An art history professor assigns letter grades on a test according to the following scheme. A: Top 13% of scores B: Scores below the top

Question

An art history professor assigns letter grades on a test according to the following scheme. A: Top 13% of scores B: Scores below the top 13% and above the bottom 56% C: Scores below the top 44% and above the bottom 21% D: Scores below the top 79% and above the bottom 9% F: Bottom 9% of scores Scores on the test are normally distributed with a mean of 79.7 and a standard deviation of 8.4. Find the numerical limits for a B grade. Round your answers to the nearest whole number, if necessary.

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Thành Công 4 years 2021-08-07T18:17:40+00:00 1 Answers 35 views 0

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  1. Maris
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    2021-08-07T18:19:12+00:00
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    Answer:

    The numerical limits for a B grade are 81 and 89, that is, a score between 81 and 89 gets a B grade.

    Step-by-step explanation:

    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.

    Scores on the test are normally distributed with a mean of 79.7 and a standard deviation of 8.4.

    This means that \mu = 79.7, \sigma = 8.4

    B: Scores below the top 13% and above the bottom 56%

    So between the 56th percentile and the 100 – 13 = 87th percentile.

    56th percentile:

    X when Z has a p-value of 0.56, so X when Z = 0.15. Then

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

    0.15 = \frac{X - 79.7}{8.4}

    X - 79.7 = 0.15*8.4

    X = 81

    87th percentile:

    X when Z has a p-value of 0.87, so X when Z = 1.13.

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

    1.13 = \frac{X - 79.7}{8.4}

    X - 79.7 = 1.13*8.4

    X = 89

    The numerical limits for a B grade are 81 and 89, that is, a score between 81 and 89 gets a B grade.

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