The average height of a current NBA player is 79 inches with a standard deviation of 3.4 inches. A random sample of 35 current NBA players i

Question

The average height of a current NBA player is 79 inches with a standard deviation of 3.4 inches. A random sample of 35 current NBA players is taken. What is the probability that the mean height of the 35 NBA players will be more than 80 inches?

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Thông Đạt 3 months 2021-07-26T02:17:30+00:00 1 Answers 8 views 0

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    2021-07-26T02:18:39+00:00

    Answer:

    0.0409 = 4.09% probability that the mean height of the 35 NBA players will be more than 80 inches.

    Step-by-step explanation:

    To solve this question, we need to understand the normal probability distribution and the central limit theorem.

    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.

    Central Limit Theorem

    The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

    For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

    The average height of a current NBA player is 79 inches with a standard deviation of 3.4 inches.

    This means that \mu = 79, \sigma = 3.4

    A random sample of 35 current NBA players is taken.

    This means that n = 35, s = \frac{3.4}{\sqrt{35}}

    What is the probability that the mean height of the 35 NBA players will be more than 80 inches?

    This is 1 subtracted by the p-value of Z when X = 80. So

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

    By the Central Limit Theorem

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

    Z = \frac{80 - 79}{\frac{3.4}{\sqrt{35}}}

    Z = 1.74

    Z = 1.74 has a p-value of 0.9591

    1 – 0.9591 = 0.0409

    0.0409 = 4.09% probability that the mean height of the 35 NBA players will be more than 80 inches.

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