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

• Last Updated : 11 Feb, 2021

### (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

Calculating Mean Deviation:

= 1/7 Ã— 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

Calculating Mean Deviation:

= 1/10 Ã— 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

Calculating Mean Deviation:

= 1/10 Ã— 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

Calculating Mean Deviation:

= 1/10 Ã— 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

Calculating Mean Deviation:

= 1/10 Ã— 84

= 8.4

âˆ´ The Mean Deviation is 8.4.

### (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, |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

MD = 1/8 * 24

= 3

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

Since,

Mean Deviation,

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

Now,

MD = 1/12 Ã— 28

= 2.33

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

We know that,

Mean Deviation,

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

MD = 1/10 Ã— 84

= 8.4

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

Mean Deviation,

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

MD = 1/10 Ã— 72

= 7.2

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

Mean Deviation,

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

MD = 1/10 Ã— 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,

MD(I) = 1/5 Ã— 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(II) = 1/7 Ã— 3200

= 457.14

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

### (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

MD = 1/10 * 167.4

=16.74

(ii) Mean deviation from the mean also.

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

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, |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

MD = 1/10 Ã— 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, |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

MD = 1/10 Ã— 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, |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

MD = = 1/10 Ã— 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:

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