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
When tested, the lives (in hours) of 5 bulbs were noted as follows
$$\text { 1357, 1090, 1666, 1494, } 1623$$
The mean deviations (in hours) from their mean is