Question

What are the mean and standard deviation of a binomial probability distribution with `n=100` and `p=1/10`?

Answer 1

`mu = np=100*1/10=10`
`sigma=sqrt(npq)=sqrt(100*1/10*9/10)=3`

Explanation:

For Binomial Probability Distribution:
Mean, `mu = np` and Variance, `sigma^2 = n*p*q; sigma=sqrt(npq)`
Given that `p=1/10 => q=9/10` thus,
`mu = np=100*1/10=10`
`sigma=sqrt(npq)=sqrt(100*1/10*9/10)=3`