Question

The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population.

Sampling distribution: B(n = 25, p = 0.08).

a) what is the probability that more than 5 will be colour blind?

Where does the number for P(X>5) come from??

Answer 1

Solution :

Given that n = 25 , p = 0.08

=> q = 1 - p = 0.92

for binomial distribution , P(x = r) = nCr*p^r*q^(n - r)
  a)   
=> P(x > 5) = 1 - P(x <= 5)

= 1 - [P(x = 5) + P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0)]

= 1 - [25C5*0.08^5*0.92^20 + 25C4*0.08^4*0.92^21 + 25C3*0.08^3*0.92^22 + 25C2*0.08^2*0.92^23 + 25C1*0.08^1*0.92^24 + 25C0*0.08^0*0.92^25]

= 1 - [0.0329 + 0.0899 + 0.1881 + 0.2821 + 0.2704 + 0.1244]

= 1 - 0.9878

= 0.0122