 # What are the mean and standard deviation of a binomial untchick04tm 2022-01-17 Answered
What are the mean and standard deviation of a binomial probability distribution with n=15 and $p=\frac{7}{17}$?
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 enhebrevz
The mean is $\mu =np=\frac{105}{17}\approx 6,176$ and the standard deviation is
$\sigma =\sqrt{np\left(1-p\right)}=\frac{5\sqrt{42}}{17}\approx 1.906$.
Explanation: If X is a binomial random variable, counting the number of successes in n independent trials (where the only two outcomes are success and failure), with constant probability of success p on each trial, the mean of X is $\mu =np$ and the standard deviation is $\sqrt{np\left(1-p\right)}$.
In the present case, note that
$\sqrt{np\left(1-p\right)}=\sqrt{15\cdot \frac{7}{17}\cdot \frac{10}{17}}=\sqrt{\frac{1050}{289}}=\frac{\sqrt{25\cdot 42}}{\sqrt{289}}=\frac{5\sqrt{42}}{17}$
###### Not exactly what you’re looking for? Jillian Edgerton
$\mu =n\cdot p$
$\frac{7}{17}\approx 0.41$
=15*0.41
=6.15
${\sigma }^{2}=np\left(1-p\right)$
=15*0.41(1-0.41)
=3.6
$\sigma =\sqrt{{\sigma }^{2}}=\sqrt{3.6}=1.9$