A normally-distributed population has a mean μ = 132 and a standard deviation σ = 20. For a set of samples of size 60 taken from this population,

Question

A normally-distributed population has a mean μ = 132 and a standard deviation σ = 20. For a set of samples of size 60 taken from this population, what is the z-value for the proportion of the sample means that will be less than 122?

Answer 1

For this, there are three main steps to calculate: (1) the difference in the target value of 122 and the mean value of 132, and (2) the sample standard deviation of this population. Then, (3) you divide the difference over the sample standard deviation to get the z-score.

The ultimate equation we're trying to fill in is:

z = (X - µ) / (σ/√n)

Here, X = 122; µ = 132; σ = 20; n = 60.

z = (122 - 132) / (20/√60) = -10/ (20/7.746) = -10 / 2.582 = -3.873

Just to a "gut check" if this makes sense, since we're investigating a target X value that is already less than our population mean of 132, we know that the z-score should be negative to indicate that we are less than the mean value. Additionally, we know we need to take into consideration the sample size to get the specific sample standard deviation (s = σ/√n) because the question specifies we are looking in this sample, not the total population. If it indicated we wanted the z-score in the "true population" or "total population," then we would have used the standard deviation they had provided.