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.
$$\begin{aligned} & \text { Given, } \quad n=10, \bar{x}=45 \text { and } \sigma^2=16 \\ & \therefore \quad \bar{x}=45 \Rightarrow \frac{\Sigma x_i}{n}=45 \\ & \Rightarrow \quad \frac{\Sigma x_i}{10}=45 \Rightarrow \Sigma x_i=450 \\ & \text { Corrected } \Sigma x_i=450-52+25=423 \\ \therefore\quad & \bar{x}=\frac{423}{10}=42.3 \\ & \sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2 \\ & 16=\frac{\Sigma x_i^2}{10}-(45)^2 \\ & \Sigma x_i^2=10(2025+16) \\ & \Sigma x_i^2=20410 \\ & \therefore \quad \text { Corrected } \Sigma x_i^2=20410-(52)^2+(25)^2=18331 \\ & \text { and } \quad \text { corrected } \sigma^2=\frac{18331}{10}-(42.3)^2=43.81 \end{aligned} $$
The mean deviation of the data $3,10,10,4,7,10,5$ from the mean is
Mean deviation for $n$ observations $x_1, x_2, \ldots, x_n$ from their mean $\bar{x}$ is given by