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

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

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

$MD=\frac{1}{n}\sum _{i=1}^n\left|d_i\right|,where\left|d_i\right|=\left|x_i-\overline{x}\right|$

$x_i$	$\left|d_i\right|=\left|x_i-45.5\right|$
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
Total	90

$MD=\frac{1}{10}\times 90=9$

$$\begin{gathered} \overline{x}-M.D.=45.5-9=36.5 \\ \mathrm{Also},\overline{x}+M.D.=45.5+9=54.5 \end{gathered}$$

Hence, there are 6 observations between 36.5 and 54.5.

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

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

$x_i$	$\left|d_i\right|=\left|x_i-29.9\right|$
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
Total	48.8

$MD=\frac{1}{10}\times 48.8=4.88$

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

There are 5 observations between 25.02 and 34.78.

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

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

$x_i$	$\left|d_i\right|=\left|x_i-49.4\right|$
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
Total	86.8

$MD=\frac{1}{10}\times 86.6=8.68$

$$\begin{gathered} \overline{x}-MD=49.4-8.68=40.72 \\ and\overline{x}+MD=49.4+8.68=58.08 \end{gathered}$$

There are 6 observations between 40.72 and 58.08.