Let ( W1, W2, ldots ) be an infinite sequence of independent...

Let ( W_{1}, W_{2}, ldots ) be an infinite sequence of independent random variables, each with the same distribution. Suppose that ( Eleft[W_{i} ight]=1 ) and ( operatorname{Var}left(W_{i} ight)=9 ) for all ( i geq 1 ). For any integer ( n ), let ( ar{W}^{n}=frac{1}{n} sum_{i=1}^{n} W_{i} ). What is ( lim _{n ightarrow infty} Pleft(ar{W}^{n} geq 1.96 ight) ) ? 1 0.05 0 0.025

Transcript

00:01 So in this question we're told that w1, w2, etc., is an infinite sequence, infinite sequence of independent and identically distributed random variables.

00:24 So the expected value of wi is equal to 1, and the variance of wi is equal to 9 for all i greater than a equal to 1.

00:39 Now w bar n is 1 over n times the sum from i equals 1 to n of w i.

00:52 So first of all, we need to use the central limit theorem to tell us that the limit as n goes to infinity of w bar n is, so well first of all let's just write down what the central limit theorem tells us.

01:14 So the central limit theorem tells us that when we have an infinite sequence of independent identically distributed random variables with mean equals mu and standard deviation equals or a variance equals sigma squared, then the limit as n goes to infinity, and let's call these variables xi, the limit as n goes to infinity of a, uh, a limit as n goes to infinity of, uh, 1 over n sum from i equals 1 to n x i minus mu divided by sigma is normally distributed with mean zero and variance 1 as long as they as long as there is a mean and a well -defined mean and variance so what does that mean? well that means that the limit as n goes to infinity of w bar n minus 1 divided by the square rate of 9 which is 3 is normally distributed with means zero and variance 1.

02:36 So what we are looking at is the limit as n goes to infinity of the probability that w bar n is greater than or equal to 1 .96.

02:53 So this is equal to the limit as n goes to infinity of the probability that w bar n minus 1 over 3 is greater than or equal to 1 .96 minus 1 over 3, which is 0 .32.

03:20 Oh, sorry, this should be the standard deviation on the mean, and this standard deviation on the mean is the standard deviation divided by the square root of n.

03:34 So that means that the standard deviation is 3, so we need to divide by the square root of n, which means multiplying by root n up here.

03:50 So this what we have here is the probability that wm minus 1 over 3 times root n so we can divide by the square root of n...

Question

Please give Ace some feedback