Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8% What percentage of students receive between a 70% and 94%

Question

Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8% What percentage of students receive between a 70% and 94% enter the value of the percentage without the percent sign

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Linh Đan 5 months 2021-08-27T13:10:12+00:00 1 Answers 4 views 0

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    2021-08-27T13:11:15+00:00

    Answer:

    49.87

    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.

    Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8%.

    This means that \mu = 70, \sigma = 8

    What percentage of students receive between a 70% and 94%

    The proportion is the p-value of Z when X = 94 subtracted by the p-value of Z when X = 70. So

    X = 94

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

    Z = \frac{94 - 70}{8}

    Z = 3

    Z = 3 has a p-value of 0.9987.

    X = 70

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

    Z = \frac{70 - 70}{8}

    Z = 0

    Z = 0 has a p-value of 0.5.

    0.9987 – 0.5 = 0.4987.

    0.4987*100% = 49.87%.

    So the percentage is 49.87%, and the answer, without the percent sign, is 49.87.

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