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    Determining the Sampling Distribution of a Proportion

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    According to a well-known newspaper journal, 58.3% of the 222,900 households in a particular county get the Sunday edition of the local newspaper. Suppose that random samples of 200 households are selected.

    In what proportion of the samples will between 55% and 60% of the households get the Sunday edition of the newspaper?

    P(0.55 < Ps <0.60) = P(-0.95 < Z < 0.49) = 0.5168

    0.49 in the Cumulative Standardized Normal Distribution Table (= 0.6879)

    What are the specific steps involved in converting P(0.55 < Ps <0.60) to P(-0.95 < Z < 0.49)?

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    https://brainmass.com/statistics/normal-distribution/determining-sampling-distribution-proportion-4465

    Solution Preview

    We should change the normal distribution into Standardized Normal Distribution.
    <br>If the normal distribution is X ~ (u, s)
    <br>Where u is the mean of X and s is its standard ...

    Solution Summary

    The solution addresses the fact that according to a well-known newspaper journal, 58.3% of the 222,900 households in a particular county get the Sunday edition of the local newspaper. Suppose that random samples of 200 households are selected.

    In what proportion of the samples will between 55% and 60% of the households get the Sunday edition of the newspaper?

    P(0.55 < Ps <0.60) = P(-0.95 < Z < 0.49) = 0.5168

    0.49 in the Cumulative Standardized Normal Distribution Table (= 0.6879)

    What are the specific steps involved in converting P(0.55 < Ps <0.60) to P(-0.95 < Z < 0.49)?

    $2.49

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