Question

Suppose that only 65% of all drivers in a certain state wear a seat belt. a random sample of 80 drivers is selected. what is the probability that more than forty-two drivers wear a seat belt?

Answer 1

Pr(X >42) = Pr( Z > -2.344)

= Pr( Z< 2.344) = 0.9905

Explanation:

The scenario presented can be modeled by a binomial model;

The probability of success is, p = 0.65

There are n = 80 independent trials

Let X denote the number of drivers that wear a seat belt, then we are to find the probability that X is greater than 42;

Pr(X > 42)

In this case we can use the normal approximation to the binomial model;

mu = n*p = 80(0.65) = 52

sigma^2 = n*p*(1-p) = 18.2

Pr(X >42) = Pr( Z > -2.344)

= Pr( Z< 2.344) = 0.9905