Let the mean and variance of 12 observations be 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 ?, where and are coprime, then is equal to

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Solution

Q.
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 [2023]

1 314

2 315

3 317

4 316

Ans.
(3)

Incorrect mean = $\frac{9}{2}$

$\Rightarrow Incorrect \sum x_{i} = \frac{9}{2} \times 12 = 54$

$So, correct mean = \frac{54 + 7 + 14 - 9 - 10}{12} = \frac{56}{12} = \frac{14}{3}$

$And 4 = Incorrect \frac{\sum x_{i}^{2}}{12} - {(\frac{9}{2})}^{2} \Rightarrow Incorrect \sum x_{i}^{2} = 291$

$Now, correct \sum (x_{i}^{2}) = 291 + 7^{2} + 14^{2} - 9^{2} - 10^{2} = 355$

$So, correct variance = \frac{355}{12} - {(\frac{14}{3})}^{2} = \frac{281}{36} = \frac{m}{n}$

$∴$ $m + n = 281 + 36 = 317$

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