**In a population μ _{Y} = 100 and σ_{ Y}^{2} = 43. In a random sample of size n = 64, what is Pr (101 < Ȳ < 103)?**

The sample variance = (σ_{ Y}^{2} / n) = 43/64 = 0.671875

Therefore, the Standard Error (SE) = sqrt(0.671875) = 0.81968.

Normalizing this to a Standard Normal Distribution,

Z = ((103 – μ_{Y}) / 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 – μ_{Y}) / 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