Question

In question 1 (iii), (iv), (v) find the number of observations lying between $\overline{X}-M.D.$ and $\overline{X}-M.D,$ where M.D. is the mean deviation from the mean.

Answer 1

Let $\overline{x}$ be the mean of the data set.

$\overline{x}=\frac{34+66+30+38+44+50+40+60+42+51}{10}=45.5$

$M.D.=\frac{1}{n}{\sum }_{i=1}^n|d_i|$ where $|d_i|=|x_i-\overline{x}|$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-45.5|\\ 34& 11.5\\ 66& 20.5\\ 30& 15.5\\ 38& 7.5\\ 44& 1.5\\ 50& 4.5\\ 40& 5.5\\ 60& 14.5\\ 42& 3.5\\ 51& 5.5\\ \text{Total}& 90\end{array}$

$M.D.=\frac{1}{10}\times 90=9$

$\overline{x}-M.D.=45.5-9=36.5$

Also, $\overline{x}+M.D.=45.5+9=54.5$

Hence, there are 6 observation between 36.5 and 54.4

(ii) Let $\overline{x}$ be the mean of the data set

$\overline{x}=\frac{22+24+30+27+29+31+25+28+41+42}{10}=29.9$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-45.5|\\ 22& 7.9\\ 24& 5.9\\ 30& 0.1\\ 27& 2.9\\ 29& 0.9\\ 31& 1.1\\ 25& 4.9\\ 28& 1.9\\ 41& 11.9\\ 42& 12.1\\ \text{Total}& 48.8\end{array}$

$M.D.=\frac{1}{10}\times 48.8=4.88$

$\overline{x}-M.D.=29.9-4.88=25.02,$ and $\overline{x}+M.D.=29.9+4.88=34.78$

There are 5 observations between 25.02 and 34.78

(v) Let $\overline{x}$ be the mean of the data set.

$\overline{x}=\frac{38+70+48+34+63+42+55+44+53+47}{10}=49.4$

$\begin{array}{cc}{x}_i& |d_i|=|x_i-45.5|\\ 38& 11.4\\ 70& 20.6\\ 48& 1.4\\ 34& 15.4\\ 63& 13.6\\ 42& 7.4\\ 55& 5.6\\ 44& 5.4\\ 53& 3.6\\ 47& 2.4\\ \text{Total}& 86.8\end{array}$

$M.D.=\frac{1}{10}\times 86.6=8.68$

$\overline{x}-M.D=49.4-8.68=40.72$

and $\overline{x}+M.D.=49.4+8.68=58.08$

There are 6 observations between 40.72' and and 58.08.