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.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 0.7173(round to four decimal places as needed.)(d) would it be unusual for a car to spend more than 3 minutes in the restaurants drive - through? why?the probability that a car spends more than 3 minutes in the restaurants drive - through is (square), so it (square)be unusual, since the probability is (square) than 0.05(round to four decimal places as needed.)

Explanation:

Step1: Convert 3 minutes to seconds

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

Step2: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 136.4\) (mean), \(\sigma=24\) (standard deviation), and \(x = 180\) (value).

$$z=\frac{180 - 136.4}{24}=\frac{43.6}{24}\approx1.8167$$

Step3: Find the probability

We want \(P(X>180)\). Using the property \(P(X > x)=1 - P(X\leq x)\).
Looking up the z - score of \(z = 1.82\) (rounded to two decimal places for standard normal table use) in the standard normal table, \(P(Z\leq1.82)=0.9656\).
So \(P(X>180)=1 - 0.9656=0.0344\)

Answer:

The probability that a car spends more than 3 minutes in the restaurant's drive - through is \(0.0344\), so it would be unusual, since the probability is less than \(0.05\).