probability density function
Excellent!!! Thanks Dana.
"Dana DeLouis" wrote:
...from the peer reviewed literatu
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
mean = 5.21
st dev = 1.02
--
HTH :)
Dana DeLouis
Windows XP & Office 2003
"Ariel" wrote in message
...
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 literatu
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
"Jerry W. Lewis" wrote:
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
"Ariel" wrote:
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
|