# The central limit theorem

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• October 9th 2009, 07:47 AM
Mahonroy
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! • October 9th 2009, 10:36 AM CaptainBlack Quote: Originally Posted by Mahonroy 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. Quote: 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
• October 11th 2009, 06:51 PM
Mahonroy
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=?