The number of defects in a piece of wire that is x yards long is $X\sim\operatorname{Poisson}(ux)$ for any $x>0$ The number of defects in a piece of wire that is x yards long is $X\sim\operatorname{Poisson}(ux)$ for any $x>0$.


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*If five defects are found in a 100-yard roll of wire, find a conservative one-sided upper 95% confidence limit for the mean number of defects in such a roll.

*If a total of 15 defects are found in five 100-yard rolls of wire,
find a conservative one-sided upper 95% confidence limit for u.


I think I'm supposed to use Chi-Square. The upper limit is $u_u =$ $X^2(2\sum_{i=0}^n X_i+2)/2n$
For part 1, what I did was just let n=1 and sum of x = 5, so I had $X^2(2(5)+2)/2 = 10.515$
Is this right? Any help is appreciated. Thanks!
Edit: POI means Poisson distribution, I'll edit that.  ux is just the poisson parameter u*x where x is the length of the wire in yards.
 A: One way to check your work is with simulation.  For instance, your answer to (1) can be checked in R with a command like 
mean(rpois(1e6, 10.515) > 5)

which simulates one million (1e6) trials where the true mean number of defects is $10.515$ and reports the proportion with more than $5$ defects.  The idea behind a confidence limit is that when the true mean is greater than this, you are unlikely to see as few as just $5$ defects.  This situation should be right on the borderline where there is a 95% chance of observing more than $5$ defects.  Compare the output with the intended confidence level of $0.95$.  (I obtain $95.0005\%$, which is satisfactorily close.)
As another example, general rules of thumb suggest the UCL for $u$ should be somewhere around
$$\frac{15 + 2\sqrt{15}}{5} \approx 4.55.$$
A comparable simulation confirms this value is a little low, but not by much, because its coverage of $94.24\%$ is a tiny bit lower than the intended $95\%$.
u <- (15 + 2*sqrt(15))/5
mean(rpois(1e6, u*5) > 15)

After you work out a more precise value of $u$, plug that into this simulation to see what the coverage is.
