00:01
Okay, so for a, we're going to use the fact that ui is equal to 1 when i -th population member is in the sample, and 0 otherwise.
00:21
So we have the population sample mean, or the sample mean of x bar as 1 over n times sum of xi, and this is equal to 1 over n, i, 1, through n, ui times xi.
00:44
And just rewriting this as n inverse, we have the final equation.
00:58
Now for b, we have probability ui equals 1 is equal to the probability of member is selected in sample from population, which is equal to n, small n over large n.
01:37
So that means that p ui equals 0 is equal to 1 minus n over large n.
01:50
So the expectation of ui, e of ui, is equal to sum of u times p u, which equals 1 times n over n plus 0 times 1 minus n over large n, which is equal to n over large n.
02:22
Now for c, the variance of ui is equal to expected value of ui squared minus expected value of ui squared, which is u squared p u minus n squared over large n squared, which is 1 squared times n over large n plus 0 squared times 1 minus n over large n minus n squared over large n squared, which is equal to n over large n squared times large n minus n...