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nerdjyack
April 5th 2005, 12:27 PM
Hi! Please help me. I'm reviewing for Exam P and I came across these questions which I can't solve (or rather, the answers that I get are wrong)

1. The claim amount on a certain insurance contract has a normal distribution with mean \$3,300 and standard deviation \$575. Given 25 independent claims, what is the probability that the number of claims less than \$2700 is less than or equal to 4?
--> The answer key says 0.69 is the correct answer

2. An insurance company has a large block of one year term policies, which is divided into three classes of policies. The distribution of policies is 40% in class A, 50% in class B, and 10% in class C. The probability of death is 20% in class A, and 10% in classes B and C. Five policies are selected from the same class; it is observed that one of the five policyholders dies in the year. What is the probability that these policies are from class A?
--> The answer key says 0.571 is the correct answer (but I keep on getting 0.454)

3. What is the average age to which a 90-year old man will live, given that the probability a person age x will die before age x+1 is (x - 89) / 5?
--> The answer key says 95.1 is the correct answer

could you please tell me how to solve these problems correctly? Thank you very much!!

wat
April 5th 2005, 01:57 PM
1. The claim amount on a certain insurance contract has a normal distribution with mean \$3,300 and standard deviation \$575. Given 25 independent claims, what is the probability that the number of claims less than \$2700 is less than or equal to 4?
--> The answer key says 0.69 is the correct answer

I'm not quite sure whether this is the desired solution, but it gets you the answer.

For one claim, E(X) = 3,300 and St.Dev(X) = 575. So, P(X<2700) = P(Z<(2700-3300)/575) = P(Z<-1.043) ~ 0.15.

Since we have 25 independent claims, if we treat each one like a Bernoulli random variable with p = 0.15, we get that the total number of claims (let it be T) is a Binomial random variable, with n=25 and p = 0.15.

To get the probability of the total number of claims to be less than or equal to 4, we have to calculate P(T<=4) = P(T=0) + P(T=1) + P(T=2) + P(T=3) + P(T=4).

Use the Binomial random variable to get P(T<= 4) = 0.682106967. This is not the answer you gave, but p=0.15 was an estimate of the individual probability that a claim was less than \$2,700. If you change that, you'll probably get the answer you want.

wat
April 5th 2005, 02:21 PM
2. An insurance company has a large block of one year term policies, which is divided into three classes of policies. The distribution of policies is 40% in class A, 50% in class B, and 10% in class C. The probability of death is 20% in class A, and 10% in classes B and C. Five policies are selected from the same class; it is observed that one of the five policyholders dies in the year. What is the probability that these policies are from class A?
--> The answer key says 0.571 is the correct answer (but I keep on getting 0.454)

You want to find P(A | 1 death) = P(A & 1 death) / P(1 death)

P(A & 1 death) = probability that a person from class A dies = P(death | A)*P(A) = (0.2)(0.4) = 0.08.

P(death) = probability a person dies from any class = P(death | A)*P(A) + P(death | B)*P(B) + P(death | C)*P(C) = (0.2)(0.4) + (0.1)(0.5) + (0.1)(0.1) = 0.14.

So, P(A | 1 death) = (0.08)/(0.14) = 0.571428571

wat
April 5th 2005, 02:27 PM
3. What is the average age to which a 90-year old man will live, given that the probability a person age x will die before age x+1 is (x - 89) / 5?
--> The answer key says 95.1 is the correct answer

Using tools from Exam M, I can solve this, but I don't want to introduce anything you haven't seen before. I'll have to think about it for a little bit.

nerdjyack
April 5th 2005, 10:37 PM
You want to find P(A | 1 death) = P(A & 1 death) / P(1 death)

P(A & 1 death) = probability that a person from class A dies = P(death | A)*P(A) = (0.2)(0.4) = 0.08.

P(death) = probability a person dies from any class = P(death | A)*P(A) + P(death | B)*P(B) + P(death | C)*P(C) = (0.2)(0.4) + (0.1)(0.5) + (0.1)(0.1) = 0.14.

So, P(A | 1 death) = (0.08)/(0.14) = 0.571428571

What's the use of the given FIVE polyciholders ("FIVE policies are selected from the same class; it is observed that one of the FIVE policyholders dies in the year")?

Thank you very much!!!

wat
April 6th 2005, 02:20 PM
What's the use of the given FIVE polyciholders ("FIVE policies are selected from the same class; it is observed that one of the FIVE policyholders dies in the year")?

Thank you very much!!!

Well, replace the five with five thousand. Does it make a difference? Nope. You're given that one policyholder dies, and that all the others belong to that same group. Thus, you just need to determine the probability of that one death belonging to group A.

nerdjyack
May 6th 2005, 06:44 AM
Hi. I have another batch of problems which I can't seem to solve correctly (sigh). I would really appreciate it if someone would tell me how to solve these. Thanks. (Btw, these problems are from the book "John E. Freund's Mathematical Statistics" by Miller, which is one of the recommended textbooks of SOA for course 1).

