Homework Help: Probability w/ continuous R.V

1. Jan 2, 2014

DotKite

1. The problem statement, all variables and given/known data

A commercial water distributor supplies an office with gallons of water once
a week. Suppose that the weekly supplies in tens of gallons is a random
variable with pdf

f(x) = 5(1-x)^4 , 0 < x <1

f(x) = 0 , elsewhere

Approx how many gallons should be delivered in one week so that the probability of the supply is 0.1?
2. Relevant equations

3. The attempt at a solution

I started out by finding the cdf

F(y) = 0 , y < 0

F(y) = -(1-y)^5 + 1 , 0 ≤ y < 1

F(y) = 1 , y ≥ 1

Here is where i get lost. So y is our gallons (in tens). We want to find the amount of gallons that yields a probability of 0.1

p(y≤t) = F(t) = 0.1 thus

-(1-y)^5 + 1 = 0.1

When I solve for y I do not get the right answer which is apparently 4 gallons

2. Jan 2, 2014

vela

Staff Emeritus
Is that exact wording of the problem statement as given to you?

3. Jan 2, 2014

haruspex

Clearly the probability of any specific quantity is zero, so is the question asking for a y such that the probability of at least y is .1 or the probability of at most y is .1? you have interpreted it the second way.

4. Jan 3, 2014

HallsofIvy

I am not clear on what "the probability of the supply is 0.1" means. I suspect you mean "the probability that the amount actually supplied is at least y is 0.1". That means that you want to find y such that $$\int_0^y 5(1- x)^4 dx$$.

To integrate $\int_0^y 5(1- x)^4 dx= 0.1$ let u= 1- x so that du= -dx, when x= 0, u= 1, and when x= y, u= 1- y so that the integral becomes $5\int_1^{1- y} u^4(-du)= \int_{1- y}^1 u^4 du$

5. Jan 3, 2014

Ray Vickson

He did that integration already in his OP. The real issue seems to be that he/she does not know whether or not to solve F(y) = 0.1 or 1-F(y) = 0.1.

6. Jan 3, 2014

NasuSama

Hello, DotKite.

You did the integration correctly. The problem is that you solved for the variable incorrectly.

By definition of probability of continuous variable, if $F(y)$ is cdf, we mean:

$F(y) = P(Y \leq y)$

With a lower bound $a$, $P(a \leq Y \leq y) = F(y) - F(a)$

You also wrote out the expression correctly, but the wrong value. Try to solve for $y$ again, and you should get the correct answer.