The $Fcdf($ returns a probability, not the actual number. For example $Fcdf(0,3.33,5,10) = 0.9502$ For this type of question, you need to find where $Fcdf(lower, upper, df1, df2)=P$ where you are given lower, df1, df2, and P. If we look at the problem you want to find the upper five percent. This is equivalent to the whole, minus 5%. So P is 1-0.05 which is 0.95. We then substitute into the function: $Fcdf(0,U,5,10)=.95$

If we wanted to do the calculus, it would look something like this:

(1)
\begin{align} \int_0^U \frac{\left( \frac{5x}{5x+10} \right)^{5/2} \left(1-\frac{5x}{5x+10}\right)^{5}}{x \operatorname{B}(5/2,5)} = .95 \end{align}

Needless to say, that calculation is not pretty nor is it fun. Thankfully, we can create a program to find this for us. Here is the code I would use. Its not very efficient, as it brute forces it, but it works. The way to use it is to input your degrees of freedom, the probability you want (in this case, .95), the step value (how much you want each iteration to increase by), and the value to start testing at. If you start testing at a value which is above the correct answer, it will run once and exit. If you start below, it will test each value until it reaches the probability you want. It does require a guess, but if your guess is way too low, you can just use On to break the program and try with a higher starting value.

```
Input "DF1:",A
Input "DF2:",B
Input "P:",P
Input "Step:",S
Input "Start:",C
ClrHome
Repeat Ans≥P
C+S→C
Fcdf(0,C,A,B
Output(1,1,C
Output(2,1,Ans
End
```