Question

trucks in a delivery fleet travel a mean of 90 miles per day with a standard deviation of 14 miles per day. the mileage per day is distributed normally. find the probability that a truck drives less than 102 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 \(x = 102\), \(\mu=90\), and \(\sigma = 14\).

$$z=\frac{102 - 90}{14}=\frac{12}{14}\approx0.86$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(X\lt102)\), which is equivalent to \(P(Z\lt0.86)\) using the standard normal distribution table or a calculator with a normal - distribution function.
Using a standard normal table or a calculator (e.g., for a TI - 84: normalcdf(-1000,0.86)), we find that \(P(Z\lt0.86)=0.8051\)

Answer:

\(0.8051\)