Normal Distribution calculation challenge

[Bob Flanagan]

I hope someone can help on this one. Its been too many years since I've
done this level of statistics.... The following illustrates what I need to
do

Assume we have a normal distribution, with mean 20 and a variance of 5. I
need to get the true mean if the minimum allowed value is 7 and the max
allowed value is 30. Unfortunately, I can't fibure out how to do the
calculation.

Bob Flanagan (email: bo...@dol.net)

[Jay Petrulis]

Hi Bob,

Mike Middleton has a paper on his website which may be of help.
http://www.usfca.edu/fac-staff/middleton/demand.pdf

This may be a good place to start.

Regards,
Jay

"Bob Flanagan" <nor...@noreply.net> wrote in message news:<YbdH7.199880$5A3.74...@news1.rdc2.pa.home.com>...

[Tushar Mehta]

Not quite sure what you are looking for...

From what I know, given a sample mean and standard deviation, there is no
way to calculate the 'true' population mean. The sample mean *is* the
best unbiased estimate for the population mean.

One could build a confidence interval on 'how likely is it that we got
this sample mean given the population mean of <whatever>.'

To build the C.I., use sample mean +/- <z-value for 90% or 95% or 99% or
any-other-custom-value confidence interval> * <sample std.
dev.>/sqrt(sample size>)

--
Regards,

Tushar Mehta

www.tushar-mehta.com

In article <YbdH7.199880$5A3.74...@news1.rdc2.pa.home.com>,
nor...@noreply.net says...

[Bob Weber]

"Bob Flanagan" <nor...@noreply.net> wrote in message
news:YbdH7.199880$5A3.74...@news1.rdc2.pa.home.com...

Assume we have a normal distribution, with mean 20 and a variance of 5. I

need to get the true mean if the minimum allowed value is 7 and the max
allowed value is 30. Unfortunately, I can't fibure out how to do the
calculation.

Bob -

Perhaps you want the conditional expected value of a draw from your
specified normal distribution, given that the draw lies between 7 and 30. If
so, try the following (where X is your draw):

E[X|X<a] = mu-sigma^2*NORMDIST(a,mu,sigma,FALSE)/NORMDIST(a,mu,sigma,TRUE).
E[X|X>b] =
mu+sigma^2*NORMDIST(b,mu,sigma,FALSE)/(1-NORMDIST(b,mu,sigma,TRUE)).

Then solve:

mu = Pr(X<a)*E[X|X<a] + Pr(X>b)*E[X|X>b] + Pt(a<X<b)*desired_answer,

yielding

desired_answer =
mu+sigma^2*(NORMDIST(a,mu,sigma,FALSE)-NORMDIST(b,mu,sigma,FALSE))/(NORMDIST
(b,mu,sigma,TRUE)-NORMDIST(a,mu,sigma,TRUE))

Take mu = 20, sigma = sqrt(5), a = 7, b = 30. You'll get 19.99995954. This
is close to 20, of course, since your limits cut off only two tiny tails.

Regards, Bob Weber