Suppose that you work for ChickfilA and your manager has asked you to look at the wait times in the DriveThru. Supposing that wait times are normally distrib
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1. User Answers joaobezerra
Answer:
0.0082 = 0.82% probability that a randomly selected person will wait more than 42 seconds.
Stepbystep explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the zscore formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X  \mu}{\sigma}[/tex]
The Zscore measures how many standard deviations the measure is from the mean. After finding the Zscore, we look at the zscore table and find the pvalue associated with this zscore. This pvalue 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.
Mean wait time of 30 seconds and a population standard deviation of 5 seconds.
This means that [tex]\mu = 30, \sigma = 5[/tex]
What is the probability that a randomly selected person will wait more than 42 seconds?
This is 1 subtracted by the pvalue of Z when X = 42. So
[tex]Z = \frac{X  \mu}{\sigma}[/tex]
[tex]Z = \frac{42  30}{5}[/tex]
[tex]Z = 2.4[/tex]
[tex]Z = 2.4[/tex] has a pvalue of 0.9918
1  0.9918 = 0.0082
0.0082 = 0.82% probability that a randomly selected person will wait more than 42 seconds.