Can someone help me with sampling distribution of the sample mean?

Question

A population has mean 1,542 and standard deviation 246.

a) Find the mean and standard deviation of Xbar for samples of size 100.

b)Find the probability that the mean of a sample of size 100 will be within 100 units of the population mean, that is, between 1,442 and 1,642.

Answer 1

a) mean is 1542 and std dev = 246/sqrt(100) = 246/10 = 24.6 = standard error (SE)

b) want to find P(1442 < xbar < 1642)

z = (xbar - pop mean)/SE = (xbar - 1542)/24.6

P(1442 < xbar < 1642) =

P( (1442 -1542)/24/6 < z < (1642-1542)/24.6) =

P(-4.06 < z < 4.06) =

P(z <4.06) - P(z < -4.06) ~ 100%