Question

A variable takes the values 0, 1, 2, 3, ....n with frequencies proportional to the binomial coefficients ${}^n{c}_0{,}^n{c}_1{,}^n{c}_2{,}^n{c}_3,......{.}^n{c}_n$ then variance ${\sigma }^2=$

• A. $n$
• B. $\frac{n}{2}$
• C. $\frac{n}{4}$
• D. $\frac{n}{6}$

Answer 1

The correct option is C $\frac{n}{4}$

The probability of success, thus become $\frac{1}{2}$
${\sigma }^2=Variance=np(1-p)$
$=n\left(\frac{1}{2}\right)(1-\frac{1}{2})$

$=n\frac{1}{4}$

$=\frac{n}{4}$