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.

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 and standard deviation , the zscore of a measure X is given by:

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:

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

has a pvalue of 0.8599

X = 43

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.

maya’s expression will generate the wrong sum because she violated the commutative property.

step-by-step explanation:

ok so what's the question?

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