00:01
And this question we're told that the mean lifetime of a bulb is 1 ,200 hours, and its lifetime is exponentially distributed.
00:06
So x is going to be exponentially distributed with a rate parameter of 1 over the mean, so that's 1 over 1 ,200.
00:14
So that means that the pdf of x, fx of x, is 1 over 1 ,200, e to the minus x over 1 ,200 for x greater than 0.
00:26
So first of all, we're asked to find the probability that the bulb will last less than it's guaranteed lifetime over a thousand hours.
00:40
So the probability that x is less than 1 ,000 is going to be the integral from 0 to 1 ,200, 1 over 1 ,200, e to the minus x over 1 ,200, dx, which is minus e to the minus x over 1 ,000, which is minus e to the minus x over 1 ,200, from 0 to 1 ,000, so that's going to be 1 minus e to the minus 1 ,000 divided by 1 ,200.
01:07
So 1 minus e to the minus 1 over 1 .2.
01:14
Oh, sorry, e to the minus 1 over 1 .2 gives me 0 .5654.
01:24
So that's the probability x is less than 1 ,000.
01:32
Okay.
01:34
So now let's find out, so in a batch of lightbox, what's the expected time until half the light bulbs in the batch fail? so if we have, if we have n light bulbs in a batch, then what's the, well, what's the expected time until half of them have failed? well that's just the time, that's just the median time...