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

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  • Last Updated : 11 Feb, 2021
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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:

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

= 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:

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

= 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:

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

= 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:

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

= 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:

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

= 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, 

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

Where, |di| = |xi – 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, MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

Where, |di| = |xi – 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, MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

Where, |di| = |xi – 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, 

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

Where, |di| = |xi – 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, 

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

Where, |di| = |xi – 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:

I

Income in ₹

II

Income in ₹
4000 3800
4200 4000
4400 4200
4600 4400
4800 4600
  4800
  5800

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, 

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

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, 

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

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,

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

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.

MD = \frac{1}{n}\sum_{i=1}^{n}|d_i|

Where, |di| = |xi – 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 between\bar{X}-M.D and \bar{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 = \frac{1}{n}\sum_{i=1}^{n}|d_i|

Where, |di| = |xi – 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,

\overline{X}-M D = 45.5 - 9 = 36.5 \\ \overline{X}+MD = 45.5 + 9 = 54.5

So, There are total 6 observation between \overline{X}-MD  and \overline{X}+MD

(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, |di| = |xi – 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,

\overline{X}-M D = 29.9 - 4.88 = 25.02 \\ \overline{X}+MD = 29.9 + 4.88 = 34.78

So, there are 5 observations in between. 

(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, |di| = |xi – 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, 

\overline{X}-M D = 49.4 - 8.68 = 40.72 \\ \overline{X}+MD = 49.4 + 8.68 = 58.08

There are 6 observations in between. 

Question 6. Show that the two formulae for the standard deviation of the ungrouped data \sigma = \sqrt{\frac{1}{n}\sum((x_i-\overline x)^{2}}  and \sigma = \sqrt{\frac{1}{n}\sum((x_i^2 -\overline x)^{2}}  are equivalent, where \overline x = \frac{1}{n}\sum x_i

Solution:

\sigma = \sqrt{\frac{1}{n}\sum(x_i - \overline x)^2} \\ \sigma' = \sqrt{\frac{1}{n}\sum x_i^2 - \overline x^2} \\ (x_i - \overline x)^2 = x_i^2 + \overline x^2 - 2x_i\overline x \\ \sum 2x_i\overline x = 2\overline x\sum x_i = 2n\overline x^2 \\ \frac{1}{n}\sum(x_i - \overline x)^2 = \frac{\sum(x_i^2+\overline x^2-2x_i\overline x)}{n} \\= \frac{1}{n} \sum x_i^2 - \overline x^2


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