Assume that the heights of adult Caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 women are randoml

Question

Assume that the heights of adult Caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 women are randomly​ selected, find the probability that they have a mean height greater than 63.0 inches. Round to four decimal places.

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Melody 2 hours 2021-10-14T13:31:45+00:00 1 Answer 0

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    2021-10-14T13:33:29+00:00

    Answer:

    0.9918 = 99.18% probability that they have a mean height greater than 63.0 inches.

    Step-by-step explanation:

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

    Normal probability distribution:

    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.

    Central limit theorem:

    The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

    In this problem, we have that:

    \mu = 63.6, \sigma = 2.5, n = 100, s = \frac{2.5}{\sqrt{100}} = 0.25

    Find the probability that they have a mean height greater than 63.0 inches.

    This is 1 subtracted by the pvalue of Z when X = 63. So

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

    By the Central Limit Theorem

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

    Z = \frac{63 - 63.6}{0.25}

    Z = -2.4

    Z = -2.4 has a pvalue of 0.0082

    1 – 0.0082 = 0.9918

    0.9918 = 99.18% probability that they have a mean height greater than 63.0 inches.

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