The variate of a distribution takes the values 1,2,3,...n with frequencies n,n−1,n−2,...3,2,1. then mean value of the distribution is
n(n+1)(n+2)6
n(n+2)3
n+23
(n+1)(n+2)6
A
(n+1)(n+2)6
B
n(n+2)3
C
n(n+1)(n+2)6
D
n+23
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Solution
Verified by Toppr
xi=1,2,3,4...,n−1,n fi=n,n−1,n−2,n−3,...2,1 Now n∑i=1xifi=n∑i=1[n+2(n−1)+3(n−2)...+(n−2)2+n.1]=n∑r=1[(n+1)r−r2]=(n+1)n∑r=1r−n∑r=1r2=(n+1)(n)(n+1)2−n(n+1)(2n+1)6=n(n+1)(n+2)6
Also ∑fi=1+2+...+n=n(n+1)2
∴Mean=∑fixi∑fi=n(n+1)(n+2)6×2n(n+1)=n+23
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