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Sagot :
Answer:
0.1587 = 15.87% probability that a person will wait for more than 9 minutes.
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.
The mean waiting time is 6 minutes and the variance of the waiting time is 9.
This means that [tex]\mu = 6, \sigma = \sqrt{9} = 3[/tex]
Find the probability that a person will wait for more than 9 minutes.
This is 1 subtracted by the p-value of Z when X = 9. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{9 - 6}{3}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.8413.
1 - 0.8413 = 0.1587
0.1587 = 15.87% probability that a person will wait for more than 9 minutes.
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