Re: TI-M: Stats Question
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Re: TI-M: Stats Question
In a message dated 9/11/00 4:05:54 PM Mountain Daylight Time,
danb2k@hotmail.com writes:
> I think it's 1/sqrt(2*pi*e^(x^2))
That would be the one ;)
> However, iff you have an 83 or 83+, you can just use normalpdf()
I have an 89...anyone know if there's a function to go between z-scores and
percentages (of area of the normal curve) on the 89?
I was trying to figure out what the function was for a normal curve was
yesterday and tried solving for equations like y = e^(-a*x^(2b)) and y =
a/(cosh(bx)^2)...didn't occur to me that it was y = a*e^(-b*x^2) ;) Oh
well...
JayEll
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