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Actuary Anyone?


Guest ThePhilosopher

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Guest ThePhilosopher

I am studying for exam P and I'm on problem 39 of 149 in a free packet I received, but I'm worried that simply solving problems isn't going to be enough. If anyone has sat for or us studying for exam P I'd like to hear your study recommendations.

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Guest ThePhilosopher

Maybe something more concrete, anyone?

Problem.jpg

I'm stuck on setting up the integral here from here - I'm not utterly familiar with this type of problem.

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I forwarded to my bother in law and this is his reply.

"I am pretty sure I solved it. The answer is D.

first find the distribution of X.

X = {

0 if x < 0

1-e^(-2/3) if x = 2

1/3 * e^(-x/3) if x > 2

Then just compute E(x).

E(x)=

integrate(x*0,x,-infinity,2) + 2*(1-e^(-2/3))

+ integrate(x/3 * e^(-x/3),x,2,infinity)

which equals 2+3e^(-2/3)

I got "1-e^(-2/3)" by using the CDF of T solved at x=2. This will give me the probability that the device failed during the first two years.

let me know if I got the right answer. It was fun working on it."

What do you think? Did he get it right?

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Guest ThePhilosopher

I forwarded to my bother in law and this is his reply.

"I am pretty sure I solved it. The answer is D.

first find the distribution of X.

X = {

0 if x < 0

1-e^(-2/3) if x = 2

1/3 * e^(-x/3) if x > 2

Then just compute E(x).

E(x)=

integrate(x*0,x,-infinity,2) + 2*(1-e^(-2/3))

+ integrate(x/3 * e^(-x/3),x,2,infinity)

which equals 2+3e^(-2/3)

I got "1-e^(-2/3)" by using the CDF of T solved at x=2. This will give me the probability that the device failed during the first two years.

let me know if I got the right answer. It was fun working on it."

What do you think? Did he get it right?

No - nowhere is the Poisson distribution nor the payment. You cannot integrate because the Poisson distribution is discrete and line segments have 0 area. I've got it figured out finally though.

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