# Find the mean deviation about the median for the data 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

**Solution:**

The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Here, the numbers of observations are 10, i.e., even.

Arranging the above data in ascending order,

we obtain 36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Median of the data

M = [(10/2)^{th} observation + (10/2 + 1)^{th} observation]/2

= [5^{th} observation + 6^{th} observation]/2

= (46 + 49)/2

= 95/2

= 47.5

The deviations of the respective observations from the median, i.e., x_{i} - M are

- 11.5, - 5.5, - 2.5, - 1.5, - 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The absolute values of the deviations, |x_{i} - M| are

11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

The required mean deviation about the median is

M.D. (x) = ^{10}∑_{i = 1} (|x_{i} - x|)/10

= (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5)/10

= 70/10

= 7

NCERT Solutions Class 11 Maths Chapter 15 Exercise 15.1 Question 4

## Find the mean deviation about the median for the data 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

**Summary:**

The mean deviation about the median for the given data is 7

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