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.
--
Jeroen

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/Probabi...nsity_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" wrote:

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.
--
Jeroen

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


"Jerry W. Lewis" wrote:

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/Probabi...nsity_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" wrote:

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.
--
Jeroen

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

"jeroen van dijk" wrote in message
...
Thank you for your answer. Is there a simple way to go from the pdf to the
actual probability?
--
Jeroen


"Jerry W. Lewis" wrote:

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/Probabi...nsity_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" wrote:

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.
--
Jeroen

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

"jeroen van dijk" wrote:

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


"Jerry W. Lewis" wrote:

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/Probabi...nsity_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" wrote:

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.
--
Jeroen