probability density function

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Guest

Can someone provide me with the formula for calculating the probaility
density function. Here is what I have, and what I need to get to:

x value = 60
mean = 40
st dev = 15

thus:

z score = 1.33 (using "standardize" function)
Area under normal curve = 0.09 (1-NORMSDIST(1.33))

The value I am trying to get to is 0.16 (this is supposed to be the
probability density function for the above value)

Thanks

Ariel
 
David Biddulph said:
=EXP(-A1*A1/2)/SQRT(2*PI()) with your 1.33 in A1
Click to expand...

.... and if you've been reading another thread regarding precedence of
operators, you'll realise that you can't use
=EXP(-A1^2/2)/SQRT(2*PI())
 
The simplest way is
=NORMDIST(A3,B3,C3,FALSE)
which avoids issues of operator precedence.

Note however that the value of the pdf is 0.011, not 0.16. Assuming that
your values are rounded, the pdf for the unrounded values is no more than
0.012.

Jerry
 
It may give you the answer that you were looking for, but Jerry has pointed
out that it's the wrong answer. The pdf from the normalised distribution in
my formula would need dividing by the standard deviation (15 in your case).

Or more generally =EXP(-(X-MEAN*(X-MEAN)/(2*SDEV^2))/(SDEV*SQRT(2*PI()))
 
Hi Jerry and David,

The product must be 0.16, not 0.012 (this is a case from ther peer reviewed
literature). Neither of these formulae provide that answer, however David's
earlier formula in fact does provide that. I have another case below, also
from the peer reviewed literature:

x value = 6.50
mean = 5.21
st dev = 1.02

z score = 1.26 (using "standardize" function)
Area under normal curve = 0.1038 (1-NORMSDIST(1.26))

Here the value I am trying to get to is 0.1804 (this is supposed to be the
probability density function for the above value)

Thanks

Ariel
 
...from the peer reviewed literature:

Hi. You are using a normal mean of 0, and std of 1.

=NORMDIST(1.33,0,1,FALSE)
=NORMDIST(1.26,0,1,FALSE)

These return your .16 & .18

However, you give
mean = 40
st dev = 15
Click to expand...
 
The pdf for an x value of 60 for a normal distribution of mean 40 and
standard deviation 15 is 0.0109

The value of 0.164 is the pdf of a normalised distribution (hence at an x
value of 60/15 on a distribution of mean 40/15 and a standard deviation of
1).
[I couldn't immediately get 0.1804 by an equivalent process for your second
set of figures. The only way you get 0.1804 is by rounding the (x -
mean)/sdev value to 3 significant figures, and hence you shouldn't quote the
answer to four significant figures.]

You need to be careful as to what quantity you want.
 
Excellent!!! Thanks Dana.

Dana DeLouis said:
Hi. You are using a normal mean of 0, and std of 1.

=NORMDIST(1.33,0,1,FALSE)
=NORMDIST(1.26,0,1,FALSE)

These return your .16 & .18

However, you give
Click to expand...
 
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