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Q.

Consider a set of 3n 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 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

x1+x2…………+x2n2n=6⇒x1+……x2n=12nx2n+1+………….+x3nn=3⇒x2n+1+……+x3n=3n⇒x1,x2……+x3n=15n∴ Mean =15n3n=5and  Variance =4∑xi23n-(x¯)2=4⇒∑i=13xxi2=87n Now mean x¯=x1+1+...+x2n+1+x2n+1-1+...x3n-13n=163 and variance =∑i=12n(xi+1)2+∑i=2n+13n(xi-1)23n-(x¯)2=689=k⇒9k=68
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