Question

a study found that the mean amount of time cars spent in drive-throughs of a certain fast food restaurant was 164 seconds. assuming drive-through times are normally distributed with a standard deviation of 24 seconds. complete parts (a) through (d) below
(a) what is the probability that a randomly selected car will get through the restaurants drive-through in less than 104 seconds?
the probability that a randomly selected car will get through the restaurants drive-through in less than 104 seconds is 0.0085
(round to four decimal places as needed.)
(b) what is the probability that a randomly selected car will spend more than 173 seconds in the restaurants drive through?
the probability that a randomly selected car will spend more than 173 seconds in the restaurants drive-through is 0.6636
(round to four decimal places as needed.)
(c) what proportion of cars spend between 2 and 3 minutes in the restaurants drive-through?
the proportion of cars that spend between 2 and 3 minutes in the restaurants drive-through is
(round to four decimal places as needed.)

Explanation:

Step1: Convert minutes to seconds

2 minutes = \(2\times60 = 120\) seconds, 3 minutes = \(3\times60=180\) seconds.

Step2: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 164\) seconds and \(\sigma = 24\) seconds.
For \(x = 120\): \(z_1=\frac{120 - 164}{24}=\frac{- 44}{24}\approx - 1.83\)
For \(x = 180\): \(z_2=\frac{180 - 164}{24}=\frac{16}{24}\approx0.67\)

Step3: Find probabilities using the standard normal distribution table

\(P(Z\lt - 1.83)=0.0336\), \(P(Z\lt0.67) = 0.7486\)

Step4: Calculate the proportion

\(P(-1.83\lt Z\lt0.67)=P(Z\lt0.67)-P(Z\lt - 1.83)\)
\(P(-1.83\lt Z\lt0.67)=0.7486 - 0.0336=0.7150\)

Answer:

\(0.7150\)