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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.

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cossmo @at gmail . com
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?
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 change-of-variable/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.
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 change-of-variable/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.
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?
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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