# What are the mean and standard deviation of a binomial

What are the mean and standard deviation of a binomial probability distribution with n=210 and $p=\frac{12}{21}$?
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Explanation:
Binomial distribution is a frequency distribution of possible number of successful outcomes in given number of trials n, in each of which there is the probability of success is p (and q=1-p is probability of failure).
The mean and standard deviation of a binomial distribution are given by np and $\sqrt{npq}$.
Here n = 210, $p=\frac{12}{21}$ and $q=1-\frac{12}{21}=\frac{9}{21}$.
Hence Mean is $210×\frac{12}{21}=120$
Standard Deviation is $\sqrt{210×\frac{12}{21}×\frac{9}{21}}$
$=\frac{1}{21}×\sqrt{22680}=\frac{150.6}{21}=7.17$

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usumbiix
The formula for the population mean is $\mu =n\cdot p$, so therefore we get:
$\mu =n\cdot p$
$\frac{12}{21}\approx 0.57$
=210*0.57
=119.7
the variance is computed using the following formula: ${\sigma }^{2}=np\left(1-p\right)$.
${\sigma }^{2}=np\left(1-p\right)$
=210*0.57(1-0.57)
=51.4
$\sigma =\sqrt{{\sigma }^{2}}=\sqrt{51.4}=7.17$