
2 
6 
15 
4 
20 
3 
12 
Total = 20 
Total = 25 
N = 20
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/20 Ã— 25
= 1.25
âˆ´ The mean deviation is 1.25
EXERCISE 32.3 PAGE NO: 32.16
1. Compute the mean deviation from the median of the following distribution:
Class 
010 
1020 
2030 
3040 
4050 
Frequency 
5 
10 
20 
5 
10 
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
Median is the middle term of the X_{i},
Here, the middle term is 25
So, Median = 25
Class Interval 
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M 
f_{i} d_{i} 
010 
5 
5 
5 
20 
100 
1020 
15 
10 
15 
10 
100 
2030 
25 
20 
35 
0 
0 
3040 
35 
5 
91 
10 
50 
4050 
45 
10 
101 
20 
200 
Total = 50 
Total = 450 
= 1/50 Ã— 450
= 9
âˆ´ The mean deviation is 9
2. Find the mean deviation from the mean for the following data:
(i)
Classes 
0100 
100200 
200300 
300400 
400500 
500600 
600700 
700800 
Frequencies 
4 
8 
9 
10 
7 
5 
4 
3 
Solution:
To find the mean deviation from the mean, firstly let us calculate the mean.
By using the formula,
= 17900/50
= 358
Class Interval 
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M 
f_{i} d_{i} 
0100 
50 
4 
200 
308 
1232 
100200 
150 
8 
1200 
208 
1664 
200300 
250 
9 
2250 
108 
972 
300400 
350 
10 
3500 
8 
80 
400500 
450 
7 
3150 
92 
644 
500600 
550 
5 
2750 
192 
960 
600700 
650 
4 
2600 
292 
1168 
700800 
750 
3 
2250 
392 
1176 
Total = 50 
Total = 17900 
Total = 7896 
N = 50
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/50 Ã— 7896
= 157.92
âˆ´ The mean deviation is 157.92
(ii)
Classes 
95105 
105115 
115125 
125135 
135145 
145155 
Frequencies 
9 
13 
16 
26 
30 
12 
Solution:
To find the mean deviation from the mean, firstly let us calculate the mean.
By using the formula,
= 13630/106
= 128.58
Class Interval 
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M 
f_{i} d_{i} 
95105 
100 
9 
900 
28.58 
257.22 
105115 
110 
13 
1430 
18.58 
241.54 
115125 
120 
16 
1920 
8.58 
137.28 
125135 
130 
26 
3380 
1.42 
36.92 
135145 
140 
30 
4200 
11.42 
342.6 
145155 
150 
12 
1800 
21.42 
257.04 
N = 106 
Total = 13630 
Total = 1272.6 
N = 106
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/106 Ã— 1272.6
= 12.005
âˆ´ The mean deviation is 12.005
3. Compute mean deviation from mean of the following distribution:
Marks 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
8090 
No. of students 
8 
10 
15 
25 
20 
18 
9 
5 
Solution:
To find the mean deviation from the mean, firstly let us calculate the mean.
By using the formula,
= 5390/110
= 49
Class Interval 
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M 
f_{i} d_{i} 
1020 
15 
8 
120 
34 
272 
2030 
25 
10 
250 
24 
240 
3040 
35 
15 
525 
14 
210 
4050 
45 
25 
1125 
4 
100 
5060 
55 
20 
1100 
6 
120 
6070 
65 
18 
1170 
16 
288 
7080 
75 
9 
675 
26 
234 
8090 
85 
5 
425 
36 
180 
N = 110 
Total = 5390 
Total = 1644 
N = 110
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/110 Ã— 1644
= 14.94
âˆ´ The mean deviation is 14.94
4. The age distribution of 100 lifeinsurance policy holders is as follows:
Age (on nearest birthday) 
1719.5 
2025.5 
2635.5 
3640.5 
4150.5 
5155.5 
5660.5 
6170.5 
No. of persons 
5 
16 
12 
26 
14 
12 
6 
5 
Calculate the mean deviation from the median age.
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
N = 96
So, N/2 = 96/2 = 48
The cumulative frequency just greater than 48 is 59, and the corresponding value of x is 38.25
So, Median = 38.25
Class Interval 
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ M 
f_{i} d_{i} 
1719.5 
18.25 
5 
5 
20 
100 
2025.5 
22.75 
16 
21 
15.5 
248 
3635.5 
30.75 
12 
33 
7.5 
90 
3640.5 
38.25 
26 
59 
0 
0 
4150.5 
45.75 
14 
73 
7.5 
105 
5155.5 
53.25 
12 
85 
15 
180 
5660.5 
58.25 
6 
91 
20 
120 
6170.5 
65.75 
5 
96 
27.5 
137.5 
Total = 96 
Total = 980.5 
N = 96
\(MD = \frac{1}{n}\sum_{i=1}^{n}d_{i}\)= 1/96 Ã— 980.5
= 10.21
âˆ´ The mean deviation is 10.21
5. Find the mean deviation from the mean and from a median of the following distribution:
Marks 
010 
1020 
2030 
3040 
4050 
No. of students 
5 
8 
15 
16 
6 
Solution:
To find the mean deviation from the median, firstly let us calculate the median.
N = 50
So, N/2 = 50/2 = 25
The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28
So, Median = 28
By using the formula to calculate Mean,
= 1350/50
= 27
Class Interval 
x_{i} 
f_{i} 
Cumulative Frequency 
d_{i} = x_{i} â€“ Median 
f_{i} d_{i} 
F_{i}X_{i} 
X_{i} â€“ Mean 
F_{i }X_{i} â€“ Mean 
010 
5 
5 
5 
23 
115 
25 
22 
110 
1020 
15 
8 
13 
13 
104 
120 
12 
96 
2030 
25 
15 
28 
3 
45 
375 
2 
30 
3040 
35 
16 
44 
7 
112 
560 
8 
128 
4050 
45 
6 
50 
17 
102 
270 
18 
108 
N = 50 
Total = 478 
Total = 1350 
Total = 472 
Mean deviation from Median = 478/50 = 9.56Â
And,Â Mean deviation from Median = 472/50 = 9.44Â
âˆ´ The Mean Deviation from the median is 9.56 and from mean is 9.44.
EXERCISE 32.4 PAGE NO: 32.28
1. Find the mean, variance and standard deviation for the following data:
(i) 2, 4, 5, 6, 8, 17
(ii) 6, 7, 10, 12, 13, 4, 8, 12
2. The variance of 20 observations is 4. If each observation is multiplied by 2, find the variance of the resulting observations.
Solution:
3. The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.
Solution:
4. The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.
Solution:
5. The mean and standard deviation of 6 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:
âˆ´ The mean of new observation is 24 and Standard deviation of new observation is 12.
6. The mean and variance of 8 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:
EXERCISE 32.5 PAGE NO: 32.37
1. Find the standard deviation for the following distribution:
x: 
4.5 
14.5 
24.5 
34.5 
44.5 
54.5 
64.5 
f: 
1 
5 
12 
22 
17 
9 
4 
Solution:
= 100 [1.857 – 0.0987]
= 100 [1.7583]
Var (X) = 175.83
2. Table below shows the frequency f with which â€˜xâ€™ alpha particles were radiated from a diskette
x: 
0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
f: 
51 
203 
383 
525 
532 
408 
273 
139 
43 
27 
10 
4 
2 
Calculate the mean and variance.
Solution:
3. Find the mean, and standard deviation for the following data:
(i)
Year render: 
10 
20 
30 
40 
50 
60 
No. of persons (cumulative) 
15 
32 
51 
78 
97 
109 
Solution:
(ii)
Marks: 
2 
3 
4 
5 
6 
7 
8 
9 
10 
11 
12 
13 
14 
15 
16 
Frequency: 
1 
6 
6 
8 
8 
2 
2 
3 
0 
2 
1 
0 
0 
0 
1 
Solution:
4. Find the standard deviation for the following data:
(i)
x: 
3 
8 
13 
18 
23 
f: 
7 
10 
15 
10 
6 
Solution:
(ii)
x: 
2 
3 
4 
5 
6 
7 
f: 
4 
9 
16 
14 
11 
6 
Solution:
EXERCISE 32.6 PAGE NO: 32.41
1. Calculate the mean and S.D. for the following data:
Expenditure (in â‚¹): 
010 
1020 
2030 
3040 
4050 
Frequency: 
14 
13 
27 
21 
15 
Solution:
2. Calculate the standard deviation for the following data:
Class: 
030 
3060 
6090 
90120 
120150 
150180 
180210 
Frequency: 
9 
17 
43 
82 
81 
44 
24 
Solution:
3. Calculate the A.M. and S.D. for the following distribution:
Class: 
010 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
Frequency: 
18 
16 
15 
12 
10 
5 
2 
1 
Solution:
4. A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure is 40. Find the correct mean and S.D.
Solution:
5. Calculate the mean, median and standard deviation of the following distribution
Classinterval 
3135 
3640 
4145 
4650 
5155 
5660 
6165 
6670 
Frequency: 
2 
3 
8 
12 
16 
5 
2 
3 
Solution:
EXERCISE 32.7 PAGE NO: 32.47
1. Two plants A and B of a factory show the following results about the number of workers and the wages paid to them
Plant A 
Plant B 

