Question

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

Answer 1

Answer:

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

now, we have to arrange it in ascending order,

e.g., 36, 42, 45, 46, 46, 49, 51,53, 60, 72.

here, number of observation ( N) = 10 ( an even number )

$$\begin{lgathered}Median(M)=\frac{\frac{Nth}{2} observation+(\frac{N}{2}+1)th\: observation}{2}\\\\=\frac{\frac{10}{2}th\: observation+(\frac{10}{2}+1)th\: observation}{2}\\\\=\frac{5th\: observation+6th\: observation}{2}\end{lgathered}$$

now, see in arrangement of ascending order,

5th position = 46

6th position = 49

now,

M = ( 46 + 49)/2 = 47.5

now, mean deviation about Median

$$\begin{lgathered}=\frac{\sum{|x_i-M|}}{n}\\\\=\frac{70}{10}\\\\=7\end{lgathered}$$

Answer 2

Answer:

47.5

your mean division is 47.5