Question

a study found that the mean amount of time cars spent in drive - throughs of a certain fast - food restaurant was 136.4 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 0.0885
(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 0.0636
(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

Since \(1\) minute \( = 60\) seconds, \(2\) minutes \(=2\times60 = 120\) seconds and \(3\) minutes \(=3\times60=180\) seconds.

Step2: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 136.4\) (mean) and \(\sigma=24\) (standard deviation).
For \(x = 120\):
\(z_1=\frac{120 - 136.4}{24}=\frac{- 16.4}{24}\approx - 0.68\)
For \(x = 180\):
\(z_2=\frac{180 - 136.4}{24}=\frac{43.6}{24}\approx1.82\)

Step3: Find the probabilities using the standard normal distribution

We know that \(P(Z\lt z_1)\) and \(P(Z\lt z_2)\) can be found from the standard - normal table.
\(P(Z\lt - 0.68)=0.2483\)
\(P(Z\lt1.82) = 0.9656\)

Step4: Calculate the probability between the two z - scores

Using the formula \(P(z_1\lt Z\lt z_2)=P(Z\lt z_2)-P(Z\lt z_1)\)
\(P(-0.68\lt Z\lt1.82)=0.9656 - 0.2483=0.7173\)

Answer:

\(0.7173\)