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.

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

Step-by-step explanation:

Normal Probability Distribution:

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

In a set with mean and standard deviation , the z-score 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 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

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

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

has a pvalue of 0.7823

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

there is no

42%

step-by-step explanation:

we know,

total people = 60

total males = 25

percentage of males

= 25/60 × 100

= 25/6 × 10

= 25/3 × 5

= 125 / 3

= 41.

rounding of = 42%

happy to

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