Solving PDF with set boundary values

[Ein Krieger]

I am give probability distribution function f(x)=(e(-x/1000))/1000 of the time to failure of an electronic component in a copier

The question is to determine the number of hours at which 10% of all components have failed.

My solution:
1) PDF was integrated to obtain: f(x)= e(-x/1000)

2) Then, I used e(-x/1000)=0.1 with upper boundary x, and lower boundary is 0 to find x as the number of hours at which all 10% of components have failed. However, entering it in calculator, I couldn't obtain solution. What did I wrong here?

 

[Stephen Tashi]

I am give probability distribution function f(x)=(e(-x/1000))/1000 of

1) PDF was integrated to obtain: f(x)= e(-x/1000)

You're using "f(x)" inconsistently to stand for two different things and your antiderivative is missing a negative sign.

$\int \frac{e^{-x/1000}}{1000} dx = - e^{-x/1000} + C$

You can't compute a deterministic answer for the time when 10% of the components have failed since that time is a random variable. Perhaps you want to compute the time at which the probability that a component has failed then or earlier reaches .10. Your description of what you did with the calculator isn't clear.

 

[awkward]

Ein Krieger,

I am pretty sure you are leaving out one critical part of the definition of f(x). The pdf is

$$f(x) =\frac{1}{1000} e^{-x / 1000}$$
for $x \ge 0$, zero otherwise.

So you should integrate f(x) from 0 to x; you will get a different answer for the cdf than you got before.

 

[Ein Krieger]

You're using "f(x)" inconsistently to stand for two different things and your antiderivative is missing a negative sign.

$\int \frac{e^{-x/1000}}{1000} dx = - e^{-x/1000} + C$

You can't compute a deterministic answer for the time when 10% of the components have failed since that time is a random variable. Perhaps you want to compute the time at which the probability that a component has failed then or earlier reaches .10. Your description of what you did with the calculator isn't clear.

Yes. Sure. You are right. Time is continuous variable so it is inconsistent to try to define exact probability. All we need is to get probability for time range.

Ein Krieger,

I am pretty sure you are leaving out one critical part of the definition of f(x). The pdf is

$$f(x) =\frac{1}{1000} e^{-x / 1000}$$
for $x \ge 0$, zero otherwise.

So you should integrate f(x) from 0 to x; you will get a different answer for the cdf than you got before.

I have already calculated, and I got 105 hours. is it right?

 

[blue_raver22]

It seems that you have made a small error in your integration. The correct integration of f(x) would be 1 - e(-x/1000). So, your equation for solving for x would be (1 - e(-x/1000)) = 0.1. This can be solved using logarithms to find the value of x, which would be approximately 1054.8 hours. It is important to double check your integration and equations when solving problems like this, as small errors can lead to incorrect solutions. Additionally, it is always helpful to have a second set of eyes review your work to catch any mistakes. Keep up the good work in solving complex problems like this!