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, easy-to-understand 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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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=1|di|
|
xi |
|di| = |xi – 3020| |
|
3011 |
9 |
|
2780 |
240 |
|
3020 |
0 |
|
2354 |
666 |
|
3541 |
521 |
|
4150 |
1130 |
|
5000 |
1980 |
|
Total |
4546 |
MD=1n∑ni=1|di|
= 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=1|di|
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
Where, |di| = |xi – 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
|
xi |
|di| = |xi – 10| |
|
4 |
6 |
|
7 |
3 |
|
8 |
2 |
|
9 |
1 |
|
10 |
0 |
|
12 |
2 |
|
13 |
3 |
|
17 |
7 |
|
Total |
24 |
MD=1n∑ni=1|di|
= 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=1|di|
Where, |di| = |xi – 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
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
Where, |di| = |xi – 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
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
Where, |di| = |xi – 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
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
Where, |di| = |xi – 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
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
Where, |di| = |xi – M|
|
xi |
|di| = |xi – 4400| |
|
4000 |
400 |
|
4200 |
200 |
|
4400 |
0 |
|
4600 |
200 |
|
4800 |
400 |
|
Total |
1200 |
MD=1n∑ni=1|di|
= 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=1|di|
Where, |di| = |xi – M|
|
xi |
|di| = |xi – 4400| |
|
3800 |
600 |
|
4000 |
400 |
|
4200 |
200 |
|
4400 |
0 |
|
4600 |
200 |
|
4800 |
400 |
|
5800 |
1400 |
|
Total |
3200 |
MD=1n∑ni=1|di|
= 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=1|di|
Where, |di| = |xi – M|
The number of observations are Even then Median = (40+52.3)/2 = 46.15
Median = 46.15
Number of observations, ‘n’ = 10
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
Where, |di| = |xi – 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
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
Where, |di| = |xi – 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
|
xi |
|di| = |xi – 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=1|di|
= 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=1|di|
Where, |di| = |xi – 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
|
xi |
|di| = |xi – 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=1|di|
= 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, |di| = |xi – 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
|
xi |
|di| = |xi – 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=1|di|
= 1/10 × 86.8
= 8.68
Now
Statistics Ex 32.1 Q6
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FAQs on RD Sharma Solutions Class 11 Maths Chapter 32 Exercise 32.1
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