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#101
02-25-2007, 03:22 PM
 Sandimashischool Member Join Date: Feb 2007 Posts: 547

Quote:
 Originally Posted by Gandalf Maybe it said the company would pay the greater of x and y; in that case it is asking E(max(x,y))
I am 100% positive that that question said the insurance company would pay the greater of the two
#102
02-25-2007, 03:30 PM
 Sandimashischool Member Join Date: Feb 2007 Posts: 547

Quote:
 Originally Posted by Generalshamu I too had the problem which was (lambda / (lambda - t))^r and got r/(lambda^2) Another problem was find E(x) if the f(x,y) = 2|x|y for -1W|H>5)...I got 1-(15^2 /2)/(15*20) Where 15*20 is the area of the conditional region, i.e. H>5... The last question I will mention is one that I thought was defective... It dealt with the cost of apendeoctamies and how a company will only cover reasonable expenses (x) from the surgery. The costs are normally distributed with mean 60 and standard deviation of 25. The max on reasonable expenses was the 70th, percentile. Find the maximum payment... I set it up as .7 = phi((x-60)/25)) and got x = 73 or so but the only answers close to it were 72 and 75...
i too got 0, because |x| is - x when x < 0, so you get +2/9 + -2/9 (it is not ironic that 2/9 is one of the answers, they alwasy give you the most common erros for the answers, thats what makes it so hard to pass, you see a choice and dont check your work, and the moment genrating function was a gamma distribution so i got the same answer as you on that one as well
#103
02-25-2007, 03:36 PM
 Sandimashischool Member Join Date: Feb 2007 Posts: 547

Quote:
 Originally Posted by Generalshamu Damn...I don't know how I could have messed that one up...damnit....
damn, i cant believe i oversaw that, when you rush though these you make the dummbest mistakes
#104
02-25-2007, 03:42 PM
 Sandimashischool Member Join Date: Feb 2007 Posts: 547

Quote:
 Originally Posted by daaaave Except that $\int_{x=-1}^0 -x^2 \, dx = (-0)-\left(-\frac{(-1)^3}{3}\right)=-1/3$ so we get 0. Alternatively, it is easier to note that the density of x is symmetric about the y-axis, and hence E[X]=0.
I i did not make a mistake after all, its crazy all the similiar questions with small differneces, regardless, i have this feeling in 6 weeks i will get a 5/10 and cry (my birthday is march 27, i will probably get my results around then haha)
#105
02-25-2007, 03:54 PM
 Sandimashischool Member Join Date: Feb 2007 Posts: 547

Quote:
 Originally Posted by atomic If P denotes the random profit, then $P \sim {\rm Normal}(\mu = 100, \sigma = 20)$ and $Z = (P - \mu)/\sigma \sim {\rm Normal}(0,1)$. $\Pr[P > x] = \Pr\left[\frac{P - \mu}{\sigma} > \frac{x - 100}{20}\right] = 1 - F\left(\frac{x - 100}{20}\right) = F\left(\frac{100 - x}{20}\right)$. So the probability that P > 0 is simply F(5), and the probability that P > 60 is F(2). Therefore $\Pr[P > 60 | P > 0] = F(2)/F(5)$.
do you remeber what the other 2 choices were for this question?
#106
02-25-2007, 04:04 PM
 Sandimashischool Member Join Date: Feb 2007 Posts: 547

Did anyone get a question where you had to compute that standard deviation. I do not remember the question too well but there was a max payment of 18000 and that after 20 inches of snow (or something) they will pay 5000 for each extra 2 inches of snow. And they gave the probability of the snow being 0<x<2, 2<x<4, 4<x<6, .... 20<x<22, 22<x<24. I remebr standard deviation was the SQRT(summation of (X-M))/n) but none of the answers matched.
#107
02-25-2007, 04:31 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,459

Quote:
 Originally Posted by Sandimashischool Did anyone get a question where you had to compute that standard deviation. I do not remember the question too well but there was a max payment of 18000 and that after 20 inches of snow (or something) they will pay 5000 for each extra 2 inches of snow. And they gave the probability of the snow being 0
No matter what, you would need to be summing squares, and you wouldn't get a "/n" factor as the denominator.

