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Math Help - Comparing two different methods

  1. #1
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    Comparing two different methods

    Two different means of assessing the students have been suggested. First using a normal distribution and secondly using fixed percentages for each rating.

    Using the normal distribution
    A - > 2a
    B - > a and < 2a
    C - > -a and < a
    D - > -2a and < -a
    E - < -2a
    Comparing two different methods-bellcurve.gif

    Using the fixed percentage
    A - > 80%
    B - > 60% and < 80%
    C - > 40% and < 60%
    D - > 20% and < 40%
    E - < 20%

    The task is to compare and contrast the two different assessment procedures. You should consider the outcome if a teacher sets an easier test or a harder test.
    You many assume that the standard deviation remains the same for each of the tests.

    I have no idea where to start. Any help would be greatly appreciated.

    NOTE: I have used "a" because i could not find sigma.

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  2. #2
    MHF Contributor matheagle's Avatar
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    You should find the probability of each of those intervals using the normal distribution.

    Like P(Z>2)

    P(1<Z<2)

    P(-1<Z<1)

    P(-2<Z<-1)=P(1<Z<2)

    P(Z<-2)=P(Z>2)
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  3. #3
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    P(Z>2) = 0.4772 therefore P(Z<-2) = 0.4772
    P(1<Z<2) = P(Z>2) - P(Z>1) = 0.4772 -0.3413 = 0.1359 therefore P(-2<Z<-1) = 0.1359
    P(-1<Z<1) = P(Z<1) + P(Z>1) = 0.3413 + 0.3413 = 0.6826
    Last edited by mortalcyrax; August 10th 2010 at 09:56 PM. Reason: Formatting
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  4. #4
    MHF Contributor matheagle's Avatar
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    P(Z>2) is not close to .5, it's about .02
    without looking at the tables I'm sure it's .5-.4772.
    Similarly, your third line is off too.
    Your second line is wrong too.
    P(Z>1) is larger than P(Z>2), which means you have a negative probabilty.
    Think about what you wrote and look at the intervals.
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  5. #5
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    P(Z>2) = 0.5 - 0.4772 = 0.0228
    P(Z>1) = 0.5 - 0.3413 = 0.1587
    P(1<Z<2) = P(Z>1) - P(Z>2) = 0.1359
    P(-1<Z<1) = P(Z>-1) - P(Z>1) = (0.5 + 0.3413) - (0.5 - 0.3413) = 0.6826

    This means P(Z>2) = 2.2% A
    P(1<Z<2) = 13.6% B
    P(-1<Z<1) = 68.3% C
    P(-2<Z<-1) = 13.6% D
    P(Z<-2) = 2.2% E

    meaning majority of students would receive a "C" using normal distribution
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