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Math Help - The central limit theorem

  1. #1
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    The central limit theorem

    Hello,
    I was working this problem, and was curious how I add the "10% of the forms" portion to the problem.
    Here is the problem:
    Among all the income-tax forms filed in a certain year, the mean tax paid was $2000 and the standard deviation was $500. In addition, for 10% of the forms, the tax paid was greater than $3000. A random samlpe of 625 tax forms is drawn.
    a. What is the probability that the average tax paid on the sample form is greater than $1980?

    So I did P(X > 1980)
    X = N(2000,400)
    Z = (1980 - 2000)/sqrt(400) = -1
    Z-table gives me a value of 0.1587 for -1. So 1 - 0.1587 = 0.8413 probability. But how do you factor in the 10% thing?

    b. What is the probability that more than 60 of the sampled forms have a tax of greater than $3000?
    Was not sure how to go about this one.

    Any help is greatly appreciated, thanks!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Mahonroy View Post
    Hello,
    I was working this problem, and was curious how I add the "10% of the forms" portion to the problem.
    Here is the problem:
    Among all the income-tax forms filed in a certain year, the mean tax paid was $2000 and the standard deviation was $500. In addition, for 10% of the forms, the tax paid was greater than $3000. A random samlpe of 625 tax forms is drawn.
    a. What is the probability that the average tax paid on the sample form is greater than $1980?

    So I did P(X > 1980)
    X = N(2000,400)
    Z = (1980 - 2000)/sqrt(400) = -1
    Z-table gives me a value of 0.1587 for -1. So 1 - 0.1587 = 0.8413 probability. But how do you factor in the 10% thing?
    You don't you use that in the next part,

    Also you have not included enough text, you should tell us in words what you are doing.


    b. What is the probability that more than 60 of the sampled forms have a tax of greater than $3000?
    Was not sure how to go about this one.
    Now we know that 10\% of forms have a tax greater than or equal \$3000, so in a sample of 625 the actual number of such forms has a binomial distribution B(625,0.1), which is approximately N(62.5, 625\times 0.1 \times 0.9).

    CB

    CB
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  3. #3
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    Thanks for the reply! I was trying to put the values into a Z-formula to get a z-value I can use on the look up table, but was confused as to how to go from N(62.5,56.25) and Bin(625,01) to a z=?
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