Use the following data including ages of $\mathbf{100}$ men to find the mean deviation about the median.
Age | Number of men |
$\mathbf{16}-\mathbf{20}$ | $\mathbf{5}$ |
$\mathbf{21}-\mathbf{25}$ | $\mathbf{6}$ |
$\mathbf{26}-\mathbf{30}$ | $\mathbf{12}$ |
$\mathbf{31}-\mathbf{35}$ | $\mathbf{14}$ |
$\mathbf{36}-\mathbf{40}$ | $\mathbf{26}$ |
$\mathbf{41}-\mathbf{45}$ | $\mathbf{12}$ |
$\mathbf{46}-\mathbf{50}$ | $\mathbf{16}$ |
$\mathbf{51}-\mathbf{55}$ | $\mathbf{9}$ |
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Ans.
Observe that the provided data is not continuous. So, to convert these discrete data to continuous data, subtract $0.5$ from the lower limit and add $0.5$ to the upper limit of the ages.
So, consider the following table of data.
Age | Number of men ${{f}_{i}}$ | Cumulative frequency (c.f.) | Mid-point ${{x}_{i}}$ | $\left| {{x}_{i}}-Med. \right|$ | ${{f}_{i}}\left| {{x}_{i}}-Med. \right|$ |
$15.5-20.5$ | $5$ | $5$ | $18$ | $20$ | $100$ |
$20.5-25.5$ | $6$ | $11$ | $23$ | $15$ | $90$ |
$25.5-30.5$ | $12$ | $23$ | $28$ | $10$ | $120$ |
$30.5-35.5$ | $14$ | $37$ | $33$ | $5$ | $70$ |
$35.5-40.5$ | $26$ | $63$ | $38$ | $0$ | $0$ |
$40.5-45.5$ | $12$ | $75$ | $43$ | $5$ | $60$ |
$45.5-50.5$ | $16$ | $91$ | $48$ | $10$ | $160$ |
$50.5-55.5$ | $9$ | $100$ | $53$ | $15$ | $135$ |
Total | $100$ | $735$ |
Therefore, here $N=100$ and it is an even number.
So, $\dfrac{N}{2}=\dfrac{100}{2}=50$. That is, the cumulative frequency greater than $50$ is $63$. Thus, the median class $=$$35.5-40.5$.
Now, we known that,
Median, $M=l+\dfrac{\dfrac{N}{2}-C}{f}\times h$
We have $l=35.5$, $C=37$, $f=26$, $h=5$. Also, $N=100$.
Therefore,
$M=35.5+\dfrac{50-37}{26}\times 5=35.5+\dfrac{13\times 5}{26}=35.5+2.5=38$.
Hence, the mean deviation about the median for the given data is
$M.D.\left( M \right)=\dfrac{1}{N}\sum\limits_{i=1}^{8}{{{f}_{i}}\left| {{x}_{i}}-M \right|} \\ $
$=\dfrac{1}{100}\times 735 \\ $
$=7.35 \\ $