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