Find the mean deviation about the median of the following distribution.
| Marks obtained | 10 | 11 | 12 | 14 | 15 |
|---|---|---|---|---|---|
| Number of students | 2 | 3 | 8 | 3 | 4 |
| Marks obtained | $f_i$ | $cf$ | $d_i=|x_i-M_e|$ | $f_id_i$ |
|---|---|---|---|---|
| 10 | 2 | 2 | 2 | 4 |
| 11 | 3 | 5 | 1 | 3 |
| 12 | 8 | 13 | 0 | 0 |
| 14 | 3 | 16 | 2 | 6 |
| 15 | 4 | 20 | 3 | 12 |
| Total | $\Sigma f_i=20$ | $\Sigma f_i d_i=25$ |
$$\begin{array}{ll} \text { Now, } & M_e=\left(\frac{20+1}{2}\right) \text { th item }=\left(\frac{21}{2}\right)=10.5 \text { th item } \\ \therefore & M_e=12 \\ \therefore & M D=\frac{\sum f_i d_i}{\sum f_i}=\frac{25}{20}=1.25 \end{array}$$
3 Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an odd number.
Consider first natural number when $n$ is an odd i.e., 1, 2, 3, 4, ...n, [odd].
$$\begin{aligned} & \text { Mean } \bar{x}= \frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ \therefore \quad & \mathrm{MD}= \left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\left|3-\frac{n+1}{2}\right|+\ldots+\left|n-\frac{n+1}{2}\right| \end{aligned}$$
$$=\frac{\left|-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\ldots+\left|\frac{n-1}{2}-\frac{n+1}{2}\right|+\left|\frac{n+1}{2}-\frac{n+1}{2}\right|+\left|\frac{n+3}{2}-\frac{n+1}{2}\right|+\ldots+\left|\frac{n-2}{2}-\frac{n+1}{2}\right|+\left|n-\frac{n+1}{2}\right|}{n}$$
$$\begin{aligned} & =\frac{2}{n}\left[1+2+\ldots+\frac{n-3}{2}+\frac{n-1}{2}\right] \quad\left(\frac{n-1}{2}\right) \text { terms } \\ & =\frac{2}{n}\left[\frac{\left(\frac{n-1}{2}\right)\left(\frac{n-1}{2}+1\right)}{2}\right] \quad\left[\because \text { sum of first } n \text { natural numbers }=\frac{n(n+1)}{2}\right] \\ & =\frac{2}{n} \cdot \frac{1}{2}\left[\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)\right]=\frac{1}{n}\left(\frac{n^2-1}{4}\right)=\frac{n^2-1}{4 n} \end{aligned}$$
Calculate the mean deviation about the mean of the set of first $n$ natural numbers when $n$ is an even number.
Consider first $n$ natural number, when $n$ is even i.e., $1,2,3,4,\quad$ [even]
$$\begin{aligned} & \therefore \quad \text { Mean } \bar{x}=\frac{1+2+3+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n+1}{2} \\ & \mathrm{MD}=\frac{1}{n}\left[\left|1-\frac{n+1}{2}\right|+\left|2-\frac{n+1}{2}\right|+\left|3-\frac{n+1}{2}\right|\right]+\left|\frac{n-2}{2}-\frac{n+1}{2}\right|+\left|\frac{n}{2}-\frac{n+1}{2}\right| \\ &+\left|\frac{n+2}{2}-\frac{n+1}{2}\right|+\ldots+\left|n-\frac{n+1}{2}\right| \\ &=\frac{1}{n}\left[\left|\frac{1-n}{2}\right|+\left|\frac{3-n}{2}\right|+\left|\frac{5-n}{2}\right|+\ldots .+\left|\frac{-3}{2}\right|+\left|\frac{1}{2}\right|+\ldots+\left|\frac{n-1}{2}\right|\right] \\ &=\frac{2}{n}\left[\frac{1}{2}+\frac{3}{2}+\ldots+\frac{n-1}{2}\right]\left(\frac{n}{2}\right) \text { terms } \\ &=\frac{1}{n} \cdot\left(\frac{n}{2}\right)^2 \quad\left[\because \text { sum of first } n \text { natural numbers }=n^2\right]\\ &=\frac{1}{n} \cdot \frac{n^2}{4}=\frac{n}{4} \\ \end{aligned}$$
Find the standard deviation of first n natural numbers.
| $x_i$ | 1 | 2 | 3 | 4 | 5 | ... | ... | $n$ |
|---|---|---|---|---|---|---|---|---|
| $x^2_i$ | 1 | 4 | 9 | 16 | 25 | ... | ... | $n^2$ |
Now, $\quad \sum x_i=1+2+3+4+\ldots+n=\frac{n(n+1)}{2}$
and $\quad \sum x_i^2=1^2+2^2+3^2+\ldots+n^2=\frac{n(n+1)(2 n+1)}{6}$
$$\begin{aligned} \therefore & =\sqrt{\frac{\sum x_i^2}{N}-\left(\frac{\sum x_i}{N}\right)^2} \\ & =\sqrt{\frac{n(n+1)(2 n+1)}{6 n}-\frac{n^2(n+1)^2}{4 n^2}} \\ & =\sqrt{\frac{(n+1)(2 n+1)}{6}-\frac{(n+1)^2}{4}} \\ & =\sqrt{\frac{2\left(2 n^2+3 n+1\right)-3\left(n^2+2 n+1\right)}{12}} \\ & =\sqrt{\frac{4 n^2+6 n+2-3 n^2-6 n-3}{12}} \\ & =\sqrt{\frac{n^2-1}{12}} \end{aligned}$$
The mean and standard deviation of some data for the time taken to complete a test are calculated with the following results Number of observation $=25$, mean $=18.2 \mathrm{~s}$, standard, deviation $=3.25 \mathrm{~s}$ Further, another set of 15 observations $x_1 x_2 \ldots x_{15}$, also in seconds, is now available and we have $\sum_\limits{i=1}^{15} x_i=279$ and $\sum_\limits{i=1}^{15} x_i^2=5524$. Calculate the standard derivation based on all 40 observations.
Given,
$$\begin{aligned} & n_i=25, \bar{x}_i=18.2, \sigma_1=3.25 \\ & n_2=15, \sum_{i=1}^{15} x_i=279 \text { and } \sum_{i=1}^{15} x_i^2=5524 \end{aligned}$$
For first set, $$\sum x_i=25 \times 18.2=455$$
$\therefore \sigma_1^2=\frac{\Sigma x_i^2}{25}-(18.2)^2$
$$\begin{aligned} \Rightarrow \quad(3.25)^2 =\frac{\Sigma x_i^2}{25}-331.24 \\ \Rightarrow \quad 10.5625+331.24 =\frac{\Sigma x_i^2}{25} \\ \Rightarrow \quad \Sigma x_i^2 =25 \times(10.5625+331.24) \\ =25 \times 341.8025 \\ =8545.0625 \end{aligned}$$
For combined SD of the 40 observations $n=40$,
$$\begin{aligned} & \begin{aligned} \text { Now }\quad\Sigma x_i^2 & =5524+8545.0625=14069.0625 \\ \text { and } \quad\Sigma x_i & =455+279=734 \end{aligned} \\ & \qquad \begin{aligned} \therefore \quad \mathrm{SD} & =\sqrt{\frac{14069.0625}{40}-\left(\frac{734}{40}\right)^2} \\ & =\sqrt{351.726-(18.35)^2} \\ & =\sqrt{351.726-336.7225} \\ & =\sqrt{15.0035}=3.87 \end{aligned} \end{aligned}$$