A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that a randomly

Question

A set of exam scores is normally distributed and has a mean of 80.2 and a standard deviation of 11. What is the probability that a randomly selected score will be greater than 63.7

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Delilah 6 days 2021-11-23T23:08:42+00:00 1 Answer 0 views 0

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    2021-11-23T23:09:50+00:00

    Answer:

    93.32% probability that a randomly selected score will be greater than 63.7.

    Step-by-step explanation:

    Problems of normally distributed samples are solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

    In this problem, we have that:

    \mu = 80.2, \sigma = 11

    What is the probability that a randomly selected score will be greater than 63.7.

    This is 1 subtracted by the pvalue of Z when X = 63.7. So

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

    Z = \frac{63.7 - 80.2}{11}

    Z = -1.5

    Z = -1.5 has a pvalue of 0.0668

    1 – 0.0668 = 0.9332

    93.32% probability that a randomly selected score will be greater than 63.7.

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