anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
f(x) is the function sine. Ok... How do you do that?
anonymous
  • anonymous
sorry, it's 2e^-2
anonymous
  • anonymous
so \[f(x)=2e ^{-2}\]

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anonymous
  • anonymous
and 0, yes
anonymous
  • anonymous
But, it's basically just what you said.
anonymous
  • anonymous
Oh no! Wait. It's e^-2x
anonymous
  • anonymous
f(x)=2e^-2x
anonymous
  • anonymous
Then integrate it from 1000 to infinty \[\int\limits_{1000}^{\infty} e ^{-2x}=-e ^{-2x}/2\]
anonymous
  • anonymous
putting limits you get (e^-2000)/2.
anonymous
  • anonymous
I thought it was -e?
anonymous
  • anonymous
What exactly did you think was -e?
anonymous
  • anonymous
No I mean, you wrote\[ -e^{-2x}\over2\]
anonymous
  • anonymous
\[\int\limits_{}^{}e ^{ax} = e ^{ax}/a\]
anonymous
  • anonymous
Yes, so, wouldn't it be -1/2 times e^-2x?
anonymous
  • anonymous
thats exactly what i wrote.
anonymous
  • anonymous
Yes, but when you wrote e^-2x putting in limits, would the e still carry the - sign?
anonymous
  • 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
  • anonymous
Ah Ok.
anonymous
  • anonymous
So, how do you come to the answer?
anonymous
  • anonymous
\[e ^{-\infty}=0\]
anonymous
  • anonymous
I know the answer... Um, why do you do\[ e^{-\infty}=0?\]
anonymous
  • anonymous
I just don't know how to get to the answer.
anonymous
  • anonymous
After substitution, I got a strange answer...
anonymous
  • anonymous
the limit is infinity right, so you get e^-2x= e power minus infinty which is 0.
anonymous
  • 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
  • anonymous
Um, that's true, but the answe in my book says the P = 0.454?
anonymous
  • anonymous
Then you would need a scientific calculator
anonymous
  • anonymous
It's not manditory here... It's statistics 2 A-levels...
anonymous
  • anonymous
It's continuous random variables for the density functions.

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