# RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.1 (Updated for 2021-22)

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.

## Download RD Sharma Class 11 Solutions Chapter 32 Statistics Exercise 32.1 Free PDF

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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