How Many Gallons Should Be Delivered to Have a Probability of 0.1?

[DotKite]

Homework Statement

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?

Homework Equations

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

 

[vela]

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

 

[haruspex]

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

p(y≤t) = F(t) = 0.1 thus

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

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.

 

[HallsofIvy]

Homework Statement

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?

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 $$\int_0^y 5(1- x)^4 dx$$.

Homework Equations

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

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 $5\int_1^{1- y} u^4(-du)= \int_{1- y}^1 u^4 du$

 

[Ray Vickson]

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 $5\int_1^{1- y} u^4(-du)= \int_{1- y}^1 u^4 du$

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.

 

[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.