Question

The American Journal of Public Health (July 1995) published a study of the relationship between passive smoking and nasal allergies in Japanese female students. The study revealed that 40% of the students from heavy-smoking families showed signs of nasal allergies on physical examination. Let x denote the number of students with nasal allergies in a random sample of 6 Japanese female students exposed daily to heavy smoking.

a. What is the probability that at least four of the students will have nasal allergies?

b. What is the probability that at least two of the students will have no nasal allergies?

c. Suppose instead of 6, a random sample of 20 Japanese female students were exposed daily to heavy smoking, what is the probability that between 4 and 14 (both points inclusive) of the students will have nasal allergies?

Answer 1

a) n = 6

p = 0.4

It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X > 4) = P(X = 4) + P(X = 5) + P(X = 6)

= 6C4 * (0.4)^4 * (0.6)^2 + 6C5 * (0.4)^5 * (0.6)^1 + 6C6 * (0.4)^6 * (0.6)^0 = 0.1792

b) n = 6

p = 1 - 0.4 = 0.6

P(X > 2) = 1 - P(X < 2)

= 1 - (P(X = 0) + P(X = 1))

= 1 - (6C0 * (0.6)^0 * (0.4)^6 + 6C1 * (0.6)^1 * (0.4)^5)

= 1 - 0.04096 = 0.95904

c) n = 20

p = 0.4

P(4 < X < 14)

= P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14)

= 20C4 * (0.4)^4 * (0.6)^16 + 20C5 * (0.4)^5 * (0.6)^15 + 20C6 * (0.4)^6 * (0.6)^14 + 20C7 * (0.4)^7 * (0.6)^13 + 20C8 * (0.4)^8 * (0.6)^12 + 20C9 * (0.4)^9 * (0.6)^11 + 20C10 * (0.4)^10 * (0.6)^10 + 20C11 * (0.4)^11 * (0.6)^9 + 20C12 * (0.4)^12 * (0.6)^8 + 20C13 * (0.4)^13 * (0.6)^7 + 20C14 * (0.4)^14 * (0.6)^6 = 0.9824