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Consider a set of $3 n$ numbers having variance 4. In this set the mean of first 2n numbers is 6 and the mean of the remaining ‘n’ numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers and subtracting 1 from each of the remaining ‘n’ numbers. If the variance of the new set is ‘k’, then 9k is equal to …..

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answer is 68.

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Detailed Solution

$L e t n u m b e r b e a_{1}, a_{2}, a_{3}, ......., a_{2 n}, b_{1}, b_{2}, b_{3}, b_{n}$

$σ^{2} = \frac{\sum a^{2} + \sum b^{2}}{3 n} - {(5)}^{2} \Rightarrow \sum a^{2} + \sum b^{2} = 87 n$

Now, distribution becomes

$= \frac{\sum {(a + 1)}^{2} + \sum {(b - 1)}^{2}}{3 n} - {(\frac{12 n + 2 n + 3 n - n}{3 n})}^{2}$

$= \frac{87 n + 3 n + 2 (12 n) - 2 (3 n)}{3 n} - {(\frac{16}{3})}^{2} \Rightarrow k = \frac{108}{3} - {(\frac{16}{3})}^{2}$

$\Rightarrow 9 k = 3 (108 - {(16)}^{2} = 324 - 256 = 68)$

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