Class 11 RD Sharma Solutions - Chapter 32 Statistics - Exercise 32.1

Question 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

Calculating Median (M) of the following observation:

Arranging numbers in ascending order,

2354, 2780, 3011, 3020, 3541, 4150, 5000

Median is the middle number of all the observations.

Therefore, Median = 3020 and n = 7

xi	|di| = |xi - 3020|
3011	9
2780	240
3020	0
2354	666
3541	521
4150	1130
5000	1980
Total	4546

Calculating Mean Deviation:

= 1/7 × 4546

= 649.42

Hence, Mean Deviation is 649.42.

(ii) 38, 70, 48, 34, 42, 55, 63, 46, 54, 44

Calculating Median (M) of the following observation:

Arranging numbers in ascending order,

34, 38, 42, 44, 46, 48, 54, 55, 63, 70

Median is the middle number of all the observations.

Here, the number of observations are even, 

therefore the Median = (46 + 48)/2 = 47

Median = 47 and n = 10

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

Calculating Mean Deviation:

= 1/10 × 86

= 8.6

Hence, Mean Deviation is 8.6.

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

Calculating Median (M) of the following observation:

Arranging numbers in ascending order,

30, 34, 38, 40, 42, 44, 50, 51, 60, 66

Median is the middle number of all the observations.

Here, the number of observations are even, 

therefore the Median = (42 + 44)/2 = 43

Median = 43 and n = 10

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

Calculating Mean Deviation:

= 1/10 × 87

= 8.7

Hence, Mean Deviation is 8.7.

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

Calculating Median (M) of the following observation:

Arranging numbers in ascending order,

22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Median is the middle number of all the observations.

Here, the number of observations are even, 

therefore the Median = (28 + 29)/2 = 28.5

Median = 28.5 and n = 10

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

Calculating Mean Deviation:

= 1/10 × 47

= 4.7

Hence, Mean Deviation is 4.7.

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Calculating Median (M) of the following observation:

Arranging numbers in ascending order,

34, 38, 43, 44, 47, 48, 53, 55, 63, 70

Median is the middle number of all the observation.

Here, the number of observations are even,

therefore the Median = (47 + 48)/2 = 47.5

Median = 47.5 and n = 10

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

Calculating Mean Deviation:

= 1/10 × 84

= 8.4

∴ The Mean Deviation is 8.4.

Question 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, Mean Deviation, 

Where, |d_i| = |x_i - x|

So, let us assume x to be the mean of the given observation.

Now, 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 = 1/8 * 24

= 3

(ii) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Since,

Mean Deviation, 

Where, |d_i| = |x_i - x|

So, let us assume x to 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

Now, 

MD = 1/12 × 28

= 2.33

(iii) 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

We know that,

Mean Deviation, 

Where, |d_i| = |x_i - x|

So, let us assume x to 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 = 1/10 × 84

= 8.4

(iv) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Mean Deviation, 

Where, |d_i| = |x_i - x|

So, let us assume x to 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 = 1/10 × 72

= 7.2

(v) 57, 64, 43, 67, 49, 59, 44, 47, 61, 59

Mean Deviation, 

Where, |d_i| = |x_i - x|

So, let us assume x to 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 = 1/10 × 74

= 7.4

Question 3. Calculate the mean deviation of the following income groups of five and seven members from their medians:

Solution:

Dataset I : 

Since the data is arranged in ascending order,

4000, 4200, 4400, 4600, 4800

Median (Middle of ascending order observation) = 4400

Total observations, n = 5

Now, Mean Deviation, 

xi	|di| = |xi – 4400|
4000	400
4200	200
4400	0
4600	200
4800	400
Total	1200

MD(I) = 1/5 × 1200

= 240

Dataset II :

Since the data is arranged in ascending order,

3800, 4000, 4200, 4400, 4600, 4800, 5800

Median (Middle of ascending order observation) = 4400

Total observations, n = 7

Now, Mean Deviation, 

xi	|di| = |xi – 4400|
3800	600
4000	400
4200	200
4400	0
4600	200
4800	400
5800	1400
Total	3200

MD(II) = 1/7 × 3200

= 457.14

Therefore, the Mean Deviation of set 1, MD(I) is 240 and set 2, MD(II) is 457.14

Question 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) The mean deviation from the median

Arranging 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,

Since, the number of observations are even, 

therefore Median = (40 + 52.3)/2 = 46.15

Median = 46.15

Also, 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 = 1/10 * 167.4

=16.74

(ii) Mean deviation from the mean also.

Where, |d_i| = |x_i - x|

So, let us assume x to be the mean of the given observation.

Now, 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

And, 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 = 1/10 * 166.4

= 16.64

Question 5. In question 1(iii), (iv), (v) find the number of observations lying betweenand , 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,

Where, |d_i| = |x_i - x|

So, let us assume x to be the mean of the given observation.

x = [34 + 66 + 30 + 38 + 44 + 50 + 40 + 60 + 42 + 51]/10

= 455/10

= 45.5

And, 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 = 1/10 × 90

= 9

Now,

So, There are total 6 observation between   and 

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

We know that,

Where, |d_i| = |x_i - x|

So, let us assume x to be the mean of the given observation.

x = [22 + 24 + 30 + 27 + 29 + 31 + 25 + 28 + 41 + 42]/10

= 299/10

= 29.9

Also, 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 = 1/10 × 48.8

= 4.88

And,

So, there are 5 observations in between. 

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

We know that,

Where, |d_i| = |x_i - x|

So, let us assume x to 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 = = 1/10 × 86.8

= 8.68

Also, 

There are 6 observations in between. 

Question 6. Show that the two formulae for the standard deviation of the ungrouped data  and  are equivalent, where 

Solution:

Summary

Exercise 32.1 introduces the basics of statistics, including definitions, types of data, and importance of statistics. Students learn about primary and secondary data, quantitative and qualitative data, variability, discrete and continuous data, inferential statistics, and descriptive statistics. Understanding these concepts lays the foundation for further study.