00:01
Here we are told that the length of time that a user views a web page has a log normal distribution with theta equals 0 .5 and omega squared equals 1.
00:14
For part a, we are asked for the probability that the viewing time is greater than 10 seconds.
00:29
This is equal to 1 minus the probability that x is less than or equal to 10.
00:37
And it's equal to the probability, 1 minus the probability, that z is lessen or equal to natural log of 10 minus theta over omega.
01:03
This is equal to 1 minus the probability that z is less than or equal to 1 .8026.
01:17
This gives a final probability of approximately 0 .0357.
01:28
And then for part b we were asked by what length of time have half the users stopped viewing the web page.
01:38
So the probability that the time is less than a certain time is equal to .5.
01:48
This means that the probability that z is less than or equal to natural log of x minus .5 over 1 is equal to 0 .5.
02:08
And we know that the middle of a standard normal distribution is zero...