Question 16 If ¯x is the mean of x1,x2,……xn, then for a≠0, 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,x2……xnis¯x. ∴∑ni=1xi=n¯x……(i) Now, let the mean of (ax1+ax2+……+axn)+x1a,x2a,……xnais¯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