00:01
Hello students, for the part a, let xi is equal to 1 if p and 0 if 1 -p, where i runs to 1 to n.
00:19
Then expectation of xi is equal to summation x into probability of x, that is equal to 1 into p plus 0 into 1 -p, that is equal to p.
00:33
Now variance of xi is equal to summation x square into p of x minus summation x into p of x whole square, that is equal to 1 square into p plus 0 square into 1 -p whole square, which is equal to minus p square, p minus p square, that is equal to p into 1 -p.
01:10
Now, therefore, mean that is equal to expectation of x bar n is equal to expectation of 1 by n summation i runs to 1 to n, xi is equal to 1 by n summation i runs to 1 to n, expectation of xi that is equal to 1 by n into summation i runs to 1 to n p, that is equal to n p divided by n, which is equal to p.
01:43
Therefore, mean is p and variance of x bar n, that is equal to variance of x bar n is equal to variance of 1 by n summation i runs to 1 to n, xi, which is equal to 1 by n square into summation i runs to 1 to n variance of xi, that is equal to 1 by n square into summation i runs to 1 to n probability that 1 minus p, which is equal to p into 1 minus p divided by n, this is the variance.
02:23
Now, part b, when n is large, then by central limit theorem x n bar minus expectation of x n bar divided by variance of x n bar into whole power 1 by 2 follows normal with 0 comma 1.
03:03
Now, part c, given n is equal to 100 and p is equal to 0 .2...