# Homework Help: Probability density and continuous variables

1. Apr 21, 2013

### peripatein

Hi,
I would certainly appreciate it if you could please confirm the result I obtained to the following Statistics problem.
1. The problem statement, all variables and given/known data
A tank is supplied with fuel once a week. If the fuel (in thousands of liters) that the station sells in a week is a random variable which is distributed with the following density:
fx(x) = c(1-x)4 for 0≤x≤1 and 0 otherwise
what ought to be the capacity of the tank so that the probability that it is emptied within a week is less than 5%?

2. Relevant equations

3. The attempt at a solution
Since x is a random variable, with continuous distribution, it must fulfill the following conditions:
f(x)≥0 and ∫f(x)dx between 0 and 1 must be equal to 1. Hence, c = 5.
I then found F(x), the cumulative distribution, to be:
0 for x<0, x5 for 0≤x≤1, and 1 for x>1.
Which inequality must now be written in order to assure the required condition? Is it x5<0.05, yielding 550 liters?

2. Apr 21, 2013

### LCKurtz

Your work up until the last couple of lines is fine, but your guess of 550 liters doesn't pass the snicker test. After all, if he sold all his gas it would be 1000 liters. He usually doesn't sell all of it, so he could use a smaller tank if he is willing to run out less than 5% of the time. How much doesn't he exceed selling 95% of the time?

Last edited: Apr 21, 2013
3. Apr 21, 2013

### peripatein

4. Apr 21, 2013

### peripatein

If I am not mistaken, the answer should be 1-(0.05)^(0.2) = 450 liters. But why?

5. Apr 21, 2013

### LCKurtz

Let's step back a bit. I overlooked that your cumulative distribution function of $x^5$ doesn't seem to be correct. How did you get that?

6. Apr 21, 2013

### peripatein

I integrated 5(1-x)^4 between 0 and x.

7. Apr 21, 2013

### peripatein

My apologies, it should have been 1-(1-x)^5. How shall I proceed?

Last edited: Apr 21, 2013
8. Apr 21, 2013

### LCKurtz

That's better. What value of X would assure that you don't run out 95% of the time? That may lead you to one of your previous answers, but hopefully now you will see why.

9. Apr 21, 2013

### peripatein

I see where this is going and still am unable to fathom why it shouldn't be 1-(1-x)^5 = 0.05? I mean, let's say the capacity of the tank is r gallons (or liters). If r gallons/liters are then sold in a given week, the tank is exhausted, which is the desired outcome merely 5% of times. Right? Ergo, I would think the equation should be 1-(1-r)^5 = 0.05, instead of 1-(1-r)^5 = 0.95. Now, I KNOW I am wrong, but cannot understand why. Would you please care to explain?

10. Apr 21, 2013

### LCKurtz

You want the probability that you could have sold more than your new tank holds to be small (.05). That means you want the probability that the amount sold is less than equal to your new tank capacity to be large (.95). That's F(x)=.95. Does that help?

11. Apr 21, 2013

### peripatein

It does. Thank you!