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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, find the correct standard deviation.
Solution
Given, $n=100, \bar{x}=40$ and $\sigma=10$
$\therefore \quad \frac{\Sigma x_{i}}{n}=40$
$\Rightarrow \quad \frac{\Sigma x_{i}}{100}=40$
$\Rightarrow \quad \Sigma x_{i}=4000$
Now, Corrected $\Sigma x_{i}=4000-30-70+3+27=3930$
Corrected mean $=\frac{2930}{100}=39.3$
Now, $\sigma^{2}=\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}=\frac{\Sigma x_{i}^{2}}{n}-(40)^{2}$
$\Rightarrow \quad 100=\frac{\Sigma x_{i}^{2}}{100}-1600$
$\Rightarrow \quad \Sigma x_{i}^{2}=170000$
Now, $\quad$ Corrected $\Sigma x_{i}^{2}=170000-(30)^{2}-(70)^{2}+3^{2}+(27)^{2}=164938$
Corrected $\sigma=\sqrt{\frac{164938}{100}-(39.3)^{2}}=\sqrt{1649.38-1544.49}=\sqrt{104.9}$
$=10.24$
Similar Questions
Let the mean of the data
| $X$ | $1$ | $3$ | $5$ | $7$ | $9$ |
| $(f)$ | $4$ | $24$ | $28$ | $\alpha$ | $8$ |
be $5.$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $……….$.
