Question

"Eleven percent of U.S. adults eat fast food four to six times per week. You randomly select" 16 "U.S. adults. Find the probability that the number of U.S. adults who eat fast food four to six times per week is" (Larson & Farber) for questions a through d (round to three decimal places). Also answer question 1e. 1a) exactly 3 1b) more than 4 1c) at most one 1d) less than 3 le) Determine the mean, variance, and standard deviation of the binomial distribution. For each, round to one decimal place?

Answer 1

This is a binomial distribution

p = 0.11

n = 16

P(X = x) = 16Cx * 0.11^x * (1 - 0.11)^16-x

a) P(X = 3) = 16C3 * 0.11³ * 0.89¹³ = 0.164

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

                  = 1 - (16C0 * 0.11⁰ * 0.89¹⁶ + 16C1 * 0.11¹ * 0.89¹⁵ + 16C2 * 0.11² * 0.89¹⁴ + 16C3 * 0.11³ * 0.89¹³ + 16C4 * 0.11⁴ * 0.89¹²)

                  = 1 - 0.975

                  = 0.025

c) P(X < 1) = P(X = 0) + P(X = 1) = 16C0 * 0.11⁰ * 0.89¹⁶ + 16C1 * 0.11¹ * 0.89¹⁵ = 0.461

d) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

                  = 16C0 * 0.11⁰ * 0.89¹⁶ + 16C1 * 0.11¹ * 0.89¹⁵ + 16C2 * 0.11² * 0.89¹⁴

                  = 0.745

e) Mean = n * p = 16 * 0.11 = 1.76 = 1.8

Variance = n * p * (1 - p) = 16 * 0.11 * (1 - 0.11) = 1.5664 = 1.6

Standard deviation = sqrt(1.5664) = 1.3