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Sagot :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 90, \sigma = 36[/tex]
Find the probability that a truck drives less than 118 miles in a day.
This is the pvalue of Z when X = 118. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{118 - 90}{36}[/tex]
[tex]Z = 0.78[/tex]
[tex]Z = 0.78[/tex] has a pvalue of 0.7823
0.7823 = 78.23% probability that a truck drives less than 118 miles in a day.
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