In a population μY = 100 and σ Y2 = 43. In a random sample of size n = 64, what is Pr (101 < Ȳ < 103)?
The sample variance = (σ Y2 / n) = 43/64 = 0.671875
Therefore, the Standard Error (SE) = sqrt(0.671875) = 0.81968.
Normalizing this to a Standard Normal Distribution,
Z = ((103 – μY) / SE)
Z = ((103 – 100) / 0.81968) = 3.66
The Z value is the number of Standard Errors away from the mean that will yield the desired Ȳ value of 103.
This is a one sided hypothesis since we are interested in the probability of Ȳ being < 103.
In EXCEL, the probability that Ȳ is < 103 is:
=NORMDIST(Z-Value, Mean of 0, Standard Deviation of 1, 1 for Cumulative)
=NORMDIST(3.66, 0, 1, 1)
=0.999874
Using the same logic for the probability of Ȳ being < 101,
Z = ((101 – μY) / SE)
Z = 1 / 0.81968 = 1.22
NORMDIST(1.22, 0, 1, 1) = 0.8888
The difference between the two is the probability of Ȳ being between 101 and 103.
Answer: 0.1111