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(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 173\), \(\mu=136.4\), and \(\sigma = 24\).

$$z=\frac{173 - 136.4}{24}=\frac{36.6}{24}=1.525$$

Step2: Find the probability

We want \(P(X>173)\), which is equivalent to \(P(Z > 1.525)\). Since \(P(Z>z)=1 - P(Z\leq z)\).
Looking up \(P(Z\leq1.525)\) in the standard normal table (or using a calculator with a normal - distribution function), \(P(Z\leq1.525)\approx0.9364\)

$$P(Z > 1.525)=1 - 0.9364 = 0.0636$$

Answer:

\(0.0636\)