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LongTerm Actuarial Math Old Exam MLC Forum 

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#1




#4 Nov 2003
Can someone please work this problem out. I had a mental block on it, and still do.
Thanks. 
#2




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 >1.2E[S]) >=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 > 1.2(240t)=288t)>=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




Grrrr. Forgot the t in the frickin Variance.
$#%#$%#$%!# 
#5




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*muAmuA)/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 
#7




Quote:
E[L] is Axpi(aduex) = 0.24905(0.025)[(10.24905)(1.06)/.06] = 0.082619 for one policy, 0.082619t for t policies. Var[L] = (1+P/d)^2[2AxAx^2] = [1+(0.025)(1.06)/.06]^2[0.094760.24905^2] = 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




Quote:
For some reason, I work the same as you but keep getting the wrong ans from calculator Now the calculator works. sux. 
#9




Quote:
for each case sold Var[L]=(1+p/d)^2* 2AxAx^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 = (1Ax)/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 
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