Question

A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was

156.6 seconds. Assuming​ drive-through times are normally distributed with a standard deviation of

26 ​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

116 ​seconds?

B.) What is the probability that a randomly selected car will spend more than

189seconds in the​ restaurant's drive-through?

C.) What proportion of cars spend 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

Given,

= 156.6 , = 26

We convert this to standard normal as

P( X < x) = P( Z < x - / )

a)

P( X < 116) = P( Z < 116 - 156.6 / 26)  

= P( Z < -1.5615)

= 0.0592

b)

P( X > 189) = P( Z > 189 - 156.6 / 26)

= P( Z > 1.2462)

= 0.1063

c)

P( 2 min < X < 3 min) = P( 120 sec < X < 180 sec)

= P( X < 180) - P( X < 120)

= P( Z < 180 - 156.6 / 26) - P( Z < 120 - 156.6 / 26)

= P( Z < 0.9) - P( Z < - 1.4077)

= 0.8159 - 0.0796

= 0.7363

d)

P( X > 3 min ) = P( X > 180 sec)

= P( Z > 180 - 156.6 / 26)

= P( Z > 0.9)

= 0.1841

Since this probability is greater than 0.05, the event is not unusual.