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 | $N=20$ | $\Sigma f_i d_i=140$ |
$$\begin{aligned} &\because \quad \frac{N}{2}=\frac{20}{2}=10\\ &\text { So, the median class is 12-18. } \end{aligned}$$
$$\begin{aligned} \therefore \quad \text { Median } & =l+\frac{\frac{N}{2}-c f}{f} \times i \\ & =12+\frac{6}{3}(10-9) \\ & =12+2=14 \\ \mathrm{MD} & =\frac{\Sigma f_i d_i}{\Sigma f_i}=\frac{140}{20}=7 \end{aligned}$$
Determine the mean and standard deviation for the following distribution.
| Marks | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Frequency | 1 | 6 | 6 | 8 | 8 | 2 | 2 | 3 | 0 | 2 | 1 | 0 | 0 | 0 | 1 |
| Marks | $f_i$ | $f_i x_i$ | $d_i=x_i=\bar{x}$ | $f_i d_i$ | $f_i d^2_i$ |
|---|---|---|---|---|---|
| 2 | 1 | 2 | $2-6=-4$ | $-4$ | 16 |
| 3 | 6 | 18 | $3-6=-3$ | $18$ | 54 |
| 4 | 6 | 24 | $4-6=-2$ | $-12$ | 24 |
| 5 | 8 | 40 | $5-6=-1$ | $-8$ | 8 |
| 6 | 8 | 48 | $6-6=0$ | 0 | 0 |
| 7 | 2 | 14 | $7-6=1$ | 2 | 2 |
| 8 | 2 | 16 | $8-6=2$ | 4 | 8 |
| 9 | 3 | 27 | $9-6=3$ | 9 | 27 |
| 10 | 0 | 0 | $10-6=4$ | 0 | 0 |
| 11 | 2 | 22 | $11-6=5$ | 10 | 50 |
| 12 | 1 | 12 | $12-6=6$ | 6 | 36 |
| 13 | 0 | 0 | $13-6=7$ | 0 | 0 |
| 14 | 0 | 0 | $14-6=8$ | 0 | 0 |
| 15 | 0 | 0 | $15-6=9$ | 0 | 0 |
| 16 | 1 | 16 | $16-6=0$ | 10 | 100 |
| Total | $\Sigma f_i=40$ | $\Sigma f_i x_i=239$ | $\Sigma f_i d_i=-1$ | $\Sigma f_i x_i^2=325$ |
$$\begin{aligned} & \therefore \quad \text { Mean } \bar{x}=\frac{\Sigma f_i x_i}{\Sigma t_i}=\frac{239}{40}=5.975 \approx 6 \\ & \text { and } \\ & \sigma=\sqrt{\frac{\sum f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2}=\sqrt{\frac{325}{40}-\left(\frac{-1}{40}\right)^2} \\ & =\sqrt{8.125-0.000625}=\sqrt{8.124375}=2.85 \end{aligned}$$
The weights of coffee in 70 jars is shown in the following table
| Weight (in g) | Frequency |
|---|---|
| 200-201 | 13 |
| 201-202 | 27 |
| 202-203 | 18 |
| 203-204 | 10 |
| 204-205 | 1 |
| 205-206 | 1 |
Determine variance and standard deviation of the above distribution.
| $Cl$ | $f_i$ | $x_i$ | $d_i=x_i-\bar{x}$ | $f_i d_i$ | $f_i d_i^2$ |
|---|---|---|---|---|---|
| 200-201 | 13 | 200.5 | $-2$ | $-26$ | 52 |
| 201-202 | 27 | 201.5 | $-1$ | $-27$ | 27 |
| 202-203 | 18 | 202.5 | 0 | 0 | 0 |
| 203-204 | 10 | 203.5 | 1 | 10 | 10 |
| 204-205 | 1 | 204.5 | 2 | 2 | 4 |
| 205-206 | 1 | 205.5 | 3 | 3 | 9 |
| $\Sigma f_i=70$ | $\Sigma f_i d_i=-38$ | $\Sigma f_i d^2_i=102$ |
$$\begin{aligned} & \therefore \quad \quad \sigma^2=\frac{\Sigma f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2=\frac{102}{70}-\left(\frac{-38}{70}\right)^2 \\ & \text { Now, } \\ & =1.4571-0.2916=1.1655 \\ & \sigma=\sqrt{1.1655}=1.08 \mathrm{~g} \end{aligned}$$