Question

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

• A. $\frac{n(n+2)}{3}$
• B. $\frac{n(n+1)(n+2)}{6}$
• C. $\frac{n+2}{3}$
• D. $\frac{(n+1)(n+2)}{6}$

Answer 1

The correct option is C $\frac{n+2}{3}$

$x_i=1,2,3,4...,n-1,n$
$f_i=n,n-1,n-2,n-3,...2,1$
Now $\sum _{i=1}^n{x}_i{f}_i=\sum _{i=1}^n[n+2(n-1)+3(n-2)...$ $+(n-2)2+n.1]$ $=\sum _{r=1}^n\left[\right(n+1)r-r^2]$$=(n+1)\sum _{r=1}^{n}r-\sum _{r=1}^n{r}^2$ $=\frac{(n+1)\left(n\right)(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}$$=\frac{n(n+1)(n+2)}{6}$

Also $\sum f_i=1+2+...+n=\frac{n(n+1)}{2}$

$\therefore \text{Mean}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{n(n+1)(n+2)}{6}\times \frac{2}{n(n+1)}$ $=\frac{n+2}{3}$