While calculating the mean and variance of 10 readings,

Given 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

Now we have to find the correct mean and the variance.

As per given criteria,

Number of reading, $n=10$

Mean of the given readings before correction, $\overline{\mathrm{x}}=45$

But we know,

$\overline{\mathrm{x}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}$

Substituting the corresponding values, we get

$45=\frac{\sum x_{i}}{10}$

$\Rightarrow \sum x_{i}=45 \times 10=450$

It is said one reading 25 was wrongly taken as 52 ,

So $\sum x_{i}=450-52+25=423$

So the correct mean after correction is

$\overline{\mathrm{X}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{423}{10}=42.3$

Also given the variance of the 10 readings is 16 before correction, i.e., $\sigma^{2}=16$

But we know

$\sigma^{2}=\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}$

Substituting the corresponding values, we get

$16=\frac{\sum x_{i}^{2}}{10}-(45)^{2}$

$\Rightarrow 16=\frac{\sum x_{i}^{2}}{10}-2025$

$\Rightarrow 16+2025=\frac{\sum x_{i}^{2}}{10}$

$\Rightarrow \frac{\sum x_{i}^{2}}{10}=2041$

$\Rightarrow \Sigma x_{i}^{2}=20410$

It is said one reading 25 was wrongly taken as 52, so

$\Rightarrow \sum x_{i}^{2}=20410-(52)^{2}+(25)^{2}$

$\Rightarrow \sum x_{i}^{2}=20410-2704+625$

$\Rightarrow \sum x_{i}^{2}=18331$

So the correct variance after correction is

$\sigma^{2}=\frac{18331}{10}-\left(\frac{423}{10}\right)^{2}$

$\sigma^{2}=1833.1-(42.3)^{2}=1833.1-1789.29$

$\sigma^{2}=43.81$

Hence the corrected mean and variance is $42.3$ and $43.81$ respectively.