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 (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 \(x = 104\), \(\mu=136.4\), and \(\sigma = 24\).

$$ z=\frac{104 - 136.4}{24}=\frac{-32.4}{24}=- 1.35 $$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt104)\), which is equivalent to \(P(Z\lt - 1.35)\) in the standard normal distribution.
Looking up the value of \(P(Z\lt - 1.35)\) in the standard normal table (or using a calculator with a normal - distribution function, e.g., for a TI - 84: normalcdf(-1000,-1.35)), we get \(P(Z\lt - 1.35)=0.0885\)

Answer:

\(0.0885\)