Question

.For a Binomial Distribution Mean m = 3 and Variance = 3/2 then the value of n is​

Answer 1

$\large\underline{\sf{Solution-}}$

We know,

In Binomial Distribution,

• n represents number of independent trials

• p represents probability of success

• q represents probability of failure

Given that

☆ Mean of Binomial Distribution = 3

$\red{\rm :\longmapsto\:np = 3} - - - (1)$

and

$\sf \: Variance = \dfrac{3}{2}$

$\red{\rm :\longmapsto\:npq = \dfrac{3}{2}} - - - (2)$

$\rm :\longmapsto\:3q = \dfrac{3}{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: np = 3\bigg \}}$

$\bf\implies \:q = \dfrac{1}{2}$

We know,

$\rm :\longmapsto\:p + q = 1$

$\rm :\longmapsto\:p + \dfrac{1}{2} = 1$

$\rm :\longmapsto\:p = 1 - \dfrac{1}{2}$

$\rm :\longmapsto\:p = \dfrac{2 - 1}{2}$

$\bf\implies \:p = \dfrac{1}{2}$

☆ On substituting the value of p in equation (1), we get

$\rm :\longmapsto\:n \times \dfrac{1}{2} = 3$

$\bf\implies \: n\: = \: 6$

Additional Information :-

In Binomial Distribution, Mean > Variance.

Proof :-

Consider,

$\rm :\longmapsto\:Mean - Variance$

$\rm \: \: = \: \: np - npq$

$\rm \: \: = \: \: np(1 - q)$

$\rm \: \: = \: \: np \times p$

$\rm \: \: = \: \: {np}^{2}$

$\rm :\implies\:Mean - Variance > 0$

$\bf\implies \:Mean > Variance$