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Math Help - Mean & Standard Deviation

  1. #1
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    Mean & Standard Deviation

    In a survey of twelve Savannah State Business School graduates, the mean starting salary was $93,000, with a standard deviation of $17,000.

    The 95% confidence interval for the average starting salary among all Savannah State graduates is?
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  2. #2
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    Mean & Standard deviation, confidence interval

    I think I have solved it and it is as follows:

    R =93000 + (1.96) 17000/3.4641061513775
    R = 93000 + (1.96) (4907)
    R = 93000 + 9618
    R = 102,618


    R = 93000 - (1.96) 17000/3.4641061513775
    R = 93000 - (1.96) (4907)
    R = 93000 - 9618
    R = 83,382

    So the range of salaries would be $83,382 ; $102,618
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  3. #3
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    I like it. Do you still doubt?
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