Find the standard deviation, for the binomial distribution which has the stated values of n and p. Round your answer to the nearest...

We know for a binomial distibution with probability
function,

P(x = r) = nCr * p^r *(1-p)^(n-r), the mean = np
and variance = np(1-p).

Standard deviation, s= 
(variance)^(1/2) ={np(1-p)}^(1/2).

Standard deviation , s
 = (np(1-p)}^(1/2)....(1)

So we substitute  n = 2815, p =
0.63, in the formula at (1).

Standard deviation =
{2815*0.63*(1-0.63)}^(1/2).

Standard deviation =
{2815*0.63*0.37}^(1/2) = 25.6159 = 25.62.

So the required
standard deviation for the give binomial distribution is 25.62