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# In a binomial distribution the mean is 15 and variance is 10, then parameter $n$ is

$\begin {array} {1 1} (A)\;28 & \quad (B)\;16 \\ (C)\;45 & \quad (D)\;25 \end {array}$

Can you answer this question?

Given mean = $np = 15$
Variance $np(1-P)=10$
$1-P = \large\frac{10}{15}$
$= \large\frac{2}{3}$
$P = 1- \large\frac{2}{3}$
$= \large\frac{1}{3}$
$n = 15 \times 3$
$= 45$
Ans : (C)
answered Jan 16, 2014