Question

a study found that the mean amount of time cars spent in drive - throughs of a certain fast - food restaurant was 130.4 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

2 minutes = 2×60 = 120 seconds, 3 minutes = 3×60 = 180 seconds.

Step2: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 130.4\) and \(\sigma=24\).
For \(x = 120\), \(z_1=\frac{120 - 130.4}{24}=\frac{- 10.4}{24}\approx - 0.43\)
For \(x = 180\), \(z_2=\frac{180 - 130.4}{24}=\frac{49.6}{24}\approx2.07\)

Step3: Find probabilities using the standard normal distribution table

\(P(Z\lt z_1)=P(Z\lt - 0.43)\)
From the standard - normal table, \(P(Z\lt - 0.43)=0.3336\)
\(P(Z\lt z_2)=P(Z\lt2.07)\)
From the standard - normal table, \(P(Z\lt2.07) = 0.9808\)

Step4: Calculate the probability between the two z - scores

\(P(-0.43\lt Z\lt2.07)=P(Z\lt2.07)-P(Z\lt - 0.43)\)
\(P(-0.43\lt Z\lt2.07)=0.9808 - 0.3336=0.6472\)

Answer:

0.6472