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how to generate a normal distribution with mean=1, min=0.2, max=0.8

8 messages · Mao Jianfeng, Ravi Varadhan, Giovanni Petris +2 more

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Surely you must be joking, Mr. Jianfeng.

-------------------------------------------------------
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology School of Medicine Johns Hopkins University

Ph. (410) 502-2619
email: rvaradhan at jhmi.edu


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Mao Jianfeng
Sent: Thursday, April 28, 2011 12:02 PM
To: r-help at r-project.org
Subject: [R] how to generate a normal distribution with mean=1, min=0.2, max=0.8

Dear all,

This is a simple probability problem. I want to know, How to generate a
normal distribution with mean=1, min=0.2 and max=0.8?

I know how the generate a normal distribution of mean = 1 and sd = 1 and
with 500 data point.

rnorm(n=500, m=1, sd=1)

But, I am confusing with how to generate a normal distribution with expected
min and max. I expect to hear your directions.

Thanks in advance.

Best,
Jian-Feng,


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#
On Apr 28, 2011, at 12:09 PM, Ravi Varadhan wrote:

            
Perhaps not joking and perhaps not with correct statistical  
specification.

A truncated Normal could be simulated with:

set.seed(567)
x <- rnorm(n=50000, m=1, sd=1)
xtrunc <- x[x>=0.2 & x <=0.8]
require(logspline)
plot(logspline(xtrunc, lbound=0.2, ubound=0.8, nknots=7))
#
Well, but the original poster also refers to 0.2 and 0.8 as "expected
 min and max", in which case we are back to a joke...

Giovanni
On Thu, 2011-04-28 at 13:06 -0400, David Winsemius wrote:
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On Fri, 29 Apr 2011, Giovanni Petris wrote:

            
Well, he is a lot better with English than I am with Mandarin.  He seemed 
to like the truncated normal answers, so we'll let those be his answers.

It is possible to choose parameters for a normal distribution with 500 
observations such that the expected value of the maximum is .8 and the 
expected value of the minimum is .2.  Obviously, the mean would be .5, not 
1, but what would the variance then have to be to provide the correct 
expected max and min?  That's another legitimate question.

Mike
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On Apr 29, 2011, at 1:29 PM, Mike Miller wrote:

            
You would need to specify an N since the expected first and last order  
statistic would decrease/increase with increasing N.
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On Fri, 29 Apr 2011, David Winsemius wrote:

            
Right -- I chose N=500, as did the OP.  I think the order statistics for 
the normal are pretty complex, but it wouldn't be hard to use the density 
for order statistics for the uniform to compute the appropriate values for 
a standard normal, then rescale.

http://en.wikipedia.org/wiki/Order_statistic#The_order_statistics_of_the_uniform_distribution

You'd have to multiply the beta density times the inverse normal cdf and 
get the weighted average for a set of points.  It doesn't sound terribly 
difficult but I don't want to do it!  ;-)

Mike