Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation

Question

Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph. A new speed limit will be initiated such that approximately 10% of vehicles will be over the speed limit. What is the new speed limit based on this criterion?

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Bella 1 week 2021-11-19T14:52:07+00:00 1 Answer 0 views 0

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    2021-11-19T14:53:54+00:00

    Answer:

    Based on this criterion, the new speed limit will be 81.24 mph.

    Step-by-step explanation:

    Problems of normally distributed samples are solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

    In this problem, we have that:

    \mu = 71, \sigma = 8

    A new speed limit will be initiated such that approximately 10% of vehicles will be over the speed limit. What is the new speed limit based on this criterion?

    X when Z has a pvalue of 1-0.1 = 0.9.

    So X when Z = 1.28.

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

    1.28 = \frac{X - 71}{8}

    X - 71 = 8*1.28

    X = 81.24

    Based on this criterion, the new speed limit will be 81.24 mph.

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