Let the mean and variance of 12 observations be [(9)/(2)] and 4 respectively.

Question

Let the mean and variance of 12 observations be $$\frac{9}{2}$$ 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 $$\frac{m}{n}$$, wh...

Accepted Answer

$\begin{gathered} =\frac{9}{2} \ \Rightarrow \Sigma x=\frac{9}{2}	imes 12=54 \end{gathered}$

Also, variance $=4$

$\begin{gathered} \Rightarrow \frac{\sum {x}^{2}}{12}={[\frac{\sum {x}_{i}}{12}]}^{2}=4 \ \Rightarrow \frac{\sum {x}^{2}}{12}=4+\frac{81}{4}=\frac{97}{4} \ \Rightarrow \sum {x}^{2}=291 \end{gathered}$

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

$\begin{gathered} {\sigma }_{2}^{}=\frac{\sum {x}_{2}^{}}{12}-{(\frac{\sum {x}_{}}{12})}^{2} \ =\frac{355}{12}-{(\frac{56}{12})}^{2} \ =\frac{4260-3136}{144}=\frac{1124}{144}=\frac{281}{36} \ =\frac{m}{n} \ \Rightarrow m=281,n=36 \ \Rightarrow m+n=281+36=317 \end{gathered}$

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