A population has a mean of 180 and a standard deviation of 24. A sample of 100 observations will be taken. The probability that the mean fro

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A population has a mean of 180 and a standard deviation of 24. A sample of 100 observations will be taken. The probability that the mean from that sample will be between 183 and 186 is _____.

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Thu Hương 4 years 2021-08-22T00:27:30+00:00 1 Answers 165 views 0

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  1. dangkhoa
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    2021-08-22T00:29:01+00:00
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    Answer:

    The probability that the mean from that sample will be between 183 and 186 is 0.0994 = 9.94%.

    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 estabilishes 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.

    A population has a mean of 180 and a standard deviation of 24.

    This means that \mu = 180, \sigma = 24

    A sample of 100 observations will be taken.

    This means that n = 100, s = \frac{24}{\sqrt{100}} = 2.4

    The probability that the mean from that sample will be between 183 and 186 is:

    This is the pvalue of Z when X = 186 subtracted by the pvalue of Z when X = 183. So

    X = 186

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

    By the Central Limit Theorem

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

    Z = \frac{186 - 180}{2.4}

    Z = 2.5

    Z = 2.5 has a pvalue of 0.9938

    X = 183

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

    Z = \frac{183 - 180}{2.4}

    Z = 1.25

    Z = 1.25 has a pvalue of 0.8944

    0.9938 – 0.8944 = 0.0994

    The probability that the mean from that sample will be between 183 and 186 is 0.0994 = 9.94%.

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