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Math Help - percentiles... UGH

  1. #1
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    percentiles... UGH

    A student's score on an SAT was 700 on a standardized test known to have a mean of 500 and a standard deviation of 110.

    a) what percentile was the student in?
    b) what score must a student get to be in the 73rd percentile?
    c) another student had a z score equivalent of -0.55. what was her score on the test?
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  2. #2
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    A student's score on an SAT was 700 on a standardized test known to have a mean of 500 and a standard deviation of 110.

    a) what percentile was the student in?
    b) what score must a student get to be in the 73rd percentile?
    c) another student had a z score equivalent of -0.55. what was her score on the test?
    hey there,

    i would resort to using normal distribution curve.
    Normal distribution - Wikipedia, the free encyclopedia

    percentile:
    "A percentile is the value of a variable below which a certain percent of observations fall. So the 20th percentile is the value (or score) below which 20 percent of the observations may be found. "

    You were given the mean and standard deviation and required to find percentile.

    a) using probability on a normal curve with the information given to you,

    percentile the student is in = probability (score < 700) x 100%

    standardizing it using:

    <br />
 z = \frac {score - mean}{Standard \text{ } Deviation}<br />
    <br />
z = \frac{700-500}{110}<br />
    <br />
  z= 1.8182<br />

    Hence,
    probability (score < 700) = probability ( Z < 1.8182 )

    using the table for normal distribution, P(Z<1.8182) = 0.9654818
    Z Table

    For percentile, 0.9654818 x 100% = 96.54818%



    b) For part b) instead you will need to find the value of Z from the Z-table given the percentile.

    First we convert the percentile to probability,

    P ( Z< z) = 73/100 = 0.73

    From the Z-table, 0.73 corresponds to z = 0.615.

    Using the formula;
    <br />
 z = \frac {score - mean}{Standard \text{ } Deviation}<br />
    <br />
0.615 = \frac {score - 500}{110}<br />
    <br />
score = (0.615 \times 110) + 500<br />
    <br />
score = 567.65<br />
    This is the minimum score a person must obtain to be above the 73rd percentile.

    hence score \geqslant 567.65


    c)
    Given z = -0.55

    using the formula;

    <br />
 z = \frac {score - mean}{Standard \text{ } Deviation}<br />
    <br />
-0.55 = \frac {score - 500}{110}<br />
    <br />
score = (-0.55 \times 110) + 500<br />
    <br />
score = 439.5<br />

    Hope this helps.
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