# What are the mean and standard deviation of a binomial

Jason Yuhas 2022-01-19 Answered
What are the mean and standard deviation of a binomial probability distribution with n = 88 and $p=\frac{24}{32}$?
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## Expert Answer

Serita Dewitt
Answered 2022-01-19 Author has 41 answers
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 $=\sqrt{variance}$
$sd=\sqrt{np\left(1-p\right)}$
$sd=\sqrt{66×\frac{1}{4}}$
$sd=\sqrt{22}$
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yotaniwc
Answered 2022-01-20 Author has 34 answers
$\frac{24}{32}=\frac{3}{4}$
$\mu =n\cdot p$
$=88\cdot \frac{3}{4}$
=66
${\sigma }^{2}=np\left(1-p\right)$
$=88\cdot \frac{3}{4}\left(1-\frac{3}{4}\right)$
=16.5
$\sigma =\sqrt{{\sigma }^{2}}=\sqrt{16.5}=4.06$
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