probability mass function

[jeroen van dijk]

When I enter the value 27.2, the mean 27.20625 and the stdev 0.123798 into
the NORMDIST function, set to false, I get the probability mass function
outcome of 3.2184. Can anyone tell me what this means. I thought that the
maximum value fora pmf was 1, and in fact that the sum of the pmfs for all
possible values is 1.

 

[Jerry W. Lewis]

Continuous distributions have probability density functions (pdf), not
probability mass functions (pmf)
http://en.wikipedia.org/wiki/Probability_mass_function
http://en.wikipedia.org/wiki/Probability_density_function
In a pdf, probability (<=1) is an area under the curve, not the height of
the curve. For the Normal distribution, essentially all of the probability
occurs between mean+/-3*std_dev, which in your case is a region <3/4 wide, so
the height of the pdf must exceed 4/3 to achieve a total area of one (and
since most of the mass is concentrated near the mean, it must exceed 4/3 by a
great deal there).

Alternately, help for NORMDIST gives the formula for the normal pdf, which
you can use to calculate that the value at the mean is nearly 3.25.

Jerry

 

[jeroen van dijk]

Thank you for your answer. Is there a simple way to go from the pdf to the
actual probability?

 

[Mike Middleton]

Jeroen -

For a continuous distribution like the normal, the probability of a single
distinct value is essentially zero.

Probability is determined only for a range of values, usually by taking the
difference between two cumulative probabilities.

For example, P(min<=X<=max) = P(X<=max) - P(X<=min).

You can use the NORMDIST function with the cumulative argument equal to
TRUE.

- Mike
www.mikemiddleton.com

 

[Jerry W. Lewis]

Yes,
=NORMDIST(x,mean,sd,TRUE)
is the probability that X<=x
=NORMDIST(b,mean,sd,TRUE)-NORMDIST(a,mean,sd,TRUE)
is the probability that a<X<=b.

The probability that X exactly equals any prespecified value is zero [the
area is 0*NORMDIST(x,mean,sd,FALSE)], hence the need for a pdf instead of a
pmf.

Jerry