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 |
| $\Sigma f_i=70$ | $\Sigma f_i d_i=-38$ | $\Sigma f_i d^2_i=102$ |
$$\begin{aligned} & \therefore \quad \quad \sigma^2=\frac{\Sigma f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2=\frac{102}{70}-\left(\frac{-38}{70}\right)^2 \\ & \text { Now, } \\ & =1.4571-0.2916=1.1655 \\ & \sigma=\sqrt{1.1655}=1.08 \mathrm{~g} \end{aligned}$$
Determine mean and standard deviation of first n terms of an AP whose first term is a and common difference is d.
| $x_i$ | $x_i-a$ | $(x_i-a)^2$ |
|---|---|---|
| $a$ | 0 | 0 |
| $a+d$ | $d$ | $d^2$ |
| $a+2d$ | $2d$ | $4d^2$ |
| ... | ... | $9d^2$ |
| ... | ... | ... |
| ... | ... | ... |
| $a+(n-1)d$ $\Sigma x_i=\frac{n}{2}[2a+(n-1)]$ |
$(n-1)d$ | $(n-1)^2 d^2$ |
$$\begin{aligned} \because \quad \text { Mean } & =\frac{\Sigma x_i}{n}=\frac{1}{n}\left[\frac{n}{2}(2 a+(n-1) d]\right. \\ & =a+\frac{(n-1)}{2} d \end{aligned}$$
$$\begin{aligned} & \therefore \quad \Sigma\left(x_i-a\right)=d[1+2+3+\ldots+(n-1) d] \\ & =d \frac{(n-1) n}{2} \\ & \text { and } \\ & \Sigma\left(x_i-a\right)^2=d^2\left[1^2+2^2+3^2+\ldots+(n-1)^2\right] \\ & =\frac{d^2(n-1) n(2 n-1)}{6} \\ & \sigma=\sqrt{\frac{\left(x_i-\mathrm{a}\right)^2}{n}-\left(\frac{x_i-\mathrm{a}}{n}\right)^2} \\ & =\sqrt{\frac{d^2(n-1)(n)(2 n-1)}{6 n}-\left[\frac{d(n-1) n}{2 n}\right]^2} \\ & =\sqrt{\frac{d^2(n-1)(2 n-1)}{6}-\frac{d^2(n-1)^2}{4}} \\ & =d \sqrt{\frac{(n-1)(2 n-1)}{6}-\frac{(n-1)^2}{4}} \\ & =d \sqrt{\frac{(n-1)}{2}\left(\frac{2 n-1}{3}-\frac{n-1}{2}\right)} \\ & =d \sqrt{\frac{(n-1)}{2}\left[\frac{4 n-2-3 n+3}{6}\right]} \\ & =d \sqrt{\frac{(n-1)(n+1)}{12}}=d \sqrt{\frac{\left(n^2-1\right)}{12}} \end{aligned}$$
Following are the marks obtained, out of 100, by two students Ravi and Hashina in 10 tests.
| Ravi | 25 | 50 | 45 | 30 | 70 | 42 | 36 | 48 | 35 | 60 |
|---|---|---|---|---|---|---|---|---|---|---|
| Hashina | 10 | 70 | 50 | 20 | 95 | 55 | 42 | 60 | 48 | 80 |
Who is more intelligent and who is more consistent?
