Question

If a variable $x$ takes values $0,1,2,....n$ with frequencies proportional to the binomial coefficients ${{}^{n}C}_0,{{}^{n}C}_1,{{}^{n}C}_2,......{{}^{n}C}_n$, then mean of distribution is

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

Answer 1

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

Here, $\mu 1^{\prime }=\frac{\sum r{\frac{n}{r}}^{n-1}{C}_{r-1}}{{\sum }^n{C}_r}$

$=\frac{n{.2}^{n-1}}{2^n}=\frac{n}{2}$
and

$\mu 2^{\prime }=\frac{1}{2^n}\sum _0^n{\{(r-1)+r\}}^n{C}_r$

$=\frac{1}{2^n}\sum _0^{n}r{(r-1)}^n{C}_r+\frac{n}{2}$

$=\frac{1}{2^n}\sum _0^{n}r(r-1){\frac{n(n-1)}{r(r-1)}}^{n-2}{C}_{r-2}+\frac{n}{2}$

$=\frac{n(n-1)}{2^n}{.2}^{n-2}+\frac{n}{2}$

$=\frac{n(n-1)}{4}+\frac{n}{2}$
$\therefore$ Variance ${\sigma }^2=\mu 2^{\prime }-{\left(\mu 1^{\prime }\right)}^2$ $=\frac{n(n-1)}{4}+\frac{n}{2}-{\left(\frac{n}{2}\right)}^2=\frac{n}{4}$