Data Set 8-6 :A population mean (µ) of 5 and a standard deviation ($\sigma$) of 2.
-See Data Set 8-6. If N = 5, what proportion of the means of a sampling distribution would be greater than 6.0?
A) .4938
B) .3686
C) .3085
D) .1314
E) .0062

Question

Quizplus · Accepted Answer

To find the proportion of the means of a sampling distribution that would be greater than 6.0, we use the Central Limit Theorem. Given a population mean (µ) of 5, a standard deviation (σ) of 2, and a sample size (N) of 5, we first calculate the standard error of the mean (σx̄) = σ / sqrt(N) = 2 / sqrt(5). Then, we find the z-score for a sample mean of 6.0 using the formula z = (x̄ - µ) / σx̄. Plugging in the