# What are the mean and standard deviation of a binomial probability distribution with n=88  and p=24/32 ?

Mar 4, 2018

$E \left(X\right) = 66$

$s d = \sqrt{22}$

#### Explanation:

for a Binomial distribution

X~B(n,p)

in this case

X~B(88,24/32)

or" "X~B(88,3/4)

mean$\text{ } E \left(X\right) = n p$

in this case

$E \left(X\right) = 88 \times \frac{3}{4} = 66$

standard deviation =$\sqrt{\text{variance}}$

$s d = \sqrt{n p \left(1 - p\right)}$

$s d = \sqrt{66 \times \frac{1}{4}}$

$s d = \sqrt{22}$