Answer in Statistics and Probability for Amelia #88827

If X has a distribution with mean “mu,” and standard deviation “sigma,” and is approximately normally distributed or “n” is large, then the sample mean M is approximately normally distributed with mean “mu” and standard error “sigma/sqrt{n}.”

“z={M-mu over sigma/sqrt{n}}”

Given that “mu=100” and “sigma=10.”

a. M = 95 for a sample of n = 4 

“z={95-100 over 10/sqrt{4}}=-1”

b. M = 104 for a sample of n = 25 

“z={104-100 over 10/sqrt{25}}=2”

c. M = 103 for a sample of n = 100

“z={103-100 over 10/sqrt{100}}=3”