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Topic: More of a stats/calc question, but i couldn't find a better spot
Replies: 1   Last Post: Dec 6, 2008 1:26 PM

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 Ross Posts: 2 Registered: 12/5/08
More of a stats/calc question, but i couldn't find a better spot
Posted: Dec 5, 2008 8:19 PM

Ok, so in AP Stats we have been using "Table A" to find the probability that a value is "z" standard deviations away from the mean. That is all well and good. However, I was curious as to how "Table A" was generated.

After some search I have found a function on the TI graphing calculators (normalcdf(a,b) that gives the values, where a is -inf when finding prob. for it being less than "z" std dev. away (when counting -2 std dev as less than +1 std dev)

I wanted to create a program on my calculator that would generate these values when the "z" was entered. As the calc had a function in place, this was extremely easy. Additionally I want to be able to enter a prob. and have the calculator determine the "z" value the corresponds.

Unfortunately the two equations I have found that = normalcdf(a,b) are extremely difficult to solve backwards for z (which will be represented by x in the equations)

normalcdf(a,b) will be replaced with P in the equations as well

ok here's a better view of the equations (there should be a closing parenthesis after the (2^k part and before the (k!).

http://i224.photobucket.com/albums/dd255/Delta003/normaldistrib.jpg

P = Int e^(-(x^2)/2) dx (note: integral is from a to b)

and

P = .5 + (1/sqrt(2pi)) * Sigma[((-1)^k * x^(2k+1))/((2k+1)(2^k)(k!))] (note: summation is from k=0 to inf)

I am unsure how to "un-do" the summation when I have P and need to find x.

Message was edited by: Ross

Date Subject Author
12/5/08 Ross
12/6/08 Ross