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Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.2
• Last Updated : 11 Feb, 2021

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

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

N=197

= 1/197 × 336

= 1.70

Therefore, the mean deviation is 1.70.

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

= 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:

Solution:

Calculating the median,

We know, Number of observations, N = 50

Median = (50)/2 = 25

Therefore, the median corresponding to 25 is 13.

= 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)

(ii)

(iii)

Solution:

(i) We know,

Now, Mean = 234/26

= 9

= 88/26

= 3.3

∴ The mean deviation is 3.3

(ii) We know,

Mean = 350/25

= 14

= 158/25

= 6.32

∴ The mean deviation is 6.32

(iii) We know,

= 4000/80

= 50

= 1280/80

= 16

∴ The mean deviation is 16

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

(i)

(ii)

(iii)

Solution:

(i) We know,

Number of observations, N = 21

Median = (21)/2 = 10.5

Therefore, the median corresponding to 10.5 is 27

= 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.

= 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.

= 1/20 × 25

= 1.25

∴ The mean deviation is 1.25

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