Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 11 Feb, 2021
Geek Week

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|
30119
2780240
30200
2354666
3541521
41501130
50001980
Total4546

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|
389
7023
481
3413
425
558
6316
461
547
443
Total86

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|
3013
349
385
403
421
441
507
518
6017
6623
Total87

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|
226.5
244.5
301.5
271.5
290.5
312.5
253.5
280.5
4112.5
4213.5
Total47

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|
389.5
7022.5
480.5
3413.5
6315.5
425.5
557.5
443.5
535.5
470.5
Total84

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|
46
73
82
91
100
122
133
177
Total24

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|
131
173
162
140
113
131
104
162
113
184
122
173
Total28

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|
3812
7020
482
4010
428
555
6313
464
544
446
Total84

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|
3614
7222
464
428
6010
455
533
464
511
491
Total72

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|
572
649
4312
6712
496
594
4411
478
616
594
Total74

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 ₹

40003800
42004000
44004200
46004400
48004600
 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|
4000400
4200200
44000
4600200
4800400
Total1200

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|
3800600
4000400
4200200
44000
4600200
4800400
58001400
Total3200

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|
40.06.15
52.36.15
55.29.05
72.926.75
52.86.65
79.032.85
32.513.65
15.230.95
27.919.25
30.215.95
Total167.4

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|
40.05.8
52.36.5
55.29.4
72.927.1
52.87
79.033.2
32.513.3
15.230.6
27.917.9
30.215.6
Total166.4

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|
3411.5
6620.5
3015.5
387.5
441.5
504.5
405.5
6014.5
423.5
515.5
Total90

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|
227.9
245.9
300.1
272.9
290.9
311.1
254.9
281.9
4111.1
4212.1
Total48.8

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|
3811.4
7020.6
481.4
3415.4
6313.6
427.4
555.6
445.4
533.6
472.4
Total86.8

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

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :