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November 22, 2014

November 22, 2014

Posted by **Edward** on Sunday, March 1, 2009 at 11:50am.

- Statistics -
**drwls**, Sunday, March 1, 2009 at 12:00pmThis question was posted twice in a row. I may look at it later, if someone else doesn't do it first.

- Statistics -
**Edward**, Sunday, March 1, 2009 at 12:07pmYes please. I think I hit the refresh button while posting. Sorry. Would love to work with you on getting some hints on getting this done.

- Statistics -
**MathGuru**, Sunday, March 1, 2009 at 3:00pmIf I'm interpreting this problem correctly and since the suggestion is to approximate the distribution by using the Gaussian (or normal) distribution, I would first find the z-score equated to 1% using a z-table. That would be -2.33 (below the mean of the distribution). Next, you will need to find mean and standard deviation. Once you have those values, substitute into the z-score formula, then solve for x.

Mean = np = (2000)(.5) = ?

Standard deviation = √npq = √(2000)(.5)(.5) = ?

Note: I am using .5 since no value for p is stated. Also, q = 1 - p.

Formula for z-score:

z = (x - mean)/sd

Substitute the values for mean and standard deviation when you finish the above calculations:

-2.33 = (x - mean)/sd

Now solve for x.

I hope this will help and is what the problem was asking.

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