Question

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

• A. $4$
• B. $2.4$
• C. $3.6$
• D. $18.4$

Answer 1

The correct option is D $18.4$

Given that: $n=10,p=0.4,q=(1-p)=0.6$

${\sigma }^2=E\left(X^2\right)-\left[E\right(X){]}^2$

$⟹E\left(X^2\right)={\sigma }^2+\left[E\right(X){]}^2$

Where, variance ${\sigma }^2=npq=10\left(0.4\right)\left(0.6\right)=2.4$

And $E\left(X\right)=np=10\left(0.4\right)=4$

Hence, $\left[E\right(X^2\left)\right]=2.4+4^2$

$⟹\left[E\right(X^2\left)\right]=18.4$

Hence, answer is option-(D).