SAT Writing scores are normally distributed with a mean of 491 and a standard deviation of 113.A university plans to send letters of recogni

Question

SAT Writing scores are normally distributed with a mean of 491 and a standard deviation of 113.A university plans to send letters of recognition to students whose scores are in the top 8%. What is the minimum score required for a letter of recognition?

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Iris 2 months 2021-09-29T02:26:45+00:00 1 Answer 0 views 0

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    2021-09-29T02:28:02+00:00

    Answer:

    z=1.405<\frac{a-491}{113}

    And if we solve for a we got

    a=491 +1.405*113=649.765

    So the value of height that separates the bottom 92% of data from the top 8% is 649.765.  

    Step-by-step explanation:

    Previous concepts

    Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

    The Z-score is “a numerical measurement used in statistics of a value’s relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean”.  

    Solution to the problem

    Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

    X \sim N(491,113)  

    Where \mu=491 and \sigma=113

    For this part we want to find a value a, such that we satisfy this condition:

    P(X>a)=0.08   (a)

    P(X<a)=0.92   (b)

    Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

    As we can see on the figure attached the z value that satisfy the condition with 0.92 of the area on the left and 0.08 of the area on the right it’s z=1.405. On this case P(Z<1.405)=0.92 and P(z>0.92)=0.08

    If we use condition (b) from previous we have this:

    P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.92  

    P(z<\frac{a-\mu}{\sigma})=0.92

    But we know which value of z satisfy the previous equation so then we can do this:

    z=1.405<\frac{a-491}{113}

    And if we solve for a we got

    a=491 +1.405*113=649.765

    So the value of height that separates the bottom 92% of data from the top 8% is 649.765.  

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