“Scores on an English test are normally distributed with a mean of 37.4 and a standard deviation of 7.9. Find the score that separates the t

Question

“Scores on an English test are normally distributed with a mean of 37.4 and a standard deviation of 7.9. Find the score that separates the top 59% from the bottom 41%”

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Ella 3 hours 2021-10-14T04:58:37+00:00 2 Answers 0

Answers ( )

    0
    2021-10-14T05:00:14+00:00

    Answer:

    Score that separates the top 59% from the bottom 41% is 35.6

    Step-by-step explanation:

    We are given that Scores on an English test are normally distributed with a mean of 37.4 and a standard deviation of 7.9, i.e.; \mu = 37.4 and \sigma = 7.9

    Now, the z score probability distribution is given by;

                 Z = \frac{X - \mu}{\sigma} ~ N(0,1)

    The bottom 41% area is given by the critical z value of -0.2278 (from z% table)

    So, P(Z < \frac{X-37.4}{7.9} ) = 0.41

         which means  \frac{X-37.4}{7.9} = -0.2278

                               X – 37.4 = -0.2278 * 7.9

                               X = 37.4 – 1.79962 = 35.6

    Therefore, score of 35.6 separates the top 59% from the bottom 41%.

    0
    2021-10-14T05:00:28+00:00

    Answer:

    The score that separates the top 59% from the bottom 41% is 35.6225

    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 = 37.4, \sigma = 7.9

    Find the score that separates the top 59% from the bottom 41%”

    This is the value of X when Z has a pvalue of 0.41. So it is X when Z = -0.225.

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

    -0.225 = \frac{X - 37.4}{7.9}

    X - 37.4 = -0.225*7.9

    X = 35.6225

    The score that separates the top 59% from the bottom 41% is 35.6225

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