Question

If a variable $x$ takes values $0,1,2, \ldots, n$ with frequencies proportional to the binomial coefficients ${ }^{n} C_{0},{ }^{n} C_{1},{ }^{n} C_{2}, \ldots$ ${ }^{n} C_{n}$, then var(X) is

• A. $\frac{n^{2}-1}{12}$
• B. $\frac{n}{2}$
• C. $\frac{n}{4}$
• D. None of these

Answer 1

Answer: (C)

We have
$\bar{X}=\frac{0^{n} C_{0}+1^{n} C_{1}+2^{n} C_{2}+\dots+n^{n} C_{n}}{{ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\cdots+{ }^{n} C_{n}}$
$=\frac{\displaystyle\sum_{r=0}^{n} r^{n} C_{r}}{\displaystyle\sum_{r=0}^{n}{ }^{n} C_{r}}$
$=\frac{1}{2^{n}} \displaystyle\sum_{r=1}^{n} r \frac{n}{r}{ }^{n-1} C_{r-1} \left[\because \displaystyle\sum_{r=0}^{n}{ }^{n} C_{r}=2^{n} ;{ }^{n} C_{r}=\frac{n}{r}{ }^{n-1} C_{r-1}\right]$
$=\frac{n}{2^{n}} \displaystyle\sum_{r=1}^{n}{ }^{n-1} C_{r-1}=\frac{n}{2^{n}} 2^{n-1}=\frac{n}{2} \left[\because \displaystyle\sum_{r=1}^{n}{ }^{n-1} C_{r-1}=2^{n-1}\right]$
and $\frac{1}{N} \sum f_{i} x_{i}^{2}=\frac{1}{2^{n}} \sum r^{2 n} C_{r}=\frac{1}{2^{n}} \displaystyle\sum_{r=0}^{n}[r(r-1)+r]^{n} C_{r}$
$=\frac{1}{2^{n}}\left\{\displaystyle\sum_{r=0}^{n} r-(r-1){ }^{n} C_{r}+\displaystyle\sum_{r=0}^{n} r{ }^{n} C_{r}\right\}$
$=\frac{1}{2^{n}}\left\{\displaystyle\sum_{r=2}^{n} r(r-1) \frac{n}{r} \frac{n-1}{r-1}{ }^{n-2} C_{r-2}+\displaystyle\sum_{r=1}^{n} r \frac{n}{r}{ }^{n-1} C_{r-1}\right\}$
$=\frac{1}{2^{n}}\left\{n(n-1) 2^{n-2}+n 2^{n-1}\right\}=\frac{n(n-1)}{4}+\frac{n}{2}$
$\therefore $ Var(X) $=\frac{1}{N} \sum f_{i} x_{i}^{2}-\bar{X}^{2}=\frac{n(n-1)}{4}+\frac{n}{2}-\frac{n^{2}}{4}=\frac{n}{4}$