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Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.2

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Question 1. Calculate the mean deviation from the median of the following frequency distribution:

Heights in inches 58 59 60 61 62 63 64 65 66
No. Of Students 15 20 32 35 35 22 20 10 8

Solution:

Median is the middle term of the observation in ascending order,

So, Median = 61

Let us assume, 

xi =Heights in inches

fi = Number of students

xi fi Cumulative Frequency

|di| = |xi – M|

= |xi – 61|

fi |di|
58 15 15 3 45
59 20 35 2 40
60 32 67 1 32
61 35 102 0 0
62 35 137 1 35
63 22 159 2 44
64 20 179 3 60
65 10 189 4 40
66 8 197 5 40
  N = 197     Total = 336

N=197

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/197 × 336

= 1.70

Therefore, the mean deviation is 1.70.

Question 2. The number of telephone calls received at an exchange in 245 successive on2-minute intervals is shown in the following frequency distribution:

Number of calls 0 1 2 3 4 5 6 7
Frequency 14 21 25 43 51 40 39 12

Compute the mean deviation about the median.

Solution:

Median is the middle term of the observation in ascending order,

We know, Median is the even term, (3+5)/2 = 4

So, Median = 8

Let us assume

xi =Number of calls

fi = Frequency

N = 245

xi fi Cumulative Frequency

|di| = |xi – M|

= |xi – 61|

fi |di|
0 14 14 4 56
1 21 35 3 63
2 25 60 2 50
3 43 103 1 43
4 51 154 0 0
5 40 194 1 40
6 39 233 2 78
7 12 245 3 36
  Total = 245     Total = 366

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/245 × 336

= 1.49

Therefore, mean deviation is 1.49.

Question 3. Calculate the mean deviation about the median of the following frequency distribution:

xi 5 7 9 11 13 15 17
fi 2 4 6 8 10 12 8

Solution:

Calculating the median,

We know, Number of observations, N = 50

Median = (50)/2 = 25

Therefore, the median corresponding to 25 is 13.

xi fi Cumulative Frequency

|di| = |xi – M|

= |xi – 61|

fi |di|
5 2 2 8 16
7 4 6 6 24
9 6 12 4 24
11 8 20 2 16
13 10 30 0 0
15 12 42 2 24
17 8 50 4 32
  Total = 50     Total = 136

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/50 × 136

= 2.72

Therefore, the mean deviation is 2.72.

Question 4. Find the mean deviation from the mean for the following data:

(i)

xi 5 7 9 10 12 15
fi 8 6 2 2 2 6

(ii)

xi 5 10 15 20 25
fi 7 4 6 3 5

(iii)

xi 10 30 50 70 90
fi 4 24 28 16 8

Solution:

(i) We know,

Mean = \frac{\sum f_ix_i}{f_i}

xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di|
5 8 40 4 32
7 6 42 2 12
9 2 18 0 0
10 2 20 1 2
12 2 24 3 6
15 6 90 6 36
  Total = 26 Total = 234   Total = 88

Now, Mean = 234/26

= 9

Mean Deviation = \frac{\sum f_i|d_i|}{f_i}

= 88/26

= 3.3

∴ The mean deviation is 3.3

(ii) We know,

Mean = \frac{\sum f_ix_i}{f_i}

xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di|
5 7 35 9 63
10 4 40 4 16
15 6 90 1 6
20 3 60 6 18
25 5 125 11 55
  Total = 25 Total = 350   Total = 158

Mean = 350/25

= 14
Mean Deviation = \frac{\sum f_i|d_i|}{f_i}

= 158/25

= 6.32

∴ The mean deviation is 6.32

(iii) We know,

Mean = \frac{\sum f_ix_i}{f_i}

= 4000/80

= 50

xi fi Cumulative Frequency (xifi) |di| = |xi – Mean| fi |di|
10 4 40 40 160
30 24 720 20 480
50 28 1400 0 0
70 16 1120 20 320
90 8 720 40 320
  Total = 80 Total = 4000   Total = 1280

Mean Deviation = \frac{\sum f_i|d_i|}{f_i}

= 1280/80

= 16

∴ The mean deviation is 16

Question 5. Find the mean deviation from the median for the following data :

(i)

xi 15 21 27 30
fi 3 5 6 7

(ii)

xi 74 89 42 54 91 94 35
fi 20 12 2 4 5 3 4

(iii)

Marks obtained 10 11 12 14 15
No. of students 2 3 8 3 4

Solution:

(i) We know, 

Number of observations, N = 21

Median = (21)/2 = 10.5

Therefore, the median corresponding to 10.5 is 27

xi fi Cumulative Frequency |di| = |xi – M| fi |di|
15 3 3 15 45
21 5 8 9 45
27 6 14 3 18
30 7 21 0 0
  Total = 21 Total = 46   Total = 108

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/21 × 108

= 5.14

∴ The mean deviation is 5.14

(ii) We know, 

Number of observations, N =50

Median = (50)/2 = 25

Therefore, the median corresponding to 25 is 74.

xi fi Cumulative Frequency |di| = |xi – M| fi |di|
74 20 4 39 156
89 12 6 32 64
42 2 10 20 80
54 4 30 0 0
91 5 42 15 180
94 3 47 17 85
35 4 50 20 60
  Total = 50 Total = 189   Total = 625

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/50 × 625

= 12.5

Therefore, the mean deviation is 12.5

(iii) We know, 

Number of observations, N =20

Median = (20)/2 = 10

So, the median corresponding to 10 is 12.

xi fi Cumulative Frequency |di| = |xi – M| fi |di|
10 2 2 2 4
11 3 5 1 3
12 8 13 0 0
14 3 16 2 6
15 4 20 3 12
  Total = 20     Total = 25

MD=\frac{1}{n} \sum^{n}_{i=1}|di|

= 1/20 × 25

= 1.25

∴ The mean deviation is 1.25


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Last Updated : 11 Feb, 2021
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