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 $\frac{m}{n}$ . where m and n are co-prime then m + n is equal to

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314

315

317

316

answer is B.

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Detailed Solution

$\begin{aligned} \frac{\sum x_{i}}{12} = \frac{9}{2} \\ \sum x_{i} = 54 \end{aligned}$
Correct $\sum x_{i} = 54 + 7 + 14 - 9 - 10 = 56$
Correct $\begin{array}{l} \bar{X} = \frac{56}{12} = \frac{14}{3} \\ σ^{2} = \frac{\sum x_{i}^{2}}{12} - {(\frac{9}{2})}^{2} = 4 \\ \frac{\sum x_{i}^{2}}{12} = \frac{97}{4} \Rightarrow \sum x_{i}^{2} = 291 \end{array}$
Correct
$\begin{array}{l} \sum X_{i}^{2} = 291 + 64 = 355 \\ correct σ^{2} = \frac{\sum X_{i}^{2}}{12} - (\bar{X})^{2} = \frac{355}{12} - {(\frac{14}{3})}^{2} \\ \frac{281}{36} = \frac{m}{n} \\ m + n = 281 + 36 = 317 \end{array}$