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Probability question help?
The weekly repair cost Y for a machine has a probability density function given by f(y) = 3(1-y)^2 for 0<y<1, and 0 elsewhere.
with measurements in hundreds of dollars. how much money should be budgeted each week so that the actual repair cost will exceed the budgeted amount only 10% of the time.
I know I need to integrate, but I simply am not sure on the interval I need to integrate over.
Help?
3 Answers
- Anonymous10 years ago
You want to integrate the density function from x to 1 and set it equal to .1
Then solve for x, and x times 100 is your answer
- cidyahLv 710 years ago
Find k such that Integral (k, 1) 3(1-y)^2 dy = .10
∫ 3(1-y)^2 dy = -3(1-y)^3 /3 = -(1-y)^3
Let F(y) =-(1-y)^3
Integrating from k to 1
F(1) =0
F(k) = -(1-k)^3
F(1)-F(k) = 0 - (-(1-k)^3) = .10
(1-k)^3 = .10
3 log(1-k) = log(.10)
log(1-k)=log(0.10)/3
1-k = e^log(0.10)/3
k = 1-e^[log(0.10)/3]
k=0.5358 (in hundreds of dollars)
- Anonymous10 years ago
You're probably not going to get the correct answer here.
