Discussion Forum

If  $X\sim B(n,p)$  with n=10 , p=0.4  , then E(X)²=

Answer:

Explanation:

Given , n=10  and p=0.4

 $\therefore$ q=1-p=1-0.4=0.6

 E(x) = np=10(0.4)=4

$\sigma^{2}$ = npq= 10(0.4)(0.6)=2.4

We know that,

$\sigma^{2}= E(X^{2})-(E(X))^{2}$

$\therefore$     $ E(X^{2})=\sigma^{2}+(E(X))^{2}$

$ \Rightarrow$   $ E(X^{2})=24+(4)^{2}$

  =2.4+16=18.4