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  #1  
Old 11-15-2017, 07:12 PM
efast efast is offline
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Default p exam problem continuous random variables

Hi guys,

I'm having trouble resolving the following problem. Please help me.

A group insurance policy covers the medical claims of the employees of a small company. The value, V, of the claims made in one year is described by
V = 100000Y
where Y is a random variable with density function f(x) =k(1−y)4 0 < y < 1

0 otherwise
where k is a constant.

What is the conditional probability that V exceeds 40,000, given that V exceeds 10,000?
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  #2  
Old 11-15-2017, 08:01 PM
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Gandalf Gandalf is offline
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Can you calculate the value of k? Can you calculate the probability that V exceeds 40,000? That V exceeds 10,000?
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  #3  
Old 11-16-2017, 10:02 AM
Kemisola Kemisola is offline
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First, you have to find the value of k and this is done by equating the density function i.e. f(x) to 1. You will get k=5

Then you find P(100,000Y>40,000)|100,000Y>10,000) which is same as P(Y>0.4|Y>0.1)

I hope you can go on from here. Your final answer should be 0.132
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Old 11-16-2017, 10:27 PM
efast efast is offline
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Quote:
Originally Posted by Kemisola View Post
First, you have to find the value of k and this is done by equating the density function i.e. f(x) to 1. You will get k=5

Then you find P(100,000Y>40,000)|100,000Y>10,000) which is same as P(Y>0.4|Y>0.1)

I hope you can go on from here. Your final answer should be 0.132
Thank you, Kemisola. What do you mean by equating the density function to 1?
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  #5  
Old 11-16-2017, 10:29 PM
Z3ta Z3ta is offline
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Quote:
Originally Posted by efast View Post
Thank you, Kemisola. What do you mean by equating the density function to 1?
The integral of the density function (from 0 to 1 in this case) must be 1.
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Old 11-16-2017, 10:52 PM
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I'm making a mistake but I don't know where.
I find k by saying that the integral of k*(1-y)^4 from 0 to 1 is equal to 1.
That means that the limit of k/5*(1-y)^5 ( y goes towards 1) minus the limit of k/5*(1-y)^5 ( y goes towards 0) is equal to 1.
Since for y goes towards 1 the limit of f(x) is 0, minus the limit of k/5( 1-y)^5 with y goes towards 0 is equal to 1. That gives me k=-5.
Where did I make a mistake?
thank you for the help.

Last edited by efast; 11-16-2017 at 11:03 PM..
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Old 11-16-2017, 11:00 PM
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Sign error. You forgot that the (1-y) piece contributes a factor of -1.

I.e., if you differentiated k/5( 1-y)^5, you would get -k(1-y)^4, not k(1-y)^4.
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  #8  
Old 11-17-2017, 01:44 AM
Z3ta Z3ta is offline
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You might find this thread helpful.
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  #9  
Old 11-17-2017, 09:23 AM
efast efast is offline
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Thank you Gandalf and Z3ta.
what is the rule to integrate k*(1-y)^4?

Last edited by efast; 11-17-2017 at 11:11 AM..
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  #10  
Old 11-17-2017, 11:12 AM
efast efast is offline
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Thank God, I got it using the substitution rule.
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