# What are the mean and standard deviation of a binomial

What are the mean and standard deviation of a binomial probability distribution with n=9 and $p=\frac{11}{32}$?
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Alex Sheppard
mean =np
$s.d.=\sqrt{npq}$
Explanation:
Using the formula for mean and s.d. above ...
$mean=9×\frac{11}{32}=\frac{99}{32}\approx 3.09$
$s.d.=\sqrt{9×\frac{11}{32}×\left(1-\frac{11}{32}\right)}\approx 1.425$
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twineg4
Formula:
$\mu =n\cdot p$
$=9\cdot \frac{11}{32}$
=3.06
${\sigma }^{2}=np\left(1-p\right)$
$=9\cdot \frac{11}{32}\left(1-\frac{11}{32}\right)$
=2.01
And finally, taking square root to the variance we get that the population standard deviation is:
$\sigma =\sqrt{{\sigma }^{2}}=\sqrt{2.01}=1.42$