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    cynajae's Avatar
    cynajae Posts: 1, Reputation: 1
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    #1

    Aug 24, 2007, 07:41 AM
    mathematics, statistics
    1. Assume the average annual salary for a worker in the United States is $27500 and that the annual salaries for Americans are normally distributed with a standard deviation equal to $6250. Find a. the percentage of americans working below $18000
    b. the percentage of americans working below $40000


    2. Find the value of z such that 40% of the distribution lies between it and the mean.


    :confused:
    nicespringgirl's Avatar
    nicespringgirl Posts: 1,237, Reputation: 187
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    #2

    Aug 24, 2007, 07:55 AM
    Do you have minitab or SAS?
    Excel will work
    How far have you gone through this problem?
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
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    #3

    Aug 24, 2007, 08:10 AM
    Or just plain ol' math? :P
    nicespringgirl's Avatar
    nicespringgirl Posts: 1,237, Reputation: 187
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    #4

    Aug 24, 2007, 08:17 AM
    Quote Originally Posted by Capuchin
    or just plain ol' math? :P
    That's what we asians call it... :p ;)
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #5

    Aug 24, 2007, 10:19 AM
    for part a:

    Use the formula and then look up the probability which corresponds to the z-score you just found. I will show you this one and you do the other. Okey-doke?





    Look up -1.52 in the z-table and we find 0.064...

    A little over 6% have have a salary below $18000.


    If you have a fancy-schmancy calculator or Excel you can find the z-score without the table.

    If you want to show off, here's the formula the values are derived from:
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