00:02
Hello, so in this question we have a log normal distribution with theta equals two number squared with mine.
00:18
So the cdf of log normal distributions, let's look at that.
00:29
So the function f of x that is given by phi lymn x minus theta divided by omega.
00:44
So here we can see that omega is going to be 43 because it's not the square root of 9.
00:52
So for the first part, if you want to find the probability of x less than 5 ,000, then we just use the formula because this is the same as f of x or this is same as f5 ,000.
01:08
So we just did a substitution.
01:12
So we have linn on 5 ,000 minus data 2 over omega 3.
01:25
If you crank that out, you'll get 1 .4033.
01:34
So you can look this up from the z table, okay? and therefore, 1 .4033, you get the probability of 0 .92 from the z table.
01:51
So just assume that this is a z value, right? you look up 1 .40...