Question

Find the mean deviation from the mean and from median of the following distribution:

Marks	0-10	10-20	20-30	30-40	40-50
No. of students	5	8	15	16	6

Answer 1

Computation of mean distribution from the median:

Marks	Number of Students $f_i$	Cumulative Frequency	Midpoints $x_i$	$\left|d_i\right|=\left|x_i-28\right|$	$f_{\mathrm{i}}\left|d_{\mathit{i}}\right|$	$f_i{x}_i$	$\left|x_i-27\right|$	$f_{\mathrm{i}}\left|x_{\mathit{i}}-27\right|$
0−10	5	5	5	23	115	25	22	110
10−20	8	13	15	13	104	120	12	96
20−30	15	28	25	3	45	375	2	30
30−40	16	44	35	7	112	560	8	128
40−50	6	50	45	17	102	270	18	108
	$N=50$				$\sum _{i=1}^5{f}_i\left|d_i\right|=478$	1350		$\sum _{i=1}^5{f}_i\left|x_i-27\right|=472$

$N=50,\frac{N}{2}=25$

The cumulative frequency just greater than $\frac{N}{2}=25$ is 28 and the corresponding class is 20−30.
Thus, the median class is 20−30.

Using formula:

$$\begin{gathered} \therefore l=20,F=13,f=15,h=10 \\ \mathrm{Median}=l+\frac{{\displaystyle \frac{N}{2}}-F}{f}\times h \\ \mathrm{Substituting}\mathrm{the}\mathrm{values}: \\ \mathrm{Median}=20+\frac{25-13}{15}\times 10 \\ =20+8 \\ =28 \end{gathered}$$
$$\begin{gathered} \mathrm{Mean}\mathrm{distribution}\mathrm{from}\mathrm{the}\mathrm{median}=\frac{{\displaystyle \sum _{i=1}^5}{f}_i\left|d_i\right|}{N} \\ =\frac{478}{50} \\ =9.56 \end{gathered}$$

$$\begin{gathered} \mathrm{Mean}\left(\overline{X}\right)=\frac{{\displaystyle \sum _{i=1}^5}{f}_i{x}_i}{N} \\ =\frac{1350}{50} \\ =27 \\ \mathrm{Mean}\mathrm{deviation}\mathrm{from}\mathrm{the}\mathrm{mean}=\frac{1}{50}\times \sum _{\mathrm{i}=1}^5{f}_i\left|x_{\mathrm{i}}-27\right| \\ =\frac{472}{50} \\ =9.44 \end{gathered}$$

Mean deviation from the median and the mean are 9.56 and 9.44, respectively.