Hopefully a quite simple integration problem

**approx** · October 5th 2008, 05:01 AM

Hi! I've got this problem:

The weekly repair cost Y for a machine has pdf given by:

$f(y) = 3(1-y)^{2}, 0<y<1$
and 0 elsewhere

w. measurements in hundreds of dollars. How much money should be budgeted each week for repair cost so that the actual cost will exceed the budgeted amount only 10% of the time?

The problem is that I don't know where to start. Should I calculate
P (budgeted cost < actual cost) = .10? I don't understand how I should think and what I need to calculate. I've calculated E(Y) and that's = 1/4, but I don't know what I should use it for. Can anyone guide me in the right direction?

Thankful for any help and suggestions.

**mr fantastic** · October 5th 2008, 05:16 AM

Originally Posted by approx

Hi! I've got this problem:

The weekly repair cost Y for a machine has pdf given by:

$f(y) = 3(1-y)^{2}, 0<y<1$
and 0 elsewhere

w. measurements in hundreds of dollars. How much money should be budgeted each week for repair cost so that the actual cost will exceed the budgeted amount only 10% of the time?

The problem is that I don't know where to start. Should I calculate
P (budgeted cost < actual cost) = .10? I don't understand how I should think and what I need to calculate. I've calculated E(Y) and that's = 1/4, but I don't know what I should use it for. Can anyone guide me in the right direction?

Thankful for any help and suggestions.

You require the value of a such that $\Pr(Y > a) = 0.1$ . Therefore solve for a in the following:

$\int_{a}^{1} 3 (1 - y^2) \, dy = 0.1$ .

Thread: Hopefully a quite simple integration problem

LinkBack

Thread Tools

Display

Hopefully a quite simple integration problem

Similar Math Help Forum Discussions

Simple integration problem

Simple Integration problem: Trigonometric Integration

simple integration problem

simple integration problem#2

Simple integration problem

Search Tags