According to the College Board website, the scores on the math part of the SAT (SAT-M) in a recent year had a mean of 507 and standard devia

Question

According to the College Board website, the scores on the math part of the SAT (SAT-M) in a recent year had a mean of 507 and standard deviation of 111. Assume that SAT-M scores have a normal probability distribution. One of the criteria for admission to a certain engineering school is an SAT-M score in the 98th percentile. This means the score is in the top 2% of scores. Find the actual SAT-M score marking the 98th percentile? Show your work. Interpret your result. Note: You will need the Inverse Normal Distribution Calculator below.

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Cora 3 months 2022-02-19T02:50:27+00:00 1 Answer 0 views 0

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    2022-02-19T02:52:16+00:00

    Answer:

    The actual SAT-M score marking the 98th percentile is 735.

    Step-by-step explanation:

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

    In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

    [tex]Z = \frac{X – \mu}{\sigma}[/tex]

    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:

    [tex]\mu = 507, \sigma = 111[/tex]

    Find the actual SAT-M score marking the 98th percentile

    This is X when Z has a pvalue of 0.98. So it is X when Z = 2.054. So

    [tex]Z = \frac{X – \mu}{\sigma}[/tex]

    [tex]2.054 = \frac{X – 507}{111}[/tex]

    [tex]X – 507 = 2.054*111[/tex]

    [tex]X = 735[/tex]

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45:7+7-4:2-5:5*4+35:2 =? ( )