Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 247 feet and a standard devi

Question

Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 247 feet and a standard deviation of 41 feet. Use your graphing calculator to answer the following questions. Write your answers in percent form. Round your answers to the nearest tenth of a percent. a) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 216 feet

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Neala 1 year 2021-09-04T06:21:18+00:00 1 Answers 6 views 0

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    2021-09-04T06:22:30+00:00

    Answer:

    77.5% probability that this ball traveled fewer than 216 feet.

    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 of 247 feet and a standard deviation of 41 feet.

    This means that [tex]\mu = 247, \sigma = 41[/tex]

    What is the probability that this ball traveled fewer than 216 feet?

    The probability as a decimal is the p-value of Z when X = 216. So

    [tex]Z = \frac{X – \mu}{\sigma}[/tex]

    [tex]Z = \frac{216 – 247}{41}[/tex]

    [tex]Z = 0.756[/tex]

    [tex]Z = 0.756[/tex] has a p-value of 0.775

    0.775*100% = 77.5%

    77.5% probability that this ball traveled fewer than 216 feet.

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