Question

Trucks in a delivery fleet travel a mean of 100 miles per day with a standard deviation of 38 miles per day. The mileage per day is distributed normally. Find the probability that a truck drives between 43 and 141 miles in a day. Round your answer to four decimal places.

Answer 1

Answer:

0.7931 = 79.31% probability that a truck drives between 43 and 141 miles in a day.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 100, \sigma = 38$

Find the probability that a truck drives between 43 and 141 miles in a day.

This is the pvalue of Z when X = 141 subtracted by the pvalue of Z when X = 43. So

X = 141

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{141 - 100}{38}$

$Z = 1.08$

$Z = 1.08$ has a pvalue of 0.8599

X = 43

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{43 - 100}{38}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.8599 - 0.0668 = 0.7931

0.7931 = 79.31% probability that a truck drives between 43 and 141 miles in a day.