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
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 68.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
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