Statistics Question 4
Question 4 - 2024 (27 Jan Shift 2)
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 correct observations respectively, then $15\left(\mu+\mu^{2}+\sigma^{2}\right)$ is equal to
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Answer (2521)
Solution
Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$
We have
$\mu^{\prime}=\frac{\Sigma \mathrm{x}{\mathrm{i}}}{15}=12 \Rightarrow \Sigma \mathrm{x}{\mathrm{i}}=180$
As per given information correct $\Sigma \mathrm{x}_{\mathrm{i}}=180-10+12$
$\Rightarrow \mu($ correct mean $)=\frac{182}{15}$
Also
$\sigma^{\prime}=\sqrt{\frac{\Sigma \mathrm{x}{\mathrm{i}}{ }^{2}}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}{\mathrm{i}}{ }^{2}=2295$
Correct $\Sigma \mathrm{x}_{\mathrm{i}}{ }^{2}=2295-100+144=2339$
$\sigma^{2}($ correct variance $)=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}$
Required value
$$ \begin{aligned} & =15\left(\mu+\mu^{2}+\sigma^{2}\right) \ & =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) \ & =15\left(\frac{182}{15}+\frac{2339}{15}\right) \ & =2521 \end{aligned} $$
