## anonymous 4 years ago 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.

1. anonymous

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

2. anonymous

sorry, it's 2e^-2

3. anonymous

so $f(x)=2e ^{-2}$

4. anonymous

and 0, yes

5. anonymous

But, it's basically just what you said.

6. anonymous

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

7. anonymous

f(x)=2e^-2x

8. anonymous

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

9. anonymous

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

10. anonymous

I thought it was -e?

11. anonymous

What exactly did you think was -e?

12. anonymous

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

13. anonymous

$\int\limits_{}^{}e ^{ax} = e ^{ax}/a$

14. anonymous

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

15. anonymous

thats exactly what i wrote.

16. anonymous

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

17. 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.

18. anonymous

Ah Ok.

19. anonymous

So, how do you come to the answer?

20. anonymous

$e ^{-\infty}=0$

21. anonymous

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

22. anonymous

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

23. anonymous

After substitution, I got a strange answer...

24. anonymous

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

25. 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|

26. anonymous

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

27. anonymous

Then you would need a scientific calculator

28. anonymous

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

29. anonymous

It's continuous random variables for the density functions.