Question

If a variable takes values $0,1,2,...,n$ with frequencies $1,{}^n{C}_1,{}^n{C}_2,...,{}^n{C}_n$, then the AM is

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

Answer 1

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

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 $(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+.........+{}^n{C}_n{x}^n$ .... $\left(1\right)$
Differentiating both sides of $\left(1\right)$ w.r.t $x$, we get
$n(1+x{)}^{n-1}={}^n{C}_1+{2.}^n{C}_{2}x+.........+n{.}^n{C}_n{x}^{n-1}....(2)$
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 ${}^n{C}_1+{2.}^n{C}_2+.........+n{.}^n{C}_n=n.2^{n-1}$
$\therefore \mu =\frac{n.2^{n-1}}{2^n}=\frac{n}{2}$