The mean and standard deviation of 15 observations were found to be 12 and 3 respectively.

Question

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If $$\mu$$ and $$\sigma^2$$ denote the mean and variance of the correc...

Answer: 2521

Accepted Answer

Let the incorrect mean be ${\mu }^{'}$ and standard deviation be ${\sigma }^{'}$

We have

${\mu }^{'}=\frac{\Sigma {x}_{i}}{15}=12\Rightarrow \Sigma {x}_{i}=180$

As per given information correct $\Sigma {x}_{i}=180-10+12$

$\Rightarrow \mu ()=\frac{182}{15}$

Also

${\sigma }^{'}=\sqrt{\frac{\Sigma {x}_{2}^{i}}{15}-144}=3\Rightarrow \Sigma {x}_{2}^{i}=2295$

Correct $\Sigma {x}_{i}{}^{2}=2295-100+144=2339$

${\sigma }^{2}()=\frac{2339}{15}-\frac{182	imes 182}{15	imes 15}$

Required value

$\begin{gathered} =15(\mu +{\mu }^{2}+{\sigma }^{2}) \ =15(\frac{182}{15}+\frac{182	imes 182}{15	imes 15}+\frac{2339}{15}-\frac{182	imes 182}{15	imes 15}) \ =15(\frac{182}{15}+\frac{2339}{15}) \ =2521 \end{gathered}$

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