00:01
Okay, we're going to do 7 .76 for a.
00:08
We need to find p.
00:12
Let's denote this as p star, such that the variance of y over n is maximized.
00:25
So the way we're going to do this is to find the local maximum or the global maximum of variance of y over n by taking the derivative and setting it to 0.
00:37
So we're going to take the derivative of variance with respect to p and this gives now variance of y, since y follows binomial of n np, we have np times 1 minus p.
01:15
So we're just going to take the derivative and you can cancel these out and get 1 minus p minus p.
01:33
So this is 1 of n times 1 minus 2 p.
01:39
So when we set this equation to 0, you get 1 equals 2p, which implies p is equal to 1 of 2.
01:53
0 .5.
02:00
So in order to confirm that this is the maximizer of the variance, we need to do the second derivative test.
02:08
So this is our first derivative, and we're just going to take the derivative again.
02:15
That is the square, dp square variance of y of n.
02:23
This is equal to d, d.
02:33
D .p.
02:34
1 .n.
02:35
Minus 2p which equals minus 2 over n right and since n is always larger than 0 we have that the minus 2 of n this is less than 0 and we conclude that this p is the maximizer of the variance so now we're going to move on to b so the what is the the question asking is the value of n that guarantees that let me rewrite this one value of n that guarantees that y of n will be within point one of the true fraction of the defectives so what it means is y of n is within the true p this is the true p this is a true p minus 0 .1 and p plus 0 .1 and this probability needs to be 0 .95.
04:17
So 95 % of the time the sample, the proportion of the defectives in the sample is within the 0 .1 and 0 .1 probability, within the point 1 of the true probability...