you just draw the curve, right in your mean and segment it up into the various ranges you need to find and somewhere in your course they should have given you the forumla to find it (cant remember of top of my head) but its under either sampling distributions or estimation.
you can use excel formulas to solve these like normdist I believe. [/quote]
Yeah, but, but, but then you wouldn’t have to use numeric methods to determine the definite integral of an expression whose indefinite integral does not exists. Which way sounds more fun to you?
BTW, determining that the integral of e^(-x^2) from -infinity to +infinity = sqrt(Pi) is a really, really fun math problem. Well, as far as math problems go.
So a temp of <-2.5 occurs 0.62% of the time, while a temp of <2.5 occurs 99.38% of the time.
For between 1.05 and 2.05, you’d have:
z(2.05) - z(1.05) = .9798 - .8531 = .1267. Thus, temps would be between 1.05 and 2.05 12.67% of the time.
Another trick if you’re calculator savvy is to input the formula for the normal distribution into your calculator and then integrate between whatever boundaries you need. So you could integrate N(x) = (1/sqrt(2*pi))*e^(-x^2/2) from 1.05 to 2.05 and get .1267. That formula is the normal distribution simplified for a mean of 0 and a sigma of 1. You can use it for z-scores.
Integrate N(x) from -1.8 to 2.08 to get .9453, for example.
Um, again - so why did you need solve these problems? We at T-Nation actually like to help people rather than just give answers. Teach a man to fish, as it were.
Get a TI-84 - trust me, it will be your best friend.
These are easy when you use the calc.
To the LEFT of the curve - use the normalcdf function. Input 2nd key, dist, normalcdf(-1e99,raw score(x), mean(M),std deviation). Make sure it’s negative 1e99 for anything to the left.
To the RIGHT of the curve - normalcdf(raw score(x),1e99,mean(M),st dev) Make sure it’s positive 1e99 for anything to the right.
Between 2 Z-scores - normalcdf(smaller raw score, larger raw score, mean, st dev)
