1. Consider a population with µ=73.6 and 0 = 5.38

(A) Calculate the z-score for x =72.7 from a sample of size 45.
(B) Could this z-score be used in calculating probabilities using Table 3 in Appendix B of the text? Why or why not?

(A) I don't know if your "x" stands for an individual score or the sample mean.

If individual score, ignore sample size.

Z = (score-mean)/SD

If sample mean:

Z = (score-mean)/SEm

SEm = SD/√n

(A) We still don't have your Table 3 or know what it is.

(A) To calculate the z-score for x = 72.7 from a sample of size 45, we need to use the formula:

z = (x - µ) / (σ / √n)

Where:
- x is the observed value
- µ is the population mean (μ = 73.6 in this case)
- σ is the standard deviation of the population (σ = 5.38 in this case)
- n is the sample size (n = 45 in this case)

Substituting the given values into the formula, we get:

z = (72.7 - 73.6) / (5.38 / √45)
z = -0.9 / (5.38 / 6.708)
z = -0.9 / 0.803
z ≈ -1.12

Therefore, the z-score for x = 72.7 from a sample of size 45 is approximately -1.12.

(B) The use of Table 3 in Appendix B of the text depends on the distribution being used. If the population distribution is approximately normal (or the sample size is large enough to invoke the Central Limit Theorem), then this z-score can be used for probability calculations using Table 3.

Table 3 is typically used when you have a standard normal distribution (mean = 0 and standard deviation = 1). However, in this case, we are not dealing with a standard normal distribution, as we have specific values for the population mean and standard deviation.

To find probabilities using Table 3, we would need to standardize the values by converting them into z-scores using the population mean and standard deviation. Since we already have the z-score for the given value in this case, we would not need to use Table 3 directly. However, we can still use the z-score to calculate probabilities by referring to a z-table that includes the specific population mean and standard deviation.

So, we would need to use a z-table that corresponds to the specific population mean and standard deviation to calculate probabilities for this particular case.