Question

Trucks in a delivery fleet travel a mean of 90 miles per day with a standard deviation of 36 miles per day. The mileage per day is distributed normally. Find the probability that a truck drives less than 118 miles in a day. Round your answer to four decimal places.

Answer 1

0.7823 = 78.23% probability that a truck drives less than 118 miles in a day.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Trucks in a delivery fleet travel a mean of 90 miles per day with a standard deviation of 36 miles per day.

This means that
`\mu = 90, \sigma = 36`

Find the probability that a truck drives less than 118 miles in a day.

This is the pvalue of Z when X = 118. So

`Z = (X - \mu)/(\sigma)`

`Z = (118 - 90)/(36)`

`Z = 0.78`

`Z = 0.78` has a pvalue of 0.7823

0.7823 = 78.23% probability that a truck drives less than 118 miles in a day.