Question

In a distribution the variate $x$, assumes the values $0,1,2,.....n$ with frequencies (weights) ${}^n{C}_0{,}^n{C}_1{,}^n{C}_2,...{.}^n{C}_n,$ then mean of the distribution is

• A. $\frac{2^n-1}{n+1}$
• B. $\frac{2^{n+1}}{n+1}$
• C. $\frac{n}{2}$
• D. $\frac{2^n}{n}$

Answer 1

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

Required A.M $=\frac{\sum x_i{f}_i}{\sum f_i}=\frac{{0.}^n{C}_0{+}^n{C}_1+{2.}^n{C}_2+{3.}^n{C}_3+..............+n{.}^n{C}_n}{{}^n{C}_0{+}^n{C}_1{+}^n{C}_2{+}^n{C}_3+.......{+}^n{C}_n}=\mu$ (say)

Now consider $(1+x{)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+.........{+}^n{C}_n{x}^n....(1)$

Differentiating both side (1) w.r.t $x$

$n(1+x{)}^{n-1}{=}^n{C}_1+{2.}^n{C}_{2}x+.........+n{.}^n{C}_n{x}^{n-1}....(2)$

Now putting $x=1$ in (1) and (2) we get,
${}^n{C}_0{+}^n{C}_1{+}^n{C}_2+.........{+}^n{C}_n=2^n$
and ${}^n{C}_1+{2.}^n{C}_2+.........+n{.}^n{C}_n=n.2^{n-1}$
$\therefore \mu =\frac{n.2^{n-1}}{2^n}=\frac{n}{2}$