A report by the U.S. Centers for Disease Control (CDC) based on 2013 data stated

Question

A report by the U.S. Centers for Disease Control (CDC) based on 2013 data stated that 18% of American adults smoke. Consider a simple random sample of 12 adults selected at that time. (a). What is the distribution of the number of American adult smokers in the group of 12? ( b). What is the probability that at least 4 of the 12 are smokers?    (c). What is the probability that at least 1 of them is a smoker?    (d). If you know at least 4 are smokers, what is the probability that all 12 are smokers? (e). Find the mean and standard deviation of the number of smokers in the sample.

Explanation / Answer

A) the distribution is binomial distribution with p = 0.18, n = 12

P(X = x) = 12Cx * 0.18x * (1 - 0.18)12-x

B) P(X > 4) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))

= 1 - (12C0 * 0.180 * 0.8212 + 12C1 * 0.181 * 0.8211 + 12C2 * 0.182 * 0.8210 + 12C3 * 0.183 * 0.829)

= 1 - 0.8448

= 0.1552

C) P(X > 1) = 1 - P(X = 0)

= 1 - 12C0 * 0.180 * 0.8212

= 1 - 0.0924

= 0.9076

D) Probability = P(X = 12) / P(X > 4)

= 0.1812 / 0.1552

= 7.4538 * 10-9

E) mean = n * p = 12 * 0.18 = 2.16

Standard deviation = sqrt(n * p * (1-p)) = sqrt(12 * 0.18 * 0.82) = 1.33

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