00:01
In this question, we are given x1 to xn, are independent and identically distributed random variables from a distribution with a mean of mu and variance of sigma square.
00:18
Now, n is large and t is the sum of xi.
00:23
I want to show that probability of t greater than n times mu plus square root of n times sigma is 0 .1587.
00:40
Xi follows a distribution with mean mu, variance sigma square and t is x1 plus x2 all the way to xn.
01:00
Now, since n is large, by central limit theorem, if your n is large, usually n is greater or equal to 30, t will actually follow a normal distribution approximately.
01:17
No matter what the parent distribution is, as long as your n is large enough, the sum of the n samples from this population will follow the normal distribution.
01:35
Now, the mean will be n mu and the variance will be n sigma square.
01:48
So now, that means that the standard normal z will be t minus its mean, which is n mu, divided by the standard deviation of t and that will be the square root of this.
02:05
So that will be square root n sigma.
02:09
So this standard normal will follow the normal distribution, mean is 0, standard deviation is 1, variance is 1.
02:19
So now, let's look at probability t greater than n mu plus square root n sigma.
02:29
I'm going to minus n mu on both sides.
02:41
This balances.
02:43
You can see that this will cancel.
02:47
So leaving me with t minus n mu greater than square root n sigma...