Probability density and continuous variables

[peripatein]

Hi,
I would certainly appreciate it if you could please confirm the result I obtained to the following Statistics problem.

Homework Statement

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:
f_x(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%?

Homework Equations

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?

 

[LCKurtz]

Hi,
I would certainly appreciate it if you could please confirm the result I obtained to the following Statistics problem.

Homework Statement

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:
f_x(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%?

Homework Equations

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?

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?

 

[peripatein]

I am not sure I follow your reasoning. Would you please clarify?

 

[peripatein]

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

 

[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?

 

[peripatein]

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

 

[peripatein]

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

 

[LCKurtz]

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

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.

 

[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?

 

[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?

 

[peripatein]

It does. Thank you!