# What are the mean and standard deviation of a binomial

What are the mean and standard deviation of a binomial probability distribution with n=12 and $p=\frac{3}{5}$?
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reinosodairyshm
Explanation:
Mean of a binomial probability distribution is given by np and standard deviation is given by $\sqrt{n}pq$, where q=1-p.
In the given example, $np=12\cdot \frac{3}{5}=7.2,$
As $q=1-\frac{3}{5}=\frac{2}{5},\sqrt{n}pq=\sqrt{12\cdot \frac{3}{5}\cdot \frac{2}{5}}=\frac{1}{5}\cdot \sqrt{72}=6\frac{\sqrt{2}}{5}=1.697$
Hence mean is 7.2 and standard deviation is 1.697
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Elois Puryear
The formula for the population mean is $\mu =n\cdot p$, so therefore we get:
$\mu =n\cdot p$
$=12\cdot \frac{3}{5}$
=7.2
On the other hand, the variance is computed using the following formula: ${\sigma }^{2}=np\left(1-p\right)$. Therefore, the calculation goes like:
${\sigma }^{2}=np\left(1-p\right)$
$=12\cdot \frac{3}{5}\left(1-\frac{3}{5}\right)$
=2.88
And finally, taking square root to the variance we get that the population standard deviation is:
$\sigma =\sqrt{{\sigma }^{2}}=\sqrt{2.88}=1.697$