Question

Calculate the mean deviation about the median of the following observations :

(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

Answer 1

First arrange the given numbers in ascending order write these numbers in ascending order 3011, 2780, 3020, 2354, 3541, 4150, 5000

we get 2354, 2780, 3011, 3020, 3541, 4150, 5000

Clearly, the middle number is median, 3020

Calculation of Mean Deviations

$\begin{array}{cc}{x}_i& |d_i|=|x_i-3020|\\ 3011& 9\\ 2780& 240\\ 3020& 0\\ 2354& 666\\ 3541& 521\\ 4150& 1130\\ 5000& 1980\\ \text{Total}& d_i=|x_i-3020|=4546\end{array}$

$M.D.=\frac{\sum d_i}{n}=\frac{4546}{7}=649.428$

(ii) 38, 70, 48, 34, 42, 55,63, 46, 54, 44

Arranging the data in asceding order

34, 38, 42, 44, 46, 48, 54, 55, 63, 70

Here, n is equal to 10.

Median is the arithmetic mean of the fifth and the sixth observation.

Median, $M=\frac{46+48}{2}=47$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-M|\\ 38& 9\\ 70& 23\\ 48& 1\\ 34& 13\\ 42& 5\\ 55& 8\\ 63& 16\\ 46& 1\\ 54& 7\\ 44& 3\\ \text{Total}& 86\end{array}$

$M.D=\frac{1}{10}\times 86=8.6$

(iii) 34, 66, 30, 38, 44, 50, 40, 60, 42, 51

Arranging the data in ascending order : 30, 34, 38, 40, 42, 44, 50, 51, 60, 66

Here,

n = 10

Also, median is the AM of the fifth and the sixth observation.

Median, $M=\frac{42+44}{2}=43$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-M|\\ 34& 9\\ 66& 23\\ 30& 13\\ 38& 5\\ 44& 1\\ 50& 7\\ 40& 3\\ 60& 17\\ 42& 1\\ 51& 8\\ \text{Total}& 87\end{array}$

$M.D=\frac{1}{10}\times 87=8.7$

(iv) 22, 24, 30, 27, 29, 31, 25, 28, 41, 42

Arranging the data in ascending order

22, 24, 25, 27, 28, 29, 30, 31, 41, 42

Here, n = 10

Also median in the AM of the fifth and the sixth observation.

Median, $M=\frac{28+29}{2}=28.5$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-M|\\ 22& 6.5\\ 24& 4.5\\ 30& 1.5\\ 27& 1.5\\ 29& 0.5\\ 31& 2.5\\ 25& 3.5\\ 28& 0.5\\ 41& 12.5\\ 42& 13.5\\ \text{Total}& 47\end{array}$

$M.D=\frac{1}{10}\times 47=4.7$

(v) 38, 70, 48, 34, 63, 42, 55, 44, 53, 47

Arranging the data in ascending order.

34, 38, 42, 44, 47, 48, 53, 55, 63, 70

Here, n = 10

Also, median is the AM of the fifth and the sixth observation.

Median, $M=\frac{47+48}{2}=47.5$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-M|\\ 38& 9.5\\ 70& 22.5\\ 48& 0.5\\ 34& 13.5\\ 63& 15.5\\ 42& 5.5\\ 55& 7.5\\ 44& 3.5\\ 53& 5.5\\ 47& 0.5\\ \text{Total}& 84\end{array}$

$M.D=\frac{1}{10}\times 84=8.4$