To teach computer programming to employees, many firms use on the job training. A human resources administrator wishes to review the perform

Question

To teach computer programming to employees, many firms use on the job training. A human resources administrator wishes to review the performance of trainees on the final test of the training. The mean of the test scores is 72 with a standard deviation of 5. The distribution of test scores is approximately normal. Find the z-score for a trainee, given a score of 82.

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Tài Đức 2 months 2021-08-03T09:24:24+00:00 1 Answers 2 views 0

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    2021-08-03T09:25:34+00:00

    Answer:

    The z-score for the trainee is of 2.

    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.

    The mean of the test scores is 72 with a standard deviation of 5.

    This means that \mu = 72, \sigma = 5

    Find the z-score for a trainee, given a score of 82.

    This is Z when X = 82. So

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

    Z = \frac{82 - 72}{5}

    Z = 2

    The z-score for the trainee is of 2.

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