Question

trucks in a delivery fleet travel a mean of 100 miles per day with a standard deviation of 33 miles per day. the mileage per day is distributed normally. find the probability that a truck drives less than 172 miles in a day. round your answer to four decimal places.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 172\), \(\mu=100\), and \(\sigma = 33\).

$$z=\frac{172 - 100}{33}=\frac{72}{33}\approx2.18$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(X\lt172)\), which is equivalent to \(P(Z\lt2.18)\) using the standard normal distribution table or a calculator with a normal - distribution function.
Using a standard normal table or a calculator (e.g., in Excel: NORM.S.DIST(2.18, TRUE)), we find that \(P(Z\lt2.18)\)

Answer:

\(0.9854\)