Question

What are the mean and standard deviation of a binomial probability distribution with `n=120` and `p=3/4`?

Answer 1

Mean `90`
Standard deviation `sqrt{45/2}`

Explanation:

`mu = np`

`= (120)*(3/4)`

`= 90`

`sigma = sqrt{np(1-p)}`

`= sqrt{120*(3/4)*(1/4)}`

`= 3sqrt{5/2}`

If the random variable `X` follows the binomial distribution with parameters `n in NN` and `p in [0,1]`, the probability of getting exactly `k` successes in `n` trials, is given by

`"Pr"(X=k) = nC_k*p^k*(1-p)^{n-k}`

The mean is calculated from `"E"(X)`.

The standard deviation comes from the square root of the variance, `"Var"(X) = "E"(X^2)-["E"(X)]^2`

Where

`"E"(X) = sum_{k=0}^n k*"Pr"(X=k)`

`"E"(X^2) = sum_{k=0}^n k^2*"Pr"(X=k)`