1. Suppose that 40 percent of the drivers stopped at State Police checkpoints in

Question

1. Suppose that 40 percent of the drivers stopped at State Police checkpoints in Tuscaloosa on Football Opening Weekend show evidence of driving while intoxicated. Consider a sample of 5 drivers.

a. Find the probability that none of the drivers shows evidence of intoxication.
b. Find the probability that at least one of the drivers shows evidence of intoxication. c. Find the probability that at most two of the drivers show evidence of intoxication.
d. Find the probability that more than two of the drivers show evidence of intoxication. e. What is the expected number of intoxicated drivers?

Explanation / Answer

n = 5

p = 0.4

It is a binomial distribution.

P(X = x) = nCx * px * (1 - p)n - x

a) P(X = 0) = 5C0 * (0.4)^0 * (0.6)^5 = 0.0778

b) P(X > 1)

= 1 - P(X < 1)

= 1 - P(X = 0)

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

= 1 - 0.0778

= 0.9222

c) P(x < 2) = P(X = 0) + P(X = 1) + P(X = 2)

                  = 5C0 * (0.4)^0 * (0.6)^5 + 5C0 * (0.4)^0 * (0.6)^5 + 5C0 * (0.4)^0 * (0.6)^5

                  = 0.6826

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

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

                  = 1 - 0.6826

                  = 0.3174

e) Expected number of intoxicated drivers = n * p = 5 * 0.4 = 2

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