If mean and standard deviation of 100 items are 50 and 4 respectively, then find the sum of all the item and the sum of the squares of item.
$$\begin{aligned} &\text { Here, } \bar{x}=50, n=100 \text { and } \sigma=4\\ &\begin{array}{ll} \therefore & \frac{\Sigma x_i}{100}=50 \\ \Rightarrow & \Sigma x_i=5000 \\ \text { and } & \sigma^2=\frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\ \Rightarrow & (4)^2=\frac{\Sigma f_i x_i^2}{100}-(50)^2 \\ \Rightarrow & 16=\frac{\Sigma f_i x_i^2}{100}-2500 \\ \Rightarrow & \frac{\Sigma f_i x_i^2}{100}=16+2500=2516 \\ \therefore & \Sigma f_i x_i^2=251600 \end{array} \end{aligned}$$
If for distribution $\Sigma(x-5)=3, \Sigma(x-5)^2=43$ and total number of item is 18 . Find the mean and standard deviation.
$$\begin{aligned} & \text { Given, } \quad n=18, \Sigma(x-5)=3 \text { and } \Sigma(x-5)^2=43 \\ & \therefore \quad \text { Mean }=A+\frac{\Sigma(x-5)}{18} \\ & =5+\frac{3}{18}=5+0.1666=5.1666=5.17 \\ & \text { and } \quad \mathrm{SD}=\sqrt{\frac{\sum(x-5)^2}{n}-\left(\frac{\Sigma(x-5)}{n}\right)^2} \\ & =\sqrt{\frac{43}{18}-\left(\frac{3}{18}\right)^2} \\ & =\sqrt{2.3944-(0.166)^2}=\sqrt{2.3944-0.2755}=1.59 \end{aligned}$$
Find the mean and variance of the frequency distribution given below.
| $x$ | $1\le x \le 3$ | $3\le x \le 5$ | $5\le x \le 7$ | $7\le x \le 10$ |
|---|---|---|---|---|
| $f$ | 6 | 4 | 5 | 1 |
| $x$ | $f_i$ | $x_i$ | $f_i x_i$ | $f_i x^2_i$ |
|---|---|---|---|---|
| 1-3 | 6 | 2 | 12 | 24 |
| 3-5 | 4 | 4 | 16 | 64 |
| 5-7 | 5 | 6 | 30 | 180 |
| 7-10 | 1 | 8.5 | 8.5 | 72.25 |
| Total | $n=16$ | $\Sigma f_i x_i=66.5$ | $Sigma f_i x^2_i=340.25$ |
$\therefore \quad$ Mean $=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{66.5}{16}=4.15$
$$\begin{aligned} \text{and}\quad\text { variance } & =\sigma^2=\frac{\Sigma f_i x_i{ }^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\ & =\frac{340.25}{16}-(4.15)^2 \\ & =21.2656-17.2225=4.043 \end{aligned}$$
Calculate the mean deviation about the mean for the following frequency distribution.
| Class interval | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 |
|---|---|---|---|---|---|
| Frequency | 4 | 6 | 8 | 5 | 2 |
| Class interval | $f_i$ | $x_i$ | $f_i x_i$ | $d_i=\left|x_i-\bar{x}\right|$ | $f_i d_i$ |
|---|---|---|---|---|---|
| 0-4 | 4 | 2 | 8 | 7.2 | 28.8 |
| 4-8 | 6 | 6 | 36 | 3.2 | 19.2 |
| 8-12 | 8 | 10 | 80 | 0.8 | 6.4 |
| 12-16 | 5 | 14 | 70 | 4.8 | 24.0 |
| 16-20 | 2 | 18 | 36 | 8.8 | 17.6 |
| Total | $\Sigma f_i=25$ | $\Sigma f_i x_i=230$ | $\Sigma f_i d_i=96$ |
$$\begin{array}{ll} \therefore & \text { Mean }=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{230}{25}=9.2 \\ \text { and } & \text { mean deviation }=\frac{\Sigma f d_i}{\Sigma f_i}=\frac{96}{25}=3.84 \end{array}$$
Calculate the mean deviation from the median of the following data.
| Class interval | 0-6 | 6-12 | 12-18 | 18-24 | 24-30 |
|---|---|---|---|---|---|
| Frequency | 4 | 5 | 3 | 6 | 2 |
| Class interval | $f_i$ | $x_i$ | $cf$ | $d_i=\left|x_i-\bar{m_d}\right|$ | $f_i d_i$ |
|---|---|---|---|---|---|
| 0-6 | 4 | 3 | 4 | 11 | 44 |
| 6-12 | 5 | 9 | 9 | 5 | 25 |
| 12-18 | 3 | 15 | 12 | 1 | 3 |
| 18-24 | 6 | 21 | 18 | 7 | 42 |
| 24-30 | 2 | 27 | 20 | 13 | 26 |
| Total | $N=20$ | $\Sigma f_i d_i=140$ |
$$\begin{aligned} &\because \quad \frac{N}{2}=\frac{20}{2}=10\\ &\text { So, the median class is 12-18. } \end{aligned}$$
$$\begin{aligned} \therefore \quad \text { Median } & =l+\frac{\frac{N}{2}-c f}{f} \times i \\ & =12+\frac{6}{3}(10-9) \\ & =12+2=14 \\ \mathrm{MD} & =\frac{\Sigma f_i d_i}{\Sigma f_i}=\frac{140}{20}=7 \end{aligned}$$