Solving Exponential Distribution Problem: Find c for P(Y1+Y2+Y3+Y4+Y5>c)=0.05

[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.

 

[kurt.physics]

Yes, your answer is correct. You have correctly identified that each of the Y variables has an exponential distribution with mean 20, and then that U = 2Y/X has a chi-squared distribution with 2 degrees of freedom. Then, since the machines operate independently, you can add the five Y variables together to get a chi-squared distribution with (20)(5) = 100 degrees of freedom. Finally, you looked in the chi-squared table for alpha = 0.05 and 100 degrees of freedom to get the value of c = 124.342.