A sample of n = 4 scores is selected from a normal population with μ = 30 and σ = 8. The probability of obtaining a sample mean greater than

Question

A sample of n = 4 scores is selected from a normal population with μ = 30 and σ = 8. The probability of obtaining a sample mean greater than 34 is equal to the probability of obtaining a z-score greater than z = 2.00.

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Philomena 4 years 2021-09-02T18:56:07+00:00 1 Answers 20 views 0

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  1. thnahdat
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    2021-09-02T18:58:02+00:00
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    Answer:

    False

    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.

    Population:

    \mu = 30, \sigma = 8

    Sample of 4

    This means that n = 4, s = \frac{8}{\sqrt{4}} = 4

    Probability of obtaining a sample mean greater than 34:

    This is 1 subtracted by the p-value of Z when X = 34. So

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

    By the Central Limit Theorem

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

    Z = \frac{34 - 30}{4}

    Z = 1

    Thus, the probability of obtaining a sample mean greater than 34 is equal to the probability of obtaining a z-score greater than z = 1.00, and the statement in this question is false.

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