Data Set 8-9: A population mean (µ) of 4 and a standard deviation ($\sigma$) of 2. (Carry 3 or 4 decimal places in your calculations.)
-See Data Set 8-9. If N = 5, what proportion of the means of sampling distribution would be larger than 6?
A) .4875
B) .3413
C) .1587
D) .0125

Question

Quizplus · Accepted Answer

The proportion of the means of the sampling distribution that would be larger than 6 can be found using the standard error of the mean and the Z-score formula. The standard error (SE) of the mean is calculated as the population standard deviation (σ) divided by the square root of the sample size (N), which is SE = 2 / sqrt(5). To find the Z-score for a sample mean of 6, we use the formula Z = (X - μ) / SE, where X is the sample mean (6), and μ is the population mean (4). Plugging in the values, we get Z = (6 - 4) / (2 / sqrt(5)) = 2.236. Looking up this Z-score in a standard normal distribution table or using a calculator, we find that the proportion of the area to the right of this Z-score (which represents the proportion of sample means larger than 6) is approximately 0.0125.

Data Set 8-9: a Population Mean (µ) of 4 and a Standard