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Long Nguyen science forum beginner
Joined: 22 Jun 2006
Posts: 3

Posted: Thu Jun 22, 2006 12:28 pm Post subject:
many random variable problem



exam tomorrow, its probably late in getting a response, lets try.
just a simple question P(2X + 3Y < 30).
E(X) = 0, VAR(X) = 4, E(Y)=10, VAR(Y)=9.
ans:0.5
can someone assist me in working this out.

"Hi..."
 cOS
cossmo @at gmail . com 

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Pubkeybreaker science forum Guru
Joined: 24 Mar 2005
Posts: 333

Posted: Thu Jun 22, 2006 1:12 pm Post subject:
Re: many random variable problem



Long Nguyen wrote:
Quote:  exam tomorrow, its probably late in getting a response, lets try.
just a simple question P(2X + 3Y < 30).
E(X) = 0, VAR(X) = 4, E(Y)=10, VAR(Y)=9.
ans:0.5
can someone assist me in working this out.

What are the pdf's for X and Y? 

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Pubkeybreaker science forum Guru
Joined: 24 Mar 2005
Posts: 333

Posted: Thu Jun 22, 2006 1:24 pm Post subject:
Re: many random variable problem



Long Nguyen wrote:
Quote:  exam tomorrow, its probably late in getting a response, lets try.
just a simple question P(2X + 3Y < 30).
E(X) = 0, VAR(X) = 4, E(Y)=10, VAR(Y)=9.
ans:0.5
can someone assist me in working this out.

If X & Y are normally distributed the problem is trivial.
If you can't solve this problem when X & Y are normal, then
you have some serious knowledge gaps and will have great difficulty
on your exam.
We need to know how X & Y are distributed.
I assume you know how to make a changeofvariable/transformation
of a pdf? Let W = 2X + 3Y. Compute the joint pdf of X&Y,
multiply
by the Jacobian, then compute the marginal distribution of W by
integration.
Once you know the pdf for 2X + 3Y, you can compute the probability
that W < 30.
With no information about how X & Y are distributed, the problem seems
impossible. 

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Long Nguyen science forum beginner
Joined: 22 Jun 2006
Posts: 3

Posted: Fri Jun 23, 2006 1:18 am Post subject:
Re: many random variable problem



On 22 Jun 2006 06:24:49 0700, "Pubkeybreaker"
<Robert_silverman@raytheon.com> wrote:
Quote: 
Long Nguyen wrote:
exam tomorrow, its probably late in getting a response, lets try.
just a simple question P(2X + 3Y < 30).
E(X) = 0, VAR(X) = 4, E(Y)=10, VAR(Y)=9.
ans:0.5
can someone assist me in working this out.
If X & Y are normally distributed the problem is trivial.
If you can't solve this problem when X & Y are normal, then
you have some serious knowledge gaps and will have great difficulty
on your exam.
We need to know how X & Y are distributed.
I assume you know how to make a changeofvariable/transformation
of a pdf? Let W = 2X + 3Y. Compute the joint pdf of X&Y,
multiply
by the Jacobian, then compute the marginal distribution of W by
integration.
Once you know the pdf for 2X + 3Y, you can compute the probability
that W < 30.
With no information about how X & Y are distributed, the problem seems
impossible.

yes, there are many serious knowledge gaps, either my brain or the
teaching levels here at my uni or a combination of both.
it is assumed X & Y is normally distributed.
thanks for your reply, it has helped alot. 

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Pubkeybreaker science forum Guru
Joined: 24 Mar 2005
Posts: 333

Posted: Fri Jun 23, 2006 10:45 am Post subject:
Re: many random variable problem



Long Nguyen wrote:
Quote:  On 22 Jun 2006 06:24:49 0700, "Pubkeybreaker"
Robert_silverman@raytheon.com> wrote:
yes, there are many serious knowledge gaps, either my brain or the
teaching levels here at my uni or a combination of both.
it is assumed X & Y is normally distributed.

Are they *independent*?? The problem did not say.
Do you know how to find the distribution of a random variable
that is the sum of two normally distributed random variables?
Quote: 
thanks for your reply, it has helped alot.

What book are you using? 

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Long Nguyen science forum beginner
Joined: 22 Jun 2006
Posts: 3

Posted: Sun Jun 25, 2006 4:15 am Post subject:
Re: many random variable problem



Quote: 
yes, there are many serious knowledge gaps, either my brain or the
teaching levels here at my uni or a combination of both.
it is assumed X & Y is normally distributed.
Are they *independent*?? The problem did not say.
Do you know how to find the distribution of a random variable
that is the sum of two normally distributed random variables?
thanks for your reply, it has helped alot.
What book are you using?

yes, they are independant. I am not really sure what you are
asking me.
I am using devore, statistics for engineers, 2nd ed. it doesn't
have info on differences of joint distribution.
anyhow, the test went 'ok'. i was stuck on one particular question and
wasted alot of time on it. this is what I can recall.
this is a 5.4% risk of catching some type of disease. a sample of 2000
men was taken and 78 of them was diagnosed with the disease. 64 of
1500 women was diagnosed with the disease. it is said that men have
a higher risk of catching this disease, what test statistics would you
use?

"Hi..."
 cOS
cossmo @at gmail . com 

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