RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1: We’ll discuss about dispersion, range, and mean deviation in this exercise. RD Sharma Class 11 Maths Solutions can help students gain a better understanding of the concepts and develop a good understanding of the subject. Subject experts have prepared the RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 in straightforward, easytounderstand language to fulfil the needs of students and assist them in achieving good grades on their board exams. The answers to this exercise are available in pdf format, which can be simply downloaded from the links provided below.
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Download RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.1 Free PDF
RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1
Access answers to RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1 Important Question with Answers
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=1n∑ni=1di
x_{i} 
d_{i} = x_{i} – 3020 
3011 
9 
2780 
240 
3020 
0 
2354 
666 
3541 
521 
4150 
1130 
5000 
1980 
Total 
4546 
MD=1n∑ni=1di
= 1/7 × 4546
= 649.42
∴ 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=1n∑ni=1di
x_{i} 
d_{i} = x_{i} – 47 
38 
9 
70 
23 
48 
1 
34 
13 
42 
5 
55 
8 
63 
16 
46 
1 
54 
7 
44 
3 
Total 
86 
MD=1n∑ni=1di
= 1/10 × 86
= 8.6
∴ 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=1n∑ni=1di
x_{i} 
d_{i} = x_{i} – 43 
30 
13 
34 
9 
38 
5 
40 
3 
42 
1 
44 
1 
50 
7 
51 
8 
60 
17 
66 
23 
Total 
87 
MD=1n∑ni=1di
= 1/10 × 87
= 8.7
∴ 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=1n∑ni=1di
x_{i} 
d_{i} = x_{i} – 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 
MD=1n∑ni=1di
= 1/10 × 47
= 4.7
∴ 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=1n∑ni=1di
x_{i} 
d_{i} = x_{i} – 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 
MD=1n∑ni=1di
= 1/10 × 84
= 8.4
∴ 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=1n∑ni=1di
Where, d_{i} = x_{i} – x
So, let ‘x’ be the mean of the given observation.
x = [4 + 7 + 8 + 9 + 10 + 12 + 13 + 17]/8
= 80/8
= 10
Number of observations, ‘n’ = 8
x_{i} 
d_{i} = x_{i} – 10 
4 
6 
7 
3 
8 
2 
9 
1 
10 
0 
12 
2 
13 
3 
17 
7 
Total 
24 
MD=1n∑ni=1di
= 1/8 × 24
= 3
∴ The Mean Deviation is 3.
(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
We know that,
MD=1n∑ni=1di
Where, d_{i} = x_{i} – x
So, let ‘x’ 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, ‘n’ = 12
x_{i} 
d_{i} = x_{i} – 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 
MD=1n∑ni=1di
= 1/12 × 28
= 2.33
∴ The Mean Deviation is 2.33.
(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
We know that,
MD=1n∑ni=1di
Where, d_{i} = x_{i} – x
So, let ‘x’ 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, ‘n’ = 10
x_{i} 
d_{i} = x_{i} – 50 
38 
12 
70 
20 
48 
2 
40 
10 
42 
8 
55 
5 
63 
13 
46 
4 
54 
4 
44 
6 
Total 
84 
MD=1n∑ni=1di
= 1/10 × 84
= 8.4
∴ The Mean Deviation is 8.4.
(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
We know that,
MD=1n∑ni=1di
Where, d_{i} = x_{i} – x
So, let ‘x’ 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, ‘n’ = 10
x_{i} 
d_{i} = x_{i} – 50 
36 
14 
72 
22 
46 
4 
42 
8 
60 
10 
45 
5 
53 
3 
46 
4 
51 
1 
49 
1 
Total 
72 
MD=1n∑ni=1di
= 1/10 × 72
= 7.2
∴ The Mean Deviation is 7.2.
(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59
We know that,
MD=1n∑ni=1di
Where, d_{i} = x_{i} – x
So, let ‘x’ 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, ‘n’ = 10
x_{i} 
d_{i} = x_{i} – 55 
57 
2 
64 
9 
43 
12 
67 
12 
49 
6 
59 
4 
44 
11 
47 
8 
61 
6 
59 
4 
Total 
74 
MD=1n∑ni=1di
= 1/10 × 74
= 7.4
∴ 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 ₹ 
II Income in ₹ 
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=1n∑ni=1di
Where, d_{i} = x_{i} – M
x_{i} 
d_{i} = x_{i} – 4400 
4000 
400 
4200 
200 
4400 
0 
4600 
200 
4800 
400 
Total 
1200 
MD=1n∑ni=1di
= 1/5 × 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=1n∑ni=1di
Where, d_{i} = x_{i} – M
x_{i} 
d_{i} = x_{i} – 4400 
3800 
600 
4000 
400 
4200 
200 
4400 
0 
4600 
200 
4800 
400 
5800 
1400 
Total 
3200 
MD=1n∑ni=1di
= 1/7 × 3200
= 457.14
∴ 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) Find the mean deviation from the median.
(ii) Find the mean deviation from the mean also.
Solution:
(i) 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=1n∑ni=1di
Where, d_{i} = x_{i} – M
The number of observations are Even then Median = (40+52.3)/2 = 46.15
Median = 46.15
Number of observations, ‘n’ = 10
x_{i} 
d_{i} = x_{i} – 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 
MD=1n∑ni=1di
= 1/10 × 167.4
= 16.74
∴ The Mean Deviation is 16.74.
(ii) Find the mean deviation from the mean also.
We know that,
MD=1n∑ni=1di
Where, d_{i} = x_{i} – x
So, let ‘x’ 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, ‘n’ = 10
x_{i} 
d_{i} = x_{i} – 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 
MD=1n∑ni=1di
= 1/10 × 166.4
= 16.64
∴ TheMean Deviation is 16.64
5. In question 1(iii), (iv), (v) find the number of observations lying between X¯¯¯¯–M.D and X¯¯¯¯+M.D, where M.D. is the mean deviation from the mean.
Solution:
(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51
We know that,
MD=1n∑ni=1di
Where, d_{i} = x_{i} – x
So, let ‘x’ 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, ‘n’ = 10
x_{i} 
d_{i} = x_{i} – 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 
MD=1n∑ni=1di
= 1/10 × 90
= 9
Now
X¯¯¯¯–M.D = 45.5 – 9 = 36.5
X¯¯¯¯+M.D = 45.5 + 9 = 54.5
So, There are total 6 observation between X¯¯¯¯–M.D and X¯¯¯¯+M.D
(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42
We know that,
MD=1n∑ni=1di
Where, d_{i} = x_{i} – x
So, let ‘x’ 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, ‘n’ = 10
x_{i} 
d_{i} = x_{i} – 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 
MD=1n∑ni=1di
= 1/10 × 48.8
= 4.88
Now
So, There are total 5 observation between and
(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47
We know that,
Where, d_{i} = x_{i} – x
So, let ‘x’ 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, ‘n’ = 10
x_{i} 
d_{i} = x_{i} – 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 
MD=1n∑ni=1di
= 1/10 × 86.8
= 8.68
Now
Statistics Ex 32.1 Q6
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