For a population with a mean of µ=60 and standard deviation of Ợ=24, find the z=score corresponding to each of the following samples

a. M=63 for sample of n=16 scores
b. M=63 for sample of n=36 scores
c. M=63 for sample of n=64 scores

I assume this is for a distribution of means.

Z = (score-mean)/SEm

SEm = SD/√n

Plug in data and solve for Z.

To find the z-score for a sample, we can use the formula:

z = (X - µ) / (σ / √n)

where:
- X is the sample mean
- µ is the population mean
- σ is the population standard deviation
- n is the sample size

Let's calculate the z-scores for the given samples:

a. For a sample of n = 16 scores, with a sample mean of M = 63:
z = (X - µ) / (σ / √n)
= (63 - 60) / (24 / √16)
= 3 / (24 / 4)
= 3 / 6
= 0.5

Therefore, the z-score for a sample mean of 63 with a sample size of 16 is 0.5.

b. For a sample of n = 36 scores, with a sample mean of M = 63:
z = (X - µ) / (σ / √n)
= (63 - 60) / (24 / √36)
= 3 / (24 / 6)
= 3 / 4
= 0.75

Therefore, the z-score for a sample mean of 63 with a sample size of 36 is 0.75.

c. For a sample of n = 64 scores, with a sample mean of M = 63:
z = (X - µ) / (σ / √n)
= (63 - 60) / (24 / √64)
= 3 / (24 / 8)
= 3 / 3
= 1

Therefore, the z-score for a sample mean of 63 with a sample size of 64 is 1.