No. of workers 
5000 
6000 
Average monthly wages 
â‚¹2500 
â‚¹2500 
The variance of distribution of wages 
81 
100 
In which plant A or B is there greater variability in individual wages?
Solution:
Variation of the distribution of wages in plant A (Ïƒ^{2} =18)
So, Standard deviation of the distribution A (Ïƒ â€“ 9)
Similarly, theÂ VariationÂ of the distribution of wages in plant B (Ïƒ^{2 }=100)
So, Standard deviation of the distribution B (Ïƒ â€“ 10)
And, Average monthly wages in both the plants is 2500,
Since,Â the plant with a greater value of SD will have more variability in salary.
âˆ´ Plant B has more variability in individual wages than plant A
2. The means and standard deviations of heights and weights of 50 students in a class are as follows:
Weights 
Heights 

Mean 
63.2 kg 
63.2 inch 
Standard deviation 
5.6 kg 
11.5 inch 
Which shows more variability, heights or weights?
Solution:
3. The coefficient of variation of two distribution are 60% and 70%, and their standard deviations are 21 and 16 respectively. What is their arithmetic means?
Solution:
= 22.86
âˆ´ Means are 35 and 22.86
4. Calculate coefficient of variation from the following data:
Income (in â‚¹): 
10001700 
17002400 
24003100 
31003800 
38004500 
45005200 
No. of families: 
12 
18 
20 
25 
35 
10 
Solution:
5. An analysis of the weekly 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 
Average weekly wages 
â‚¹52.5 
â‚¹47.5 
The variance of the distribution of wages 
100 
121 
(i) Which firm A or B pays out the larger amount as weekly wages?
(ii) Which firm A or B has greater variability in individual wages?
Solution:
6. The following are some particulars of the distribution of weights of boys and girls in a class:
Boys 
Girls 

Number 
100 
50 
Mean weight 
60 kg 
45 kg 
Variance 
9 
4 
Which of the distributions is more variable?
Solution: