The life, X, of the StayBright light bulb is modelled by the probability density function\[f(x)=\left(\begin{matrix}2x^{-2} \\ 0\end{matrix}\right)\]\[x \ge 0\]otherwise.
Where X is measures in thousands of hours
Find the probability that a StayBrite bulb lasts longer than 1000 hours.

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- anonymous

- schrodinger

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- anonymous

f(x) is the function sine. Ok... How do you do that?

- anonymous

sorry, it's 2e^-2

- anonymous

so \[f(x)=2e ^{-2}\]

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- anonymous

and 0, yes

- anonymous

But, it's basically just what you said.

- anonymous

Oh no! Wait. It's e^-2x

- anonymous

f(x)=2e^-2x

- anonymous

Then integrate it from 1000 to infinty \[\int\limits_{1000}^{\infty} e ^{-2x}=-e ^{-2x}/2\]

- anonymous

putting limits you get (e^-2000)/2.

- anonymous

I thought it was -e?

- anonymous

What exactly did you think was -e?

- anonymous

No I mean, you wrote\[ -e^{-2x}\over2\]

- anonymous

\[\int\limits_{}^{}e ^{ax} = e ^{ax}/a\]

- anonymous

Yes, so, wouldn't it be -1/2 times e^-2x?

- anonymous

thats exactly what i wrote.

- anonymous

Yes, but when you wrote e^-2x putting in limits, would the e still carry the - sign?

- anonymous

Substitue the limits with the minus sign you get the answer that i wrote only if a minus sign is there then we can reverse the limits and remove the minus sign(Property of definite integral.

- anonymous

Ah Ok.

- anonymous

So, how do you come to the answer?

- anonymous

\[e ^{-\infty}=0\]

- anonymous

I know the answer... Um, why do you do\[ e^{-\infty}=0?\]

- anonymous

I just don't know how to get to the answer.

- anonymous

After substitution, I got a strange answer...

- anonymous

the limit is infinity right, so you get e^-2x= e power minus infinty which is 0.

- anonymous

e^infinity= infinity so e^-infinity=1/infinity=0 (very vrey roughly). You can also observe from graph of e^-x|dw:1328079805509:dw|

- anonymous

Um, that's true, but the answe in my book says the P = 0.454?

- anonymous

Then you would need a scientific calculator

- anonymous

It's not manditory here... It's statistics 2 A-levels...

- anonymous

It's continuous random variables for the density functions.

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