In 2012, New York Yankees baseball players earned an average salary of $6,186,321, with a standard deviation of $7,938,987. Assuming that th

Question

In 2012, New York Yankees baseball players earned an average salary of $6,186,321, with a standard deviation of $7,938,987. Assuming that these data are normally distributed, what was the salary of a player in the 53rd percentile?

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Abigail 3 hours 2021-11-24T00:37:30+00:00 1 Answer 0 views 0

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    2021-11-24T00:39:00+00:00

    Answer:

    The salary of a player in the 53rd percentile was $6,781,745.

    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 = 6,186,321, \sigma = 7,938,987

    Assuming that these data are normally distributed, what was the salary of a player in the 53rd percentile?

    This is the value of X when Z has a pvalue of 0.53. So X when Z = 0.075.

    So

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

    0.075 = \frac{X - 6,186,321}{7,938,987}

    X - 6,186,321 = 0.075*7,938,987

    X = 6,781,745

    The salary of a player in the 53rd percentile was $6,781,745.

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