**In a population μ _{Y} = 100 and σ_{ Y}^{2} = 43. In a random sample of size n = 100, what is Pr (**

**Ȳ < 101)?**

The sample variance = (σ_{ Y}^{2} / n) = 43/100 = 0.43

Therefore, the Standard Error (SE) = sqrt(0.43) = 0.6557.

Normalizing this to a Standard Normal Distribution,

Z = ((101 – μ_{Y}) / SE)

Z = ((101 – 100) / 0.6557) = 1.525

The Z value is the number of Standard Errors away from the mean that will yield the desired Y value of 101.

This is a one sided hypothesis since we are interested in the probability of Y being < 101.

In EXCEL, the probability that Y is < 101 is:

=NORMDIST(Z-Value, Mean of 0, Standard Deviation of 1, 1 for Cumulative)

=NORMDIST(1.525, 0, 1, 1)

=0.9364