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Math Help - Finding xbar and standard d...

  1. #1
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    Finding xbar and standard d...

    population of 18000.
    10% are assigned letter grade of A (given only to those 475+)
    80% are letter grade of B (ranges from 25-475)
    10% are letter grade of C (25 or less)

    Assume normal distribution...
    So, I have taken 475-25, div by 2 to get the midpoint of the distribution, which is 225. My text claims the answer is 250 though... I just assumed that with 80% in between 25-475.. the exact mid-point would be halfway between them. **Turns out if its normally distributed, the values are between 0-500.. So it xbar is 250.. but I am still stuck on the SD...

    for standard D... I looked at my Z table and took 1.0 value, which is .8413... Subtracted 50% from it(to the left of the mean) leaving with 34.13% of observations between Xbar of 250 and SD... 34.13% of 250 observations is 85.325, which when applied to both sides is 170.65.. book says answer is 175.8


    Definitely need help and clarity! Thanks!!
    Last edited by Aitrus; September 26th 2008 at 04:14 PM.
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  2. #2
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    I just assumed that with 80% in between 25-475.. the exact mid-point would be halfway between them.
    The midpoint is halfway between them, but this is because the amount more than 475 is the same as the amount less than 25. To get the midpoint, you should add not subtract: \frac{475+25}{2} = 250

    for standard D... I looked at my Z table and took 1.0 value, which is .8413... Subtracted 50% from it(to the left of the mean) leaving with 34.13% of observations between Xbar of 250 and SD... 34.13% of 250 observations is 85.325, which when applied to both sides is 170.65.. book says answer is 175.8
    I'm not sure what you are doing here. To find the standard deviation, notice that 90% are less than 475. Look at the Z-table to find out how many standard deviations from the mean 475 is. Then divide 475-250 by this value to get the standard deviation.
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