Let the mean and variance of 12 observations be 92 and 4 respectively.

Question

Let the mean and variance of 12 observations be $\dfrac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\dfrac{m}{n}$, where m and n are coprime, then $m+n$ is equal to

Accepted Answer

$$
\begin{aligned}
& \text { Since, Mean }=\frac{9}{2} \\
& \Rightarrow \Sigma x=\frac{9}{2} \times 12=54
\end{aligned}
$$

Also, variance $=4$

$$
\begin{aligned}
& \Rightarrow \frac{\sum x^2}{12}=\left[\frac{\sum x_i}{12}\right]^2=4 \\
& \Rightarrow \frac{\sum x^2}{12}=4+\frac{81}{4}=\frac{97}{4} \\
& \Rightarrow \sum x^2=291
\end{aligned}
$$

$$
\begin{aligned}
& \sum x^{\prime}=54-(9+10)+7+14 \\
& =54-19+21=56 \\
& \text { and } \sum x^2=291-(81+100)+49+196 \\
& =291-181+49+196=355
\end{aligned}
$$

$$
\begin{aligned}
& \text { So, } \sigma_{\text {new }}^2=\frac{\sum x_{\text {new }}^2}{12}-\left(\frac{\sum x_{\text {new }}}{12}\right)^2 \\
& =\frac{355}{12}-\left(\frac{56}{12}\right)^2 \\
& =\frac{4260-3136}{144}=\frac{1124}{144}=\frac{281}{36} \\
& =\frac{m}{n} \\
& \Rightarrow m=281, n=36 \\
& \Rightarrow m+n=281+36=317
\end{aligned}
$$

Solutions · 1