The incidence of cystic fibrosis, a recessive genetic disorder in the caucasian population of united states, is 1 in every 2,500 individuals. find the number of heterozygous carriers.
(p + q = 1, p2 + 2pq + q2 = 1)

a: 0.0392
b: 0.0254
c: 0.045

As hardy  Weinberg theorem formula is given below
freq. of recessive allele/ q2
 applying the values
 = 1/2500
 q2= 0.0004
now take the square root
so
q=0.02
now
q=p-1
q=0.98
As the frequency. of heterozygous individuals = 2pq
putting values we get the answer
 = 0.04
so correct option is C 
hope it helps