Mathematics

Question

Suppose that you work for Chick-fil-A and your manager has asked you to look at the wait times in the Drive-Thru. Supposing that wait times are normally distributed with a mean wait time of 30 seconds and a population standard deviation of 5 seconds. What is the probability that a randomly selected person will wait more than 42 seconds

1 Answer

  • Answer:

    0.0082 = 0.82% probability that a randomly selected person will wait more than 42 seconds.

    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.

    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.

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