Question

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.

Answer 1

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

Answer 2

sorry, it's 2e^-2

Answer 3

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

Answer 4

and 0, yes

Answer 5

But, it's basically just what you said.

Answer 6

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

Answer 7

f(x)=2e^-2x

Answer 8

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

Answer 9

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

Answer 10

I thought it was -e?

Answer 11

What exactly did you think was -e?

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

thats exactly what i wrote.

Answer 16

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

Answer 17

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.

Answer 18

Ah Ok.

Answer 19

So, how do you come to the answer?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

After substitution, I got a strange answer...

Answer 24

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

Answer 25

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