Question

A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was 136.3 seconds. Assuming​ drive-through times are normally distributed with a standard deviation of 27 ​seconds, complete parts​ (a) through​ (d) below.

(a) What is the probability that a randomly selected car will get through the​ restaurant's drive-through in less than 104 ​seconds?

(b)What is the probability that a randomly selected car will spend more than 224 seconds in the​ restaurant's drive-through?

​(c) What proportion of cars spends between 2 and 3 minutes in the​ restaurant's drive-through?

(d) Would it be unusual for a car to spend more than 3 minutes in the​ restaurant's drive-through?​ Why?

Answer 1

Solution: Given that mean = 136.3, sd = 27

a) P(X < 104) = P((X-mean)/sd < (104 - 136.3)/27)
= P(Z < -1.1963)
= 0.1151

b) P(X > 224) = P(Z > (224-136.3)/27)
= P(Z > 3.2481)
= 0.0006

c) 2 minutes = 2*60 = 120 , 3 minutes 3*60 = 180

P(120 < X < 180) = P((120-136.3)/27 < Z < (180-136.3)/27)
= P(-0.6037 < Z < 1.6185)
= 0.6731

d) P(X > 180) = P(Z > 1.6185) = 0.0526

it is not unusual to spend more than 3 minutes in the drive-through.