Question

How do you find the mean, variance and standard deviation of the binomial distribution with n = 316 and p = 0.72?

Answer 1

The mean is `=227.52`
The variance is `=63.71`
The standard deviation is `=7.98`

Explanation:

The binomial distribution with `n=316` and `p=0.72`

The mean is `mu_x=n*p=316*0.72=227.52`

The variance is `=sigma_x^2=n*p*(1-p)=316*0.72*0.28=63.71`

The standard deviation is `=sigma_x=sqrt(n*p*(1-p))`

`=sqrt(63.7056)=7.98`