Discussion Forum

 If the mean and variance of a binomial  distribution are 4 and 2, respectively . Then the probability of atleast 7 successes is 

Answer:

Explanation:

 Here, mean =4 and variance =2

 $\Rightarrow$   np=4 and npq=2

 so,  $\frac{npq}{np}=\frac{2}{4}\Rightarrow q=\frac{1}{2}$

 Then   $p=1-q=1-\frac{1}{2}=\frac{1}{2}$

 Mean= np=4

 $\Rightarrow$   $n\times\frac{1}{2}=4\Rightarrow n=8$

 $\therefore$      $P(X=r)=^{n}C_{r} p^{r}q^{n-r}$

 $=^{8}C_{r} \left(\frac{1}{2}\right)^{8}$       $\left[\because p=q=\frac{1}{2}\right]$

 The required probabilty of atleast 7 successes is

  $P(X\geq 7)=P(X=7)+P(X=8)$

 = $\left(^{8}C_{7}+^{8}C_{8}\right)\left(\frac{1}{2}\right)^{8}$

  =   $\left(\frac{8!}{7!1!}+\frac{8!}{8!0!}\right)\left(\frac{1}{2}\right)^{8}$

=  $\left(8+1\right)\left(\frac{1}{2}\right)^{8}=\frac{9}{256}$