Question

(1) According to the American Lung Association, 90% of adult smokers started smoking before turning
21 years old. Ten smokers 21 years old or older were randomly selected, and the number of smokers
who started smoking before 21 is recorded.
(a) State the distribution of the random variable of number of smoker of these 10 who started
smoking before age 21 and its two parameters.
(b) Find the probability that exactly 8 of them started smoking before 21 years of age. Do not
use statistical features of your calculator.
(c) Find the probability that fewer than 8 of them started smoking before 21 years of age.
(d) Find the probability that between 7 and 9 of them, inclusively, started smoking before 21 years
of age.
(e) Compute and interpret the mean of this random variable.
(f) Compute the standard deviation of this random variable.

Answer 1

a) this is a binomial probability distribution with p = 0.9 and n = 10

P(X = x) = 10Cx * 0.9^x * (1 - 0.9)^10-x

b) P(X = 8) = 10C8 * 0.9⁸ * 0.1² = 0.1937

c) P(X < 8) = 1 - (P(X = 8) + P(X = 9) + P(X = 10))

                  = 1 - (10C8 * 0.9⁸ * 0.1² + 10C9 * 0.9⁹ * 0.1¹ + 10C10 * 0.9¹⁰ * 0.1⁰)

                  = 1 - 0.9298

                  = 0.0702

d) P(7 < X < 9) = P(X = 7) + P(X = 8) + P(X = 9)

                         = 10C7 * 0.9⁷ * 0.1³ + 10C8 * 0.9⁸ * 0.1² + 10C9 * 0.9⁹ * 0.1¹

                         = 0.6385

e) Mean = n * p = 10 * 0.9 = 9

f) Standard deviation = sqrt(n * p * (1 - p)) = sqrt(10 * 0.9 * 0.1) = 0.95