Question

A random variable X follows binomial distribution with mean a and variance b. Then

• A. $a>b>0$
• B. $\frac{a}{b}>1$
• C. $\frac{a^2}{a-b}$ is an integer
• D. $\frac{a^2}{a+b}$ is an integer

Answer 1

Answer: (A), (B), (C)

$a>b>0$
$\frac{a}{b}>1$
$\frac{a^2}{a-b}$ is an integer

Suppose $X=B(n,p)$. Then
$a=np,b=npq$
As $0<q<1$, we get $0<npq<np$
$\Rightarrow a>b>0\Rightarrow \frac{a}{b}>1$
Also, $\frac{a^2}{a-b}=\frac{n^2{p}^2}{np-npq}=\frac{np}{1-q}=n$, which is an integer.