Statistics-And-Probability Question 619
Question: The mean deviation from the mean of the A.P. $ a,a+d,a+2d,…a,a+2nd $ is
Options:
A) $ n(n+1)d $
B) $ \frac{n(n+1)d}{2n+1} $
C) $ \frac{n(n+1)d}{2n} $
D) $ \frac{n(n-1)d}{2n+1} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The mean of the series $ \bar{X}=\frac{1}{2n+1}{a+(a+d)+(a+2d)+…+(a+2nd)} $ $ =\frac{1}{2n+1}{ \frac{2n+1}{2}(2a+2nd) }=a+nd $ Therefore, mean deviation from mean $ \frac{1}{2n+1}\sum\limits_{r=0}^{2n}{| (a+rd)-(a+nd) |} $ $ =\frac{1}{2n+1}\sum\limits_{r=0}^{2n}{| r-n |d} $ $ =\frac{[2(1+2+…+n)+0]d}{2n+1}=\frac{n(n+1)d}{2n+1} $
BETA
