Consider a normal population with μ = 36 and σ = 4.9. Calculate the z-score for an x of 47.6 from a sample of size 14. (Give your answer correct to two decimal places.)
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Z = (score-mean)/SD
If you know the mean and SD for the population, sample size is extraneous information for determining the Z score for an individual raw score.
Could this z-score be used in calculating probabilities using Table 3 in Appendix B? Why or why not?
To calculate the z-score for an x value from a sample, you can use the formula:
z = (x - μ) / (σ / √n)
Where:
x = the value of interest (47.6 in this case)
μ = the mean of the population (36)
σ = the standard deviation of the population (4.9)
n = the sample size (14)
Plugging in the values into the formula:
z = (47.6 - 36) / (4.9 / √14)
Calculating √14:
√14 ≈ 3.74
z = (47.6 - 36) / (4.9 / 3.74)
Calculating (4.9 / 3.74):
(4.9 / 3.74) ≈ 1.31
z = (47.6 - 36) / 1.31
Calculating (47.6 - 36):
(47.6 - 36) = 11.6
z = 11.6 / 1.31
Calculating (11.6 / 1.31):
(11.6 / 1.31) ≈ 8.86
Therefore, the z-score for an x of 47.6 from a sample of size 14 is approximately 8.86.