# What are the mean and standard deviation of a binomial

What are the mean and standard deviation of a binomial probability distribution with n = 88 and $p=\frac{24}{32}$?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Serita Dewitt
Explanation:
for a Binomial distribution
$X\sim B\left(n,p\right)$
in this case
$X\sim B\left(88,\frac{24}{32}\right)$
or $X\sim B\left(88,\frac{3}{4}\right)$
mean E(X)=np
in this case
$E\left(X\right)=88×\frac{3}{4}=66$
standard deviation $=\sqrt{variance}$
$sd=\sqrt{np\left(1-p\right)}$
$sd=\sqrt{66×\frac{1}{4}}$
$sd=\sqrt{22}$
###### Not exactly what you’re looking for?
yotaniwc
$\frac{24}{32}=\frac{3}{4}$
$\mu =n\cdot p$
$=88\cdot \frac{3}{4}$
=66
${\sigma }^{2}=np\left(1-p\right)$
$=88\cdot \frac{3}{4}\left(1-\frac{3}{4}\right)$
=16.5
$\sigma =\sqrt{{\sigma }^{2}}=\sqrt{16.5}=4.06$