Actuarial Outpost SOA Sample Question #284
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#1
07-14-2018, 02:39 PM
 Mitsu96 Member Non-Actuary Join Date: Jan 2017 Location: Pennsylvania Studying for FM Posts: 55
SOA Sample Question #284

Hello,

I am somewhat struggling with this question:

Losses incurred by a policyholder follow a normal distribution with mean 20,000 and standard deviation 4,500. The policy covers losses, subject to a deductible of 15,000.

Calculate the 95th percentile of losses that exceed the deductible.

I really don't know how to start this question and the solution provided confuses me.

Any help is appreciated! Thank you!
#2
07-14-2018, 02:59 PM
 R2Capital CAS Join Date: Dec 2015 Studying for MAS-I Posts: 26

Losses exceed the deductible with probability Pr(Z > [15000-20000]/4500) = 0.8665.

I would draw a picture of the normal distribution, then identify the point where losses exceed the deductible and try to graphically reason what percentile of the normal distribution equates to the 95th percentile of the claim distribution (i.e. the distribution where losses exceed the deductible).

Last edited by R2Capital; 07-14-2018 at 03:04 PM..
#3
07-14-2018, 03:44 PM
 Michael Mastroianni Member SOA Join Date: Jan 2018 Posts: 33

Quote:
 Originally Posted by Mitsu96 Hello, I am somewhat struggling with this question: Losses incurred by a policyholder follow a normal distribution with mean 20,000 and standard deviation 4,500. The policy covers losses, subject to a deductible of 15,000. Calculate the 95th percentile of losses that exceed the deductible. The answer is 27,700. I really don't know how to start this question and the solution provided confuses me. Any help is appreciated! Thank you!
Let X be your loss random variable and assume R2Capital‘s calculation is correct. Then you need to solve for x such that:

$\frac{P(1500015000)}=\frac{P(X\leq x)-P(X\leq 15000)}{P(X>15000)}=\frac{\Phi(\frac{x-20000}{4500})-(1-0.8665)}{0.8665}=0.95$

because you’re restricting your distribution to values exceeding the deductible of 15,000. So you need to solve for the value of $\frac{x-20000}{4500}$ that makes the equation hold by using the standard normal table in reverse and then solve for x. I hope that helps.
__________________
Michael Mastroianni, ASA
Video Course for Exam 1/P: www.ProbabilityExam.com
#4
07-15-2018, 06:35 AM
 Mitsu96 Member Non-Actuary Join Date: Jan 2017 Location: Pennsylvania Studying for FM Posts: 55

Quote:
 Originally Posted by R2Capital Losses exceed the deductible with probability Pr(Z > [15000-20000]/4500) = 0.8665. I would draw a picture of the normal distribution, then identify the point where losses exceed the deductible and try to graphically reason what percentile of the normal distribution equates to the 95th percentile of the claim distribution (i.e. the distribution where losses exceed the deductible).
That was really helpful. Thank you!
#5
07-15-2018, 06:36 AM
 Mitsu96 Member Non-Actuary Join Date: Jan 2017 Location: Pennsylvania Studying for FM Posts: 55

Quote:
 Originally Posted by Michael Mastroianni Let X be your loss random variable and assume R2Capital‘s calculation is correct. Then you need to solve for x such that: $\frac{P(1500015000)}=\frac{P(X\leq x)-P(X\leq 15000)}{P(X>15000)}=\frac{\Phi(\frac{x-20000}{4500})-(1-0.8665)}{0.8665}=0.95$ because you’re restricting your distribution to values exceeding the deductible of 15,000. So you need to solve for the value of $\frac{x-20000}{4500}$ that makes the equation hold by using the standard normal table in reverse and then solve for x. I hope that helps.
That certainly did help! Thank you so much!

 Tags deductible, loss, normal distribution, percentiles