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Jolin
August 7th 2007, 11:05 PM
Hi everyone,
I look through this question and hardly figure out how to do. The solution seem to be doesnt make sense to me. I dont think anyone will predict it will be done that way. Any comment???

Question:
As part of the underwritting process for insurance, each prospective policyholder is tested for high blood pressure. Let X represent the number of tests completed when the first person with high blood pressure is found. The expected value X is 12.5. Calculate the probability that the sixth person tested is the first one with high blood pressure.

The solution goes like this...
we assume that each test for high blood pressure is a bernoulli trial with probability of success p. Then X has the geometric dist. and 12.5=E(x)=1/p.
Therefore p=0.08. using this, we find
p(first 5 do not have high blood pressure)*p(the 6th does have high blood pressure)=(1-0.08)^5.0.08=0.053.
:goofy:

Jo_M.
August 7th 2007, 11:19 PM
Hi everyone,
I look through this question and hardly figure out how to do. The solution seem to be doesnt make sense to me. I dont think anyone will predict it will be done that way. Any comment???

Question:
As part of the underwritting process for insurance, each prospective policyholder is tested for high blood pressure. Let X represent the number of tests completed when the first person with high blood pressure is found. The expected value X is 12.5. Calculate the probability that the sixth person tested is the first one with high blood pressure.

The solution goes like this...
we assume that each test for high blood pressure is a bernoulli trial with probability of success p. Then X has the geometric dist. and 12.5=E(x)=1/p.
Therefore p=0.08. using this, we find
p(first 5 do not have high blood pressure)*p(the 6th does have high blood pressure)=(1-0.08)^5.0.08=0.053.
:goofy:

This is simply a question about the geometric distribution (i.e. the probability of having a number x of trials before a first success).

Let X--Geometric distribution with parameter p. Then P(X=x) = p*q^(x-1)

In this case, let success: event that person has high blood pressure

I strongly suggest that you know by heart that E(x) of a geo distr = 1/p. (The proof requires series expansion, which I will not show here) Therefore, 12.5=1/p --> p=0.08 ---> q = 1-p = 0.92.

Then it is simply a matter of replacing p=0.08, q=0.92 and x=6 in the formula above.

(edited a typo..)

jthias
August 8th 2007, 12:01 AM
This is simply a question about the geometric distribution (i.e. the probability of having a number x of failures before a first success).

Let X--Geometric distribution with parameter p. Then P(X=x) = p*q^(x-1)

In this case, let success: event that person has high blood pressure

I strongly suggest that you know by heart that E(x) of a geo distr = 1/p. (The proof requires series expansion, which I will not show here) Therefore, 12.5=1/p --> p=0.08 ---> q = 1-p = 0.92.

Then it is simply a metter of replacing p=0.08, q=0.92 and x=6 in the formula above.

If I can recall correctly there are basically two different geometric distributions.

Let X and Y represent these..

If we define X as the number of failures (out of n independent trials) until the first success, where each trial has probability p of success, then

P(X=0) = p, P(X=1) = qp, P(X=2) =q^2*p,...

so the probability function looks like px(x) = q^x*p, x=0,1,2,3,...

and has E(X) = q/p.

If, however, we define Y to be the the trial number on which the 1st success occurs, then P(Y=0) = 0 which differs from the previous case. Also we have...

P(X=1) = p, P(X=2) =qp, P(X=3) =q^2*p,...

and the corresponding probability function is py(y) = q^(y-1)*p, y=1,2,3,...

and has E(Y) = 1/p.

In this problem we're dealing with the trial number on which the first success occurs.

So the random variable is of the form py(y) = q^(y-1)*p, y=1,2,3,...

and has expected value 1/p.

Y starts at a value of 1. If we were to let U = Y-1

then pu(u) = q^u*p, u=0,1,2,3,...

Then U is of the same form as X given earlier, and so

U is the number of failures until the 1st success. If we calculate the expected value of Y we will get...

E(Y) = E(U+1) = E(U) + 1 = q/p + 1 = (1-p)/p + 1 = 1/p so that we still get the same expected value.

Jolin
August 8th 2007, 09:22 PM
Well, i understand the solution...but i am just wondering how you know that you will have to use geometric distribution???

JGET
August 8th 2007, 10:09 PM
Based on this line you should be able surmise its a geometric distribution.

" Let X represent the number of tests completed when the first person with high blood pressure is found."

JGET

jthias
August 8th 2007, 10:26 PM
Well, i understand the solution...but i am just wondering how you know that you will have to use geometric distribution???
You use a geometric distribution (one of two) whenever you have n independent trials performed in succession each with the same probability of success p (and prob. of failure 1-p), and you want to determine the probability of either "the number of trials that end in failure until you get the first success" or the probability that "kth trial will be the first that yields a successful outcome".

Let X be the number of failure until the 1st success.

px(x) = q^x*p, x=0,1,2,3,...

Let Y be number of the trial (1st, 2nd, 3rd,..) on which the first successful outcome occurs.

py(y) = q^(y-1)*p, y=1,2,3,...

Notice P(X=0) = P(Y=1), P(X=1) = P(Y=2),...

X and Y are diffenrent random variables but can be equated via the transformation

X = Y-1

p = P(X=0) = P(Y-1=0) = P(Y=1) = p

qp = P(X=1) = P(Y-1=1) = P(Y=2) = qp, and so on.

Jolin
August 8th 2007, 10:37 PM
Thanks guys...i deeply understand now...

billvp
August 8th 2007, 10:56 PM
the geometric distribution is just a variation of the binomial, and if you know how to do binomial, you should be able to do the problem anyways - IF you know to get the probability from the expected value

that is pretty intuitive: for example, on average, how many tosses do you think it will take of a fair coin to get a "heads"? it would seem logical to figure 2, and the probability of a "heads" is 1/2

on average, how many rolls of a die would it take to get a 3? since all numbers 1-6 are equally probable, it would on average take 6, so the probability is 1/6, and in this way you can see that the probability will end up 1/E(X)

then if you have the probability (of high b.p. in this case), you can do the problem using the binomial distribution - it is the probability of getting 5 with low b.p., THEN 1 with high b.p., and there is only one possible way to do that: LLLLLH = (0.92)(0.92)(0.92)(0.92)(0.92)(0.08) = 0.053

Jolin
August 8th 2007, 11:11 PM
That is a very good intuitive...i am appreciated with ur detail explaination. I wish i can think like you are...Thanks!

Ken
August 9th 2007, 01:47 AM
Well, i understand the solution...but i am just wondering how you know that you will have to use geometric distribution???

After doing many similar problems, you'll think this is a brainless question.

Jo_M.
August 9th 2007, 08:59 AM
Well, i understand the solution...but i am just wondering how you know that you will have to use geometric distribution???

It shouldn't be long before you start recognizing immediately if you should use a binomial, geometric, negative binomial or hypergeometric distribution simply with some key words in the problems.