A random sample of n = 9 scores is obtained from a population with μ = 50 and σ = 9. If the sample mean is M = 59, what is the z-score corresponding to the sample mean?
z = 3.00
z = 1.00
z = 9.00
cannot determine without additional information
Sample mean from a sample of size n from a normal population with mean μ and σ has parameters N(μ,σ²/n)
Since n=9, σ=9, σ²/n=9, so the sample mean ~ N(50, 3²).
Z-score=((M-μ)/σ)=(59-50)/3=3
To find the z-score corresponding to the sample mean, we can use the formula:
z = (M - μ) / (σ / sqrt(n))
Given that the sample mean (M) is 59, the population mean (μ) is 50, the standard deviation (σ) is 9, and the sample size (n) is 9, we can substitute these values into the formula:
z = (59 - 50) / (9 / sqrt(9))
= 9 / (9 / 3)
= 9 / 3
= 3.00
Therefore, the z-score corresponding to the sample mean is z = 3.00.
To find the z-score corresponding to the sample mean, we can use the formula:
z = (M - μ) / (σ / √n)
Given that the sample mean (M) is 59, the population mean (μ) is 50, the population standard deviation (σ) is 9, and the sample size (n) is 9, we can calculate the z-score as follows:
z = (59 - 50) / (9 / √9)
z = 9 / 3
z = 3.00
Therefore, the correct answer is:
z = 3.00