Question

A study found that the mean amount of time cars spent in​ drive-throughs of a certain​ fast-food restaurant was 154.1 seconds. Assuming​ drive-through times are normally distributed with a standard deviation of 28 ​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 115 ​seconds?

b) What is the probability that a randomly selected car will spend more than 201 seconds in the​ restaurant's drive-through?

c) What proportion of cars spend between 22 and 33 minutes in the​ restaurant's drive-through?

d) Would it be unusual for a car to spend more than 33 minutes in the​ restaurant's drive-through?​ Why?

Answer 1

Solution :

Given that ,

a) P(x < 115)

= P[(x - ) / < (115 - 154.1) / 28 ]

= P(z < -1.40)

Using z table,

= 0.0808

b) P(x > 201) = 1 - p( x< 201)

=1- p P[(x - ) / < (201 - 154.1) /28 ]

=1- P(z < 1.68)

= 1 - 0.9535

= 0.0465

c) P(120 < x < 180 ) = P[(120 - 154.1)/ 28) < (x - ) /  < (180 - 154.1) / 28) ]

= P(-1.22 < z < 0.93)

= P(z < 0.93) - P(z < -1.22)

Using z table,

= 0.8238 - 0.1112

= 0.7126

d) P(x > 180) = 1 - p( x< 180)

=1- p P[(x - ) / < (180 - 154.1) /28 ]

=1- P(z < 0.93)

= 1 - 0.8238

= 0.1762

The probability that a car spends more than 3 minutes in the​ restaurant's drive-through is 0.1762. so it not would be unusual, since the probability is more than 0.05.