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 Long-Term Actuarial Math Old Exam MLC Forum

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#1
 Msta Member Join Date: Nov 2002 Posts: 53 #4 Nov 2003

Can someone please work this problem out. I had a mental block on it, and still do.

Thanks.
#2
 Woody Member SOA Join Date: Feb 2002 Location: Andy's room Favorite beer: Tusker Posts: 393 I got 99.3 so I chose 100, which was correct. However, I've heard that others got 99.6. I probably did it wrong, but here's how I did it.

Want Pr(Agg Maint Costs &gt;1.2E[S]) &gt;=0.1

E[N]=Var[N]=3
E[X]=80; Var[X]=40000

So, E[S]=240 for one computer, 240t for t computers. We are trying to find t.
Var[S]=139,200 for one computer, 139,200t for t computers.

So, Pr(Agg Costs &gt; 1.2(240t)=288t)&gt;=0.1

So, when normalizing, (288t - 240t)/(139,200t)^(0.5) = 1.282

You end up getting 48t = 478.31(t^(0.5))

When you solve for t, you get 99.3.

Worked for me. Maybe I got lucky.
#3
 Msta Member Join Date: Nov 2002 Posts: 53 Grrrr. Forgot the t in the frickin Variance.

\$#%#\$%#\$%!#
#4
 Tommy Vercetti Member CAS Join Date: Jan 2003 Posts: 988 I work it the same way as woody.
#31 is a simular question, but i can't work it out #5
 dumples Member CAS Join Date: Sep 2003 Posts: 1,250 lambda=3*n
muS=80
VarS=200^2
first moment of severity = 80^2+200^2

muA = lambda*muS
VarA= lambda * fmos = 3*n*(80^2+200^2)
sigmaA=VarA^.5

Find n so that PHI((1.2*muA-muA)/sigmaA)=(1-.1)=.9

.2*muA/sigmaA)=1.282
48*n^.5/(3*(80^2+200^2))^.5=1.282
n^.5=1.282/48*(3*(80^2+200^2))^.5
n=(1.282/48 )^2*(3*(80^2+200^2))=99.29624167
#6
 Tommy Vercetti Member CAS Join Date: Jan 2003 Posts: 988 #31?
#7
 Woody Member SOA Join Date: Feb 2002 Location: Andy's room Favorite beer: Tusker Posts: 393 Quote:
 Originally Posted by Tommy Vercetti I work it the same way as woody. #31 is a simular question, but i can't work it out Tommy, you want Pr (L&gt;0)&lt;=0.05. So you need to calc E[L] and Var[L].

E[L] is Ax-pi(a-duex) = 0.24905-(0.025)[(1-0.24905)(1.06)/.06] = -0.082619 for one policy, -0.082619t for t policies.

Var[L] = (1+P/d)^2[2Ax-Ax^2] = [1+(0.025)(1.06)/.06]^2[0.09476-0.24905^2] = 0.0680346 for one policy, 0.0680346t for t policies.

So, when normalizing, 0-(-0.082619t)/(0.0680346t)^(0.5) = 1.645. You then get 0.42907(t^(0.5))=0.082619t. Solving for t you get 26.97. So answer is B, 27.
#8
 Tommy Vercetti Member CAS Join Date: Jan 2003 Posts: 988 Quote:
Originally Posted by Woody
Quote:
 Originally Posted by Tommy Vercetti I work it the same way as woody. #31 is a simular question, but i can't work it out Tommy, you want Pr (L&gt;0)&lt;=0.05. So you need to calc E[L] and Var[L].

E[L] is Ax-pi(a-duex) = 0.24905-(0.025)[(1-0.24905)(1.06)/.06] = -0.082619 for one policy, -0.082619t for t policies.

Var[L] = (1+P/d)^2[2Ax-Ax^2] = [1+(0.025)(1.06)/.06]^2[0.09476-0.24905^2] = 0.0680346 for one policy, 0.0680346t for t policies.

So, when normalizing, 0-(-0.082619t)/(0.0680346t)^(0.5) = 1.645. You then get 0.42907(t^(0.5))=0.082619t. Solving for t you get 26.97. So answer is B, 27.
Sh!t.
For some reason, I work the same as you but keep getting the wrong ans from calculator Now the calculator works.
sux.
#9
 dumples Member CAS Join Date: Sep 2003 Posts: 1,250 Quote:
 Originally Posted by Tommy Vercetti #31?
ok, i got this wrong but here we go.

for each case sold
Var[L]=(1+p/d)^2* 2Ax-Ax^2 (the first little 2 should be floating but i can't make that happen)
=0.068034639
Sigma(L)=0.260834505

find the annuity due = (1-Ax)/d=13.26

So E[L] for one case is 13.26*.025-.24905 = .08262 (without intermediate rounding)

E[nL]=n*E[L]=.08262N
sigma(nL)=n^.5*Sigma(l)=0.260834505

PHI((E[nL]-0)/sigma(nL))=.95
E[nL]/sigma(nL)=1.645
n^.5=1.645*sigma(L)/E[L]=5.193354214
n=26.97092799=27
#10
 dumples Member CAS Join Date: Sep 2003 Posts: 1,250 it takes too long to type this stuff

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