While preparing for their comeback tour, The Flaming Rogers find that the average time it takes their sound tech to set up for a show is 56.

Question

While preparing for their comeback tour, The Flaming Rogers find that the average time it takes their sound tech to set up for a show is 56.1 minutes, with a standard deviation of 6.4 minutes. If the band manager decides to include only the fastest 23% of sound techs on the tour, what should the cutoff time be for concert setup? Assume the times are normally distributed.

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Thu Nguyệt 3 years 2021-08-10T09:50:11+00:00 1 Answers 0 views 0

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  1. thanhha
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    2021-08-10T09:52:01+00:00
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    Answer:

    The cutoff time be for concert setup should be of 51.4 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 \mu and standard deviation \sigma, the z-score of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    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.

    Average time it takes their sound tech to set up for a show is 56.1 minutes, with a standard deviation of 6.4 minutes.

    This means that \mu = 56.1, \sigma = 6.4

    If the band manager decides to include only the fastest 23% of sound techs on the tour, what should the cutoff time be for concert setup?

    The cutoff time would be the 23rd percentile of times, which is X when Z has a p-value of 0.23, so X when Z = -0.74.

    Z = \frac{X - \mu}{\sigma}

    -0.74 = \frac{X - 56.1}{6.4}

    X - 56.1 = -0.74*6.4

    X = 51.4

    The cutoff time be for concert setup should be of 51.4 minutes.

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