For Ravi,
| $x_i$ | $d_i=x_i-45$ | $d_i^2$ |
|---|---|---|
| 25 | $-20$ | 400 |
| 50 | 5 | 25 |
| 45 | 0 | 0 |
| 30 | $-15$ | 225 |
| 70 | 25 | 625 |
| 42 | $-3$ | 9 |
| 36 | $-9$ | 81 |
| 48 | 3 | 9 |
| 35 | $-10$ | 100 |
| 60 | 15 | 225 |
| Total | $\Sigma d_i=-14$ | $\Sigma d^2_i=1699$ |
$$\begin{aligned} \sigma & =\sqrt{\frac{\Sigma d^2 i}{n}-\left(\frac{\Sigma d_i}{n}\right)^2} \\ & =\sqrt{\frac{1699}{10}-\left(\frac{-14}{10}\right)^2}=\sqrt{169.9-0.0196} \\ & =\sqrt{169.88}=13.03 \\ \text{Now,}\quad \bar{x} & =A+\frac{\Sigma d_i}{\Sigma f_i}=45-\frac{14}{10}=43.6 \end{aligned}$$
For Hashina,
| $x_i$ | $d_i=x_i-55$ | $d_i^2$ |
|---|---|---|
| 10 | $-45$ | 2025 |
| 70 | 25 | 625 |
| 50 | $-5$ | 25 |
| 20 | $-35$ | 1225 |
| 95 | 40 | 1600 |
| 55 | 0 | 0 |
| 42 | $-13$ | 169 |
| 60 | 5 | 25 |
| 48 | $-7$ | 49 |
| 80 | 25 | 625 |
| Total | $\Sigma d_i=0$ | $\Sigma d^2_i=6368$ |
$$\begin{aligned} &\begin{array}{ll} \because & \text { Mean }=55 \\ \therefore & \sigma=\sqrt{\frac{6368}{10}}=\sqrt{636.8}=25.2 \\ \text { For Ravi, } & C V=\frac{\sigma}{\bar{x}} \times 100=\frac{13.03}{43.6} \times 100=29.88 \\ \text { For Hashina, } & C V=\frac{\sigma}{\bar{x}} \times 100=\frac{25.2}{55} \times 100=45.89 \end{array}\\ &\text { Hence, Hashina is more consistent and intelligent. } \end{aligned}$$
Mean and standard deviation of 100 observations were found to be 40 and 10 , respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, then find the correct standard deviation.
$$\begin{aligned} &\begin{aligned} \text { Given, }\quad n=100, \bar{x}=40, \sigma & =10 \text { and } \bar{x}=40 \\ \therefore\quad \frac{\Sigma x_i}{n} & =40 \\ \Rightarrow\quad \frac{\Sigma x_i}{100} & =40 \\ \Rightarrow\quad\Sigma x_i & =4000 \\ \text { Corrected } \Sigma x_i & =4000-30-70+3+27 \\ \therefore\quad & =4030-100=3930 \\ \text { Corrected mean } & =\frac{2930}{100}=39.3 \end{aligned} \end{aligned}$$
$$\begin{aligned} \text{Now,}\quad & \sigma^2=\frac{\Sigma x_i^2}{n}-(40)^2 \\ \Rightarrow\quad & 100=\frac{\Sigma x_i^2}{100}-1600 \\ \Rightarrow\quad & \Sigma x_i^2=170000 \\ & \text { Now, } \quad \text { Corrected } \Sigma x_i^2=170000-(30)^2-(70)^2+3^2+(27)^2 \\ & =164939 \\ & \text { Corrected } \sigma=\sqrt{\frac{164939}{100}-(39.3)^2} \\ & =\sqrt{1649.39-39.3 \times 39.3} \\ & =\sqrt{1649.39-1544.49} \\ & =\sqrt{104.9}=10.24 \end{aligned}$$
3 While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25 . He obtained the mean and variance as 45 and 16, respectively. Find the correct mean and the variance.
$$\begin{aligned} & \text { Given, } \quad n=10, \bar{x}=45 \text { and } \sigma^2=16 \\ & \therefore \quad \bar{x}=45 \Rightarrow \frac{\Sigma x_i}{n}=45 \\ & \Rightarrow \quad \frac{\Sigma x_i}{10}=45 \Rightarrow \Sigma x_i=450 \\ & \text { Corrected } \Sigma x_i=450-52+25=423 \\ \therefore\quad & \bar{x}=\frac{423}{10}=42.3 \\ & \sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2 \\ & 16=\frac{\Sigma x_i^2}{10}-(45)^2 \\ & \Sigma x_i^2=10(2025+16) \\ & \Sigma x_i^2=20410 \\ & \therefore \quad \text { Corrected } \Sigma x_i^2=20410-(52)^2+(25)^2=18331 \\ & \text { and } \quad \text { corrected } \sigma^2=\frac{18331}{10}-(42.3)^2=43.81 \end{aligned} $$