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Thread: Standard deviation word problem

  1. #1
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    Standard deviation word problem

    Household income does not tend to follow a normal distribution in a particular state, yet average income is approximately $45,000/year in this state, with a standard deviation of about $9000. At least what percentage of household incomes in this state is likely to be between $9,000 to $81,000/year

    Not sure what formula to use?

    My husband tried to help me and he came up with 99%. Anyone?
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  2. #2
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    (1) draw a graph with 45,000 at the 0

    (2) put 81,000 and 9,000 on the graph with corresponding z scores
    z=\frac{x-\bar x}{\sigma}

    (3) integrate \frac{1}{\sqrt{2\pi}}*\int_{low \ z \ score}^{high \ z \ score}e^{\frac{-x^2}{2}}dx
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  3. #3
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    I dont understand? My book is not showing any way to work this out
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  4. #4
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    In order to find the area under the normal probability density curve, you need the lower and upper bounds traditionally called z-scores.
    A z-score is obtained by subtracting the given x value minus the mean and then dividing by the standard deviation.
    Once you have obtained you lower and upper bounds, you can consult the the t-distribution for samples over 120 but your z-scores are so far out they won't be in the table; therefore, you need to integrate.
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  5. #5
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    \bar x=45,000 \ \mbox{given}

    \sigma=9,000 \ \mbox{given}

    z_1=\frac{x-\bar x}{\sigma}=\frac{9,000-45,000}{9,000}=-4

    You can find z_2 now.

    \frac{1}{\sqrt{2\pi}}*\int_{-4}^{z_2}e^{\frac{-x^2}{2}} \mbox{dx}
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  6. #6
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    I am trying to figure out the percentage?
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  7. #7
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    The percentage = the area under the curve
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  8. #8
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    When you eventually solve this integral, you would receive a decimal.

    \frac{1}{\sqrt{2\pi}}*\int_{-4}^{z_2}e^{\frac{-x^2}{2}} \mbox{dx}

    =.999936657516*100% =99.9936657516%
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  9. #9
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    Thanks I came up with 99% for my answer.
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