# Math Help - statistics question

1. ## statistics question

Here's my question:

A plant supervisor is interested in budgeting weekly repair costs for a certain type of machine. Records over the past years indicate that these repair costs have an exponential distribution with mean 20 for each machine studied. Let Y1, Y2, Y3, Y4, Y5 denote the repair costs for five of these machines for the next week. Find a number c such that P(Y1 + Y2 + Y3 + Y4 + Y5 > c) = 0.05, assuming that the machines operate independently.

I was given in the previous problem that if Y has an exponential distribution with mean X, U = 2Y/X has a chi-squared distribution with 2 degrees of freedom.

I'm not quite sure what to do here. I think I solve for Y to get Y = UX/2, which means Y1, Y2, Y3, Y4, and Y5 each are independent chi-squared distributed random variables, each with 20 degrees of freedom. Then Y1 + Y2 + Y3 + Y4 + Y5 has a chi-squared distribution with (20)(5) = 100 degrees of freedom. Then I look at a chi-squared table for 100 d.f. and alpha = 0.05. Let c = 124.342.

Is this right and/or make sense?
Thanks for any help.

2. Originally Posted by J Flanders
Here's my question:

A plant supervisor is interested in budgeting weekly repair costs for a certain type of machine. Records over the past years indicate that these repair costs have an exponential distribution with mean 20 for each machine studied. Let Y1, Y2, Y3, Y4, Y5 denote the repair costs for five of these machines for the next week. Find a number c such that P(Y1 + Y2 + Y3 + Y4 + Y5 > c) = 0.05, assuming that the machines operate independently.

I was given in the previous problem that if Y has an exponential distribution with mean X, U = 2Y/X has a chi-squared distribution with 2 degrees of freedom.

I'm not quite sure what to do here. I think I solve for Y to get Y = UX/2, which means Y1, Y2, Y3, Y4, and Y5 each are independent chi-squared distributed random variables, each with 20 degrees of freedom. Then Y1 + Y2 + Y3 + Y4 + Y5 has a chi-squared distribution with (20)(5) = 100 degrees of freedom. Then I look at a chi-squared table for 100 d.f. and alpha = 0.05. Let c = 124.342.

Is this right and/or make sense?
Thanks for any help.

What you need to know is that the sum of $n$ exponential random variables with the same mean $\beta$ has a gamma distrinution $\mbox{Gamma}(n,1/\beta)$.

(in your case $\beta =1/20$)

The attachment shows a clip from the Wikipedia page on the exponential distribution.

RonL

3. I cannot find any tables of Gamma distributions. How would I find Gamma(5, 20)? Is there a way to turn it into a chi-squared distribution?

Thanks for the help from before.

4. Originally Posted by J Flanders
I cannot find any tables of Gamma distributions. How would I find Gamma(5, 20)? Is there a way to turn it into a chi-squared distribution?

Thanks for the help from before.
Try this (not that I understand the relation between the arguments and the parameters of the Gamma distribution there)

RonL