00:01
And this question will look at the age of doctors in a certain hospital, right? and you are told that the age which i'm going to denote by x of a doctor in this hospital is randomly distributed.
00:13
So it has a random, it has a normal distribution with a mean of 45 years and a standard variation four.
00:22
And you ask if you take a random sample of 10 doctors and you look at the, average mean age of these 10 octaves, which i'm going to call x bar, obviously must also follow as a normal distribution with the same mean, but a smaller standard variation, which should be given by 4 divided by squalidum sample size, which is 10.
00:47
And you'll find this to be given by 4d5 and 4 divided by square of 10, 1 .26.
01:00
Okay.
01:02
So that's a distribution of this mean age.
01:04
And then you asked to find the probability, right, that the mean age of this doctor less than 43 .8 years, right? so you look at distribution of x bar is something like this, right? and it's the same to the border 45.
01:20
Now you look at the 43 .8, where it's somewhere here, right, 43 .8.
01:25
Less than that is you look at the area here, right? so in other words, this is going to be px less than 43 .8.
01:32
That's going by alva, basically.
01:33
Otherwise, the area of the tail to the left of this 43 .8.
01:39
And of course, you can find alpha from the z squared.
01:41
You look at z squared alpha, and that's given by 45 minus 43 .8 divided by standard variation, which is 1 .26.
01:49
You find this to be given by 0 .95.
01:55
And then you look at the d table, you'll find this to be given by.
01:59
You'll find alpha given by 17 .1%.
02:08
So this is...