You would want to calculate E(X^2) - E(X)^2, where you must recognize that X = 0 for low snowfall and X = 18000 for large snowfall. (That's the variance; then take square root.)

The formula above is equivalent to calculating E((X-E(X))^2), but is usually an easier calculation

Last edited by Gandalf; 02-25-2007 at 04:36 PM..
#108
02-25-2007, 06:01 PM
 jy006m Member SOA Join Date: Aug 2006 Studying for FAP Posts: 924

Quote:
 Originally Posted by Sandimashischool Did anyone get a question where you had to compute that standard deviation. I do not remember the question too well but there was a max payment of 18000 and that after 20 inches of snow (or something) they will pay 5000 for each extra 2 inches of snow. And they gave the probability of the snow being 0
I got that one. It was that they pay 0 if there was less than 12 inches of snow, and 5000 for every 2 full inches of snow.

So X (payment) was distributed as follows:
0<x<14 0
14<x<16 5000
16<x<18 10000
18<x<20 15000
20<x<24 18000

I don't remember the corresponding probabilities. But yeah, this was pretty straightforward, just calculation intensive. You just needed to calculate the variance and take the square root of it.
#109
02-25-2007, 06:24 PM
 INxx_Dresen Guest Posts: n/a

Quote:
 Originally Posted by Sandimashischool Did anyone get a question where you had to compute that standard deviation. I do not remember the question too well but there was a max payment of 18000 and that after 20 inches of snow (or something) they will pay 5000 for each extra 2 inches of snow. And they gave the probability of the snow being 0
I had that question (as #27 on my exam):

A probability distribution of rainfalls (in a particular year) is given below. An insurance policy pays nothing for the first 12 inches of rain. Then after 12 inches, for every two full inches of rain, it pays 5,000, to a maximum of 18,000. Calculate the standard deviation of the payment.

X = amount of rainfall

Pr (0 < x ≤ 12) = .52 (This was given as probability of individual intervals of 2, but I summed them up.)
Pr (12 < x ≤ 14) = .18
Pr (14 < x ≤ 16) = .11
Pr (16 < x ≤ 18) = .08
Pr (18 < x ≤ 20) = .07
Pr (20 < x ≤ 22) = .04

----------------------

I hesitated over what they meant by to full inches. At first I constructed distribution for the payment Y like this:

Pr (Y = 0) = .52
Pr (Y = 5000) = .18
Pr (Y = 10000) = .11
Pr (Y = 15000) = .08
Pr (Y = 18000) = .07 + .04 = .11

But this gives a st. dev. greater than all of the answer choices. So I thought the “two full inches” must have meant they still pay nothing for the amount of rainfall greater than 12 that hasn’t accumulated to 2, i.e. they only start to pay 5000 after the rain is 14 inches full, 10000 after it’s 16 inches... Then the distribution for Y should be:

Pr (Y = 0) = .52 + .18 = .70
Pr (Y = 5000) = .11
Pr (Y = 10000) = .08
Pr (Y = 15000) = .07
Pr (Y = 18000) = .04

This gives me E(Y) = 3120, E(Y square) = 39,460,000 and so st. dev. is 5452, which is answer choice D.

I noticed 3120 is also given among the choices (as a trick, maybe?), so I thought my E(Y) must be right, which made me feel somewhat confirmed about my approach. Hopefully the gurus of this forum would agree.

INxx
#110
02-25-2007, 07:25 PM
 Generalshamu Member SOA Join Date: Oct 2006 Location: LA Favorite beer: Heineken Posts: 811

Quote:
 Originally Posted by Generalshamu Another question was with a wife and husband's lives uniformally distributed on (0,20). Find the P(H>W|H>5)...I got 1-(15^2 /2)/(15*20) Where 15*20 is the area of the conditional region, i.e. H>5...
Did anyone get this question???

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