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  • Let the mean and variance of 12 observations be  and 4 respectively. Later on, it

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 coprime, then m + n is equal to

Option: 1

316


Option: 2

317


Option: 3

315


Option: 4

314


Answers (1)

best_answer

\begin{aligned} & \frac{\sum x}{12}=\frac{9}{2} \\ & \sum \mathrm{x}=54 \\ & \frac{\sum x^2}{12}-\left(\frac{9}{2}\right)^2=4 \\ & \sum \mathrm{x}^2=291 \\ & \sum \mathrm{x}_{\text {new }}=54-(9+10)+7+14=56 \\ & \sum \mathrm{x}_{\text {new }}^2=291-(81+100)+49+196=355 \\ & \sigma_{\text {new }}^2=\frac{355}{12}-\left(\frac{56}{12}\right)^2 \\ & \sigma_{\text {new }}^2=\frac{281}{36}=\frac{m}{n} \\ & \mathrm{~m}+\mathrm{n}=317 \text { Option (2) } \end{aligned}

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