6.9.) Major League Baseball teams have become concerned about the length of games. During a recent season, games averaged 2 hours and 52 min

Question

6.9.) Major League Baseball teams have become concerned about the length of games. During a recent season, games averaged 2 hours and 52 minutes (172 minutes) to complete. Assume the length of games follows the normal distribution with a standard deviation of 16 minutes. a.) What is the probability that a randomly selected game will be completed in 1.) 200 minutes or less

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Kennedy 1 month 2021-09-16T14:59:53+00:00 2 Answers 0

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    0
    2021-09-16T15:01:17+00:00

    Answer:

    95.99% probability that a randomly selected game will be completed in 200 minutes or less.

    Step-by-step explanation:

    Problems of normally distributed samples can be 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 = 172, \sigma = 16

    What is the probability that a randomly selected game will be completed in 200 minutes or less

    This is the pvalue of Z when X = 200. So

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

    Z = \frac{200 - 172}{16}

    Z = 1.75

    Z = 1.75 has a pvalue of 0.9599

    95.99% probability that a randomly selected game will be completed in 200 minutes or less.

    0
    2021-09-16T15:01:30+00:00

    Answer:

    Probability that a randomly selected game will be completed in 200 minutes or less is 0.95994.

    Step-by-step explanation:

    We are given that during a recent season, games averaged 2 hours and 52 minutes (172 minutes) to complete. Assume the length of games follows the normal distribution with a standard deviation of 16 minutes.

    Let X = length of games

    So, X ~ N(\mu = 172, \sigma^{2} = 16^{2})

    The z score probability distribution is given by;

                   Z = \frac{X-\mu}{\sigma} ~ N(0,1)

    where, \mu = average time = 172 minutes

                \sigma = standard deviation = 16 minutes

    So, Probability that a randomly selected game will be completed in 200 minutes or less is given by = P(X \leq 200 min)

         P(X \leq 200) = P( \frac{X-\mu}{\sigma} \leq \frac{200-172}{16} ) = P(Z \leq 1.75)

                                                            = 0.95994

    Hence, Probability that a randomly selected game will be completed in 200 minutes or less is 0.95994.

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