isaac <[EMAIL PROTECTED]> wrote in sci.stat.edu:
>Hi, I'm a newbie at stats. If I have a normal distribution defined by
>mean x and stddev s, I know that x + s defines a threshold at the 86th
>percentile.
84th, I think. 68% of data are between mu-sigma and mu+sigma, so 34%
are between mu and mu+sigma. Add the 50% below mu, and you have
mu+sigma at the 84th percentile.
>I know of a chart that dumps out a relationship between
>threshold ( expressed as # of s from x ) and percentile. However, is
>there a formula for computing the threshold, given a percentile, x,
>and s? For example, how about the 93rd percentile? I'd rather not
>have to use interpolation.
Are you familiar with the concept of standard score or z score? The
formula is
z = (x-mu)/sigma or x = z*sigma + mu
(Here I'm using x to mean the measured data point, not the mean. I
find it too confusing to call the mean x. :-) The z score is the
number of standard deviations above the mean. Example: z=-1.5 is 1.5
sigma below mu; z=2.2 is 2.2 sigma above mu. You already know that
z=+1.00 is at the 84th percentile.)
So your problem reduces to this: given a percentile, what is the z
score at that percentile? If you once have the z score, mean mu, and
standard deviation sigma, finding the threshold data value x is a
picnic.
Unfortunately, I don't believe there is a nice formula to solve that
problem. Let me explain why I think that.
Now to find the height of the normal curve at any z score, the
formula is
f(z) = (1/sqrt(2*pi)) * exp(-0.5*z^2)
where "exp" is the number e raised to that power. To find the total
area to the left of a given x, you integrate the above formula from
z = -infinity to the desired z score:
area to left of z =
(1/sqrt(2*pi)) * INTEGRAL[exp(-0.5*t^2) dt , -oo, z]
and therefore
Percentile =
(100/sqrt(2*pi)) * INTEGRAL[exp(-0.5*t^2) dt , -oo, z]
That integral unfortunately can't be evaluated in a closed form (a
nice formula), as far as I'm aware.
To answer your question -- what is the formula for z given a
percentile -- you would need to evaluate the integral, which would
give you an expression in z, and then solve that expression for z.
As I say, I don't believe there is a formula for that integral. It's
just got to be done by numeric techniques.
TI-83 users will use invNorm. For instance, invNorm(0.80) gives the
z score for the 80th percentile, and if the mean and standard
deviation are 12.5 and 1.1 then invNorm(0.80,12.5,1.1) gives the raw
score at the 80th percentile.
Excel users will use =NORMINV( ) with, I think, the same arguments.
Failing a calculator or computer, I think you're stuck with tables.
I believe tables typically give the z for the area to the right, so
if you want the z for say the 80th percentile you would look in the
table for z(0.20) because 100%-80%=20% or 0.20.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://oakroadsystems.com/
"What in heaven's name brought you to Casablanca?"
"My health. I came to Casablanca for the waters."
"The waters? What waters? We're in the desert."
"I was misinformed."
.
.
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