Question

induviduals who have a particular gene have a .72 probability of contracting a certain disease suppose that 526 induviduals with the gene participate in a lifetime study what is the standard deviation of the number of people who eventually contract the disease?

Answer 1

SOLUTION:

Given that,

p = 0.72

q = 1 - p =1-0.72=0.28

n = 526

Using binomial distribution,

$\mu$ = n * p = 526*0.72=378.72

STANDARD DEVIATION $\sigma$ = $\sqrt$ n * p * q = $\sqrt$ 526*0.72*0.28=10.2976