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  #51  
Old 08-18-2006, 10:53 PM
justincarstarphen justincarstarphen is offline
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Quote:
Originally Posted by jexica View Post
I don't remember an answer of 45%-I bet you got it right!
hellz yeah.... i hope
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  #52  
Old 08-18-2006, 10:53 PM
jexica jexica is offline
 
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Originally Posted by justincarstarphen View Post
I remember the answers to the question, does anyone know what I am talking about???
a. -.006
b. -.003
c. 0
d. .003
e. .006

I believe you had to fine p-r or something like that.
I got .003
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  #53  
Old 08-18-2006, 10:54 PM
justincarstarphen justincarstarphen is offline
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Originally Posted by CaptainAwesome View Post
Yes. The probability of a fire is .3 and the probability of a flood is .1 (maybe those are flip-flopped...doesn't matter though).
Let p be the probability that neither a floor nor fire occurs assuming they're independent.
Let r be the probability that neither a floor nor fire occurs assuming they're mutually exclusive.

I remember getting .03.

uh-oh...... I got -.03
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  #54  
Old 08-18-2006, 10:59 PM
CaptainAwesome CaptainAwesome is offline
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Originally Posted by justincarstarphen View Post
uh-oh...... I got -.03
Regrettably, that means you've failed. That was the only non-pilot question.
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  #55  
Old 08-18-2006, 10:59 PM
jexica jexica is offline
 
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OK, it was the scenario when they were independent minus when they were mutually exclusive.

1-P(fire union theft)
1-(.1 +.3 - .1*.3) = .63
in first case
and
1-(.1 +.3) = .6

.63-.6 = .3
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  #56  
Old 08-18-2006, 11:00 PM
jexica jexica is offline
 
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Or maybe .63-.6 = .003 ...just maybe.
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  #57  
Old 08-18-2006, 11:00 PM
uncafe uncafe is offline
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Originally Posted by CaptainAwesome View Post
One of my questions was:

A policy is exponentially distributed with mean 2, has a policy limit of 5, and a deductible of 1.

For the life of me, I could not match my answer with what they had. The only thing that I could figure was that I forgot how to multiply (very likely).

Also, I had two questions on f(x,y) = 4y/x where 0 < y<x<1
One asked for E(Y|X=1/3) and the other Var(Y|X=2/3).

It should have been a really easy problem, but for the second one, when I was finding fx(2/3) I was getting 4/3. 4/3 is greater than 1, right? I'm pretty sure that's a problem.

I also had that 4/3 thing in my exam...damn I spent so long trying to figure out how can I got something more than 1....
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  #58  
Old 08-18-2006, 11:03 PM
jexica jexica is offline
 
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Originally Posted by uncafe View Post
I also had that 4/3 thing in my exam...damn I spent so long trying to figure out how can I got something more than 1....
It's OK if it's more than one. It's f(x) you are plugging it into...the greater than one thing would only be a problem if it was F(x). For example, f(x)=2x 0<x<1 is a common pdf. f(1)=2, but it's OK.
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  #59  
Old 08-18-2006, 11:05 PM
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Gandalf Gandalf is offline
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Quote:
Originally Posted by uncafe View Post
I also had that 4/3 thing in my exam...damn I spent so long trying to figure out how can I got something more than 1....
If you are saying the marginal density of x, evaluated at x=2/3, is > 1, that's not necessarily a problem. In fact, if X can only take on values between 0 and 1, then the density must be greater than 1 some places (or equal to one everywhere) if it is going to integrate to 1.

(i.e., what jexica said)
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  #60  
Old 08-18-2006, 11:06 PM
CaptainAwesome CaptainAwesome is offline
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Originally Posted by uncafe View Post
I also had that 4/3 thing in my exam...damn I spent so long trying to figure out how can I got something more than 1....
Yeah, I feel retarded. Would someone be so kind as to point out my mistake here?

f(x,y) = 4y/x. 0<y<x<1 Find Var(Y|X=2/3)

I found the conditional distribution:
fy|x=2/3 = f(2/3,y)/fx(2/3) = 6y/fx(2/3)

fx(x) = integral from 0 to x of 4y/x dy which i believe is 2x
so, fx(2/3) = 4/3 <--- what went wrong there??
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