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 | $22A$ | $92A^2$ |
| $n=7$ | $\Sigma f_in_i=22A$ | $\Sigma f_i n^2_i=92A^2$ |
$$\begin{array}{ll} \therefore & \sigma^2=\frac{\Sigma f_i x_i^2}{n}-\left(\frac{\Sigma f_i x_i}{n}\right)^2 \\ \Rightarrow & 160=\frac{92 A^2}{7}-\left(\frac{22 A}{7}\right)^2 \\ \Rightarrow & 160=\frac{92 A^2}{7}-\frac{484 A^2}{49} \\ \Rightarrow & 160=(644-484) \frac{A^2}{49} \\ \Rightarrow & 160=\frac{160 A^2}{49} \Rightarrow A^2=49 \\ \therefore & A=7 \end{array}$$
For the frequency distribution
| $x$ | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|
| $f$ | 4 | 9 | 16 | 14 | 11 | 6 |
Find the standard distribution.
| $x_i$ | $f_i$ | $d_i=x_i-4$ | $f_i d_i$ | $f_i d^2_i$ |
|---|---|---|---|---|
| 2 | 4 | $-$2 | $-$8 | 16 |
| 3 | 9 | $-$1 | $-$9 | 9 |
| 4 | 16 | 0 | 0 | 0 |
| 5 | 14 | 1 | 14 | 14 |
| 6 | 11 | 2 | 22 | 44 |
| 7 | 6 | 3 | 18 | 54 |
| Total | 60 | $\Sigma f_i d_i=37$ | $\Sigma f_i d^2_i=137$ |
$$\begin{aligned} \therefore \quad \mathrm{SD} & =\sqrt{\frac{\Sigma f_i d_i^2}{N}-\left(\frac{\Sigma f_i d_i}{N}\right)^2} \\ & =\sqrt{\frac{137}{60}-\left(\frac{37}{60}\right)^2} \\ & =\sqrt{2.2833-(0.616)^2} \\ & =\sqrt{2.2833-0.3794} \\ & =\sqrt{1.9037}=1.38 \end{aligned}$$
There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Frequency | $x-2$ | $x$ | $x^2$ | $(x+1)^2$ | $2x$ | $x+1$ |
where, x is positive integer. Determine the mean and standard deviation of the marks.
$\therefore$ Sum of frequencies,
$$\begin{array}{l} & x-2+x+x^2+(x+1)^2+2 x+x+1 & =60 \\ \Rightarrow \quad & 2 x-2+x^2+x^2+1+2 x+2 x+x+1 =60 \\ \Rightarrow \quad & 2 x^2+7 x =60 \\ \Rightarrow & 2 x^2+7 x-60 =0 \\ \Rightarrow & 2 x^2+15 x-8 x-60 =0 \\ \Rightarrow & x(2 x+15)-4(2 x+15) =0 \\ \Rightarrow & (2 x+15)(x-4) =0 \\ \Rightarrow & x =-\frac{15}{2}, 4 \\ \Rightarrow & x =-\frac{15}{2} \quad \text{[inaddmisible]} [\because x \in I^+] \end{array}$$
| $x_i$ | $f_i$ | $d_i=x_i-3$ | $f_i d_i$ | $f_i d^2_i$ |
|---|---|---|---|---|
| 0 | 2 | $-$3 | $-$6 | 18 |
| 2 | 4 | $-$2 | $-$8 | 16 |
| 2 | 16 | $-$1 | $-$16 | 16 |
| $A=3$ | 25 | 0 | 0 | 0 |
| 4 | 8 | 1 | 8 | 8 |
| 5 | 5 | 2 | 10 | 20 |
| Total | $\Sigma f_i=60$ | $\Sigma f_i d_i=-12$ | $Sigma f_i d^2_i=78$ |
$$\begin{aligned} \text { Mean } & =A+\frac{\Sigma f_i d_i}{\Sigma f_i}=3+\left(\frac{-12}{60}\right)=2.8 \\ \sigma & =\sqrt{\frac{\Sigma f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2}=\sqrt{\frac{78}{60}-\left(\frac{-12}{60}\right)^2} \\ & =\sqrt{1.3-0.04}=\sqrt{1.26}=1.12 \end{aligned}$$
The mean life of a sample of 60 bulbs was 650 h and the standard deviation was 8 h . If a second sample of 80 bulbs has a mean life of 660 h and standard deviation 7 h , then find the over all standard deviation.
Here, $n_1=60, \bar{x}_1=650, s_1=8$ and $n_2=80, \bar{x}_2=660, s_2=7$
$$\begin{aligned} \therefore \quad \sigma & =\sqrt{\frac{n_1 s_1^2+n_2 s_2^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2}} \\ & =\sqrt{\frac{60 \times(8)^2+80 \times(7)^2}{60+80}+\frac{60 \times 80(650-660)^2}{(60+80)^2}} \\ & =\sqrt{\frac{6 \times 64+8 \times 49}{14}+\frac{60 \times 80 \times 100}{140 \times 140}} \\ & =\sqrt{\frac{192+196}{7}+\frac{1200}{49}}=\sqrt{\frac{388}{7}+\frac{1200}{49}} \\ & =\sqrt{\frac{2716+1200}{49}}=\sqrt{\frac{3916}{49}}=\frac{62.58}{7}=8.9 \end{aligned}$$
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.
$$\begin{aligned} &\text { Here, } \bar{x}=50, n=100 \text { and } \sigma=4\\ &\begin{array}{ll} \therefore & \frac{\Sigma x_i}{100}=50 \\ \Rightarrow & \Sigma x_i=5000 \\ \text { and } & \sigma^2=\frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\ \Rightarrow & (4)^2=\frac{\Sigma f_i x_i^2}{100}-(50)^2 \\ \Rightarrow & 16=\frac{\Sigma f_i x_i^2}{100}-2500 \\ \Rightarrow & \frac{\Sigma f_i x_i^2}{100}=16+2500=2516 \\ \therefore & \Sigma f_i x_i^2=251600 \end{array} \end{aligned}$$