it is known that the population proton of utha residnet that are members of the church of jesus christ 0l6 suppose a random sample of 46 sel

Question

it is known that the population proton of utha residnet that are members of the church of jesus christ 0l6 suppose a random sample of 46 selceted and prioon of the sample that belongs to the churh is calcutated what is the problaity of obtaining a sample priton less than 0;50 g

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Thiên Di 4 years 2021-07-28T11:41:57+00:00 1 Answers 13 views 0

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  1. thaiduong
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    2021-07-28T11:43:50+00:00
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    Answer:

    0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

    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.

    For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

    Proportion of 0.6

    This means that p = 0.6

    Sample of 46

    This means that n = 46

    Mean and standard deviation:

    \mu = p = 0.6

    s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6*0.4}{46}} = 0.0722

    Probability of obtaining a sample proportion less than 0.5.

    p-value of Z when X = 0.5. So

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

    By the Central Limit Theorem

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

    Z = \frac{0.5 - 0.6}{0.0722}

    Z = -1.38

    Z = -1.38 has a p-value of 0.0838

    0.0838 = 8.38% probability of obtaining a sample proportion less than 0.5.

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