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Question 16
If ¯x is the mean of x1,x2,xn, then for a0, the mean of ax1,ax2,axn,x1a,x2a,xna is:

A) (a+1a)¯x
B) (a+1a)¯x2
C) (a+1a)¯xn
D) (a+1a)¯x2n

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Solution

The answer is B.
Given, mean of x1,x2xnis¯x.
ni=1xi=n¯x(i)
Now, let the mean of (ax1+ax2++axn)+x1a,x2a,xna is ¯z.
Then, ¯z=(ax1+ax2++axn)+(x1a+x2a+xna)n+n
=a(x1+x2++xn)+1a(x1+x2++xn)2n
=(a+1a)(x1+x2++xn)2n
=(a+1a)ni=1xi2n=(a+1a)n¯x2n [from Eq. (i)]
=(a+1a)¯x2

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