Actuary Anyone?

[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.

[Guest ThePhilosopher]

Maybe something more concrete, anyone?

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

[Member X]

I really should have paid more attention in maths at school...

I'm no help whatsoever, sorry!!

[Guest ThePhilosopher]

No worries, I figured it out (had to look up a thing or two):

Edited June 21, 2010 by ThePhilosopher

[Member X]

My head hurts just looking at that!

[Guest ThePhilosopher]

I'm stuck with this summation - I'm not allowed to use a graphing calculator at all.

[Member X]

Nope, I'm still no use to you... LoL

[StormTooper4]

Good grief, I used to be able to do that stuff.

Hence why those actuary boys get paid the big bucks.

Good luck with that lot. Be keen to hear how you go.

[cougaree]

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?

[plaifender]

soooo not cool. Why did you just do that to me!?

[cougaree]

huh?

[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.

[Brightight]

Surely if you're going to exponentially distribute with a mean of three years you're going to need a gen? A rep wouldn't last that long under normal conditions. I suggest you try a gen forum :D

[federico.fiorillo]

I really should have paid more attention in maths at school...

I'm no help whatsoever, sorry!!

me too :shock:

[mas6002mi]

'poisson distribution'

[nikki6]

WOW, I think that just destroyed my JAWS software!! lol.

Cougaree and his brother were right on their answer, the answer was for the first problem not the second.

Sixx

[Guest ThePhilosopher]

I see that now Sixx, I was confused as I had answered the first already - I feel doltish.