1. In a certain community, 8 percent of all adults over 50 have diabetes. If a health service in this community correctly diagnoses 95 percent of all persons with diabetes as having the disease and incorrectly diagnoses 2 percent of all persons without diabetes as having the disease, find the probabilities that

(a) the community health service will diagnose an adult over 50 as having diabetes (answer: 0.0944)
(b) a person over 50 diagnosed by the health service as having diabetes actually has the disease (answer: 0.8051)

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2. An art dealer receives a shipment of five old paintings from abroad, and, on the bases of past experience, she feels that the probabilities are, respectively, 0.76, 0.09, 0.02, 0.01, 0.02 and 0.10 that 0, 1, 2, 3, 4, or all 5 of them are forgeries. Since the cost of authentication is fairly high, she decides to select one of the five paintings at random and send it away for authentication. If it turns out that this painting is a forgery, what probability should she now assign to the possibility that all the other paintings are also forgeries? (answer: 0.6757)

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3. The manager of a bakery knows that the number of chocolate cakes he can sell on any given day is a random variable having the probability distribution f(x)=1/6 for x=0,1,2,3,4 and 5. He also knows that there is a profit of 1.00 for each cake that he sells and a loss (due to spoilage) of 0.40 for each cake that he does not sell.

(a)Assuming that each cake can be sold only on the day it is made, find the baker's expected profit for the day on which he bakes
(i) one of the cakes (answer: 0.77)
(ii) 2 of the cakes (answer: 1.30)
(iii) 3 of the cakes (answer: 1.60)
(iv) 4 of the cakes (answer:150)
(v) 5 of the cakes. (answer:150)

(b) How many should he bake in order to maximize his expected profit? (answer: 4)

Karen
May 6th 2005, 08:27 AM
[QUOTE=nerdjyack]Hi. I have another batch of problems which I can't seem to solve correctly (sigh). I would really appreciate it if someone would tell me how to solve these. Thanks. (Btw, these problems are from the book "John E. Freund's Mathematical Statistics" by Miller, which is one of the recommended textbooks of SOA for course 1).

1. In a certain community, 8 percent of all adults over 50 have diabetes. If a health service in this community correctly diagnoses 95 percent of all persons with diabetes as having the disease and incorrectly diagnoses 2 percent of all persons without diabetes as having the disease, find the probabilities that

(a) the community health service will diagnose an adult over 50 as having diabetes (answer: 0.0944)
(b) a person over 50 diagnosed by the health service as having diabetes actually has the disease (answer: 0.8051)

QUOTE]

1. Let D=disease, Pos=positive test result
P[D]=0.08 P[Pos|D]=0.95 P[Pos|D']=0.02
then P[D']=0.92
a. This is asking for the probability of a positive test:
P[Pos]= P[Pos|D]xP[D] + P[Pos|D']xP[D']
=0.95x0.08 + 0.02x0.92
=0.0944
This uses the law of total probability.

b. Now this is asking for a conditional probability:
P[D|Pos]= (P[Pos|D]xP[D]) / P[Pos]
We found P[Pos] is the last question.
So the desired answer is (0.95x0.08) / (0.0944) = 0.8051
This uses Bayes' Theorem.

Hope this helps!

Hermann
May 7th 2005, 01:19 AM
2. An art dealer receives a shipment of five old paintings from abroad, and, on the bases of past experience, she feels that the probabilities are, respectively, 0.76, 0.09, 0.02, 0.01, 0.02 and 0.10 that 0, 1, 2, 3, 4, or all 5 of them are forgeries. Since the cost of authentication is fairly high, she decides to select one of the five paintings at random and send it away for authentication. If it turns out that this painting is a forgery, what probability should she now assign to the possibility that all the other paintings are also forgeries? (answer: 0.6757)

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

Let P(first) mean the Probability that the first painting sent in is a forgery
Let P(#) mean that probability that there are # forgeries total.

P(first | 5) = 5/5 = 1.0
P(first | 4) = 4/5 = 0.8
P(first | 3) = 3/5 = 0.6
P(first | 2) = 2/5 = 0.4
P(first | 1) = 1/5 = 0.2

P(5 | first) = ? = (P(first | 5) P(5))/
[ (P(first | 5) P(5)+P(first | 4) P(4)+P(first | 3) P(3)+P(first | 2) ]P(2)+P(first | 1) P(1))

= 1(.10)/
[1(.10)+.8(.02)+.6(.01)+.4(.02)+.2(.09)]

=.10/.148
= 0.6757

danz
June 1st 2006, 10:05 AM

I a game of Yahtzee, five balanced dice are rolled simultaneously. Find the probabilities of getting,
(a) two pairs;
(b) three of a kind;
(c) a full house (three of a kind and a pair);
(d) four of a kind.