Discuss the exam! It's past 9!

[jy006m]

Originally Posted by Generalshamu

No I did not get this one sorry...I got one where it had to deal with a husband and wife's expected lifetimes/probability of their deaths...

I did get that one. I'm surprised that we didn't have the same exam though, given that I had at least 5 of the same questions you did. Maybe that question that I posted was so easy that you didn't spend but a minute thinking about it, so that's why you don't remember it?

[daaaave]

Originally Posted by gosuruss

my friend got a problem where it was

you roll a die until you get three sixes, calculate variance of x (amount of times you roll)

took us a while to figure it out at home but it is just negative binomial, rq/(p^2)

3(5/6)/(1/6)^2

Alternatively, the variance of the time until getting 3 sixes is 3 times the variance of the number rolls needed for the first (the number of rolls between the first six and the 2nd six is equal in distribution to the number of rolls for the first six, and is also independent. Same thing for the number of rolls between the second and third six). The number of rolls until getting the first six is a geometric and is also easier to calculate directly.

[pete2441]

I had a problem that said:

A company pays a benefit of 50 for each of its 1000 employees if they die in the next year. The probability that an employee dies in the next year is .14%. What is the amount the company needs to have in a fund or in reserve(or something of that nature) in order to ensure a 99% chance they can cover the loss.

The wording is a bit fuzzy but if anyone has any ideas that would be great.

[Gandalf]

Originally Posted by pete2441

I had a problem that said:

A company pays a benefit of 50 for each of its 1000 employees if they die in the next year. The probability that an employee dies in the next year is .14%. What is the amount the company needs to have in a fund or in reserve(or something of that nature) in order to ensure a 99% chance they can cover the loss.

The wording is a bit fuzzy but if anyone has any ideas that would be great.

Assuming independence, the number of deaths has a binomial distribution, so you can compute the mean and variance. Then, using the normal approximation, you can determine n such that probability (deaths > n) < 1%. Then you need 50n to have a 99% chance of paying them.

It is conceivable they didn't want you to use the normal approximation; you could calculate the exact probability of 0,1,2 deaths.

[SooMe]

I think the probability that someone died was 1.4%. I'm interested in the solution as well.

[Gandalf]

If 1.4%, definitely the normal approximation was expected; too many cases to do exactly.

[SooMe]

What was the strict solution? I got E(x) = 700, Var(x) = npq = 690.2, SD(x) = 26.27. So Pr[(X - 700)/26.27] = Phi(2.33) = 0.99. I know something went wrong somewhere.

[gosuruss]

i had that problem

phi (x-np)/(sqrt(npq)) = .99

figured out x, multiplied by 50

[colby2152]

I used the continuity correction for that problem and got an answer that wasn't one of the 5, so I went with 750 as the closest.

[Gandalf]

Assuming 1000 lives with probability of death 0.014, the expected number of deaths is 14. The variance of the number of deaths is 13.804. The standard deviation is 3.715. For 99% probability, you need to prepare for 14 + 2.326*3.715 = 22.64 deaths, if it were continuous. As a discrete distribution, for 99% probability you need to be prepared for 23. 23 * 50 = 1150.