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 $n=40$,
$$\begin{aligned} & \begin{aligned} \text { Now }\quad\Sigma x_i^2 & =5524+8545.0625=14069.0625 \\ \text { and } \quad\Sigma x_i & =455+279=734 \end{aligned} \\ & \qquad \begin{aligned} \therefore \quad \mathrm{SD} & =\sqrt{\frac{14069.0625}{40}-\left(\frac{734}{40}\right)^2} \\ & =\sqrt{351.726-(18.35)^2} \\ & =\sqrt{351.726-336.7225} \\ & =\sqrt{15.0035}=3.87 \end{aligned} \end{aligned}$$
The mean and standard deviation of a set of $n_1$ observations are $\bar{x}_1$ and $s_1$, respectively while the mean and standard deviation of another set of $n_2$ observations are $\bar{x}_2$ and $s_2$, respectively. Show that the standard deviation of the combined set of $\left(n_1+n_2\right)$ observations is given by
$$S D=\sqrt{\frac{n_1\left(s_1\right)^2+n_2\left(s_2\right)^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1-n_2\right)^2}}$$
$$\begin{array}{ll} \text { Let } & x_i, i=1,2,3 \ldots, n_1 \text { and } y_j, j=1,2,3, \ldots, n_2 \\ \therefore & \bar{x}_1=\frac{1}{n_1} \sum_,\limits{i=1}^{n_1} x_i \text { and } \bar{x}_2=\frac{1}{n_2} \sum_\limits{i=1}^{n_2} y_j \\ \Rightarrow & \sigma_1^2=\frac{1}{n_1} \sum_\limits{i=1}^{n_1}\left(x_i-\bar{x}_1\right)^2 \\ \text { and } & \sigma_2^2=\frac{1}{n_2} \sum_\limits{j=1}^n\left(y_j-\bar{x}_2\right)^2 \end{array}$$
Now, mean $\bar{x}$ of the given series is given by
$$\bar{x}=\frac{1}{n_1+n_2}\left[\sum_\limits{i=1}^{n_1} x_i+\sum_\limits{j=1}^{n_2} y_j\right]=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}$$
The variance $\sigma^2$ of the combined series is given by
$$\sigma^2=\frac{1}{n_1+n_2}\left[\sum_\limits{i=1}^{n_1}\left(x_i-\bar{x}\right)^2+\sum_\limits{i=1}^{n_2}\left(y_j-\bar{x}\right)^2\right]$$
Now,
$$\begin{aligned} \sum_{i=1}^{n_1}\left(x_i-\bar{x}\right)^2 & =\sum_{i=1}^{n_1}\left(x_i-\bar{x}_j+\bar{x}_j-\bar{x}\right)^2 \\ & =\sum_{i=1}^{n_1}\left(x_i-\bar{x}_j\right)^2+n_1\left(\bar{x}_j-\bar{x}\right)^2+2\left(\bar{x}_j-\bar{x}\right) \sum_{i=1}^{n_1}\left(x_i-\bar{x}_j\right)^2 \end{aligned}$$
But $$\sum_\limits{i=1}^{n_1}\left(x_i-\bar{x}_i\right)=0$$ [algebraic sum of the deviation of values of first series from their mean is zero]
Also, $$\sum_\limits{i=1}^{n_1}\left(x_i-\bar{x}\right)^2=n_1 s_1^2+n_1\left(\bar{x}_1-\bar{x}\right)^2=n_1 s_1^2+n_1 d_1^2$$
Where, $$d_1=\left(\bar{x}_1-\bar{x}\right)$$
Similarly, $\quad \sum_{j=1}^{n_2}\left(y_j-\bar{x}\right)^2=\sum_{j=1}^{n_2}\left(y_j-\bar{x}_i+\bar{x}_i-\bar{x}\right)^2=n_2 s_2^2+n_2 d_2^2$
where, $d_2=\bar{x}_2-\bar{x}$
Combined SD, $\sigma=\sqrt{\frac{\left[n_1\left(s_1^2+d_1^2\right)+n_2\left(s_2^2+d_2^2\right)\right]}{n_1+n_2}}$
where, $d_1=\bar{x}_1-\bar{x}=\bar{x}_1-\left(\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\right)=\frac{n_2\left(\bar{x}_1-\bar{x}_2\right)}{n_1+n_2}$and $d_2=\bar{x}_2-\bar{x}=\bar{x}_2-\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}=\frac{n_1\left(\bar{x}_2-\bar{x}_1\right)}{n_1+n_2}$
$\therefore$ $\sigma^2=\frac{1}{n_1+n_2}\left[n_1 s_1^2+n_2 s_2^2+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}+\frac{n_2 n_1\left(\bar{x}_2-\bar{x}_1\right)^2}{\left(n_1+n_2\right)^2}\right]$
Also, $\sigma=\sqrt{\frac{n_1 s_1^2+n_2 s_2^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}}$
Two sets each of 20 observations, have the same standard deviation 5 . The first set has a mean 17 and the second mean 22. Determine the standard deviation of the $x$ sets obtained by combining the given two sets.
$$\begin{aligned} & \text { Given, } n_1=20, \sigma_1=5, \bar{x}_1=17 \text { and } n_2=20, \sigma_2=5, \bar{x}_2=22 \\ & \text { We know that, } \sigma=\sqrt{\frac{n_1 s_1^2+n_2 s_2^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}} \\ & =\sqrt{\frac{20 \times(5)^2+20 \times(5)^2}{20+20}+\frac{20 \times 20(17-22)^2}{(20+20)^2}} \\ & =\sqrt{\frac{1000}{40}+\frac{400 \times 25}{1600}}=\sqrt{25+\frac{25}{4}}=\sqrt{\frac{125}{4}}=\sqrt{31.25}=5.59 \end{aligned}$$
The frequency distribution
| $x$ | A | 2A | 3A | 4A | 5A | 6A |
|---|---|---|---|---|---|---|
| $f$ | 2 | 1 | 1 | 1 | 1 | 1 |
where, A is a positive integer, has a variance of 160. Determine the value of A.
| $x$ | $f_i$ | $f_i x_i$ | $f_i x^2_i$ |
|---|---|---|---|
| $A$ | 2 | $2A$ | $2A^2$ |
| $2A$ | 1 | $2A$ | $4A^2$ |
| $3A$ | 1 | $3A$ | $9A^2$ |
| $4A$ | 1 | $4A$ | $16A^2$ |
| $5A$ | 1 | $5A$ | $25A^2$ |
| $6A$ | 1 | $6A$ | $36A^2$ |
| Total | 7 | $22A$ | $92A^2$ |
| $n=7$ | $\Sigma f_in_i=22A$ | $\Sigma f_i n^2_i=92A^2$ |
$$\begin{array}{ll} \therefore & \sigma^2=\frac{\Sigma f_i x_i^2}{n}-\left(\frac{\Sigma f_i x_i}{n}\right)^2 \\ \Rightarrow & 160=\frac{92 A^2}{7}-\left(\frac{22 A}{7}\right)^2 \\ \Rightarrow & 160=\frac{92 A^2}{7}-\frac{484 A^2}{49} \\ \Rightarrow & 160=(644-484) \frac{A^2}{49} \\ \Rightarrow & 160=\frac{160 A^2}{49} \Rightarrow A^2=49 \\ \therefore & A=7 \end{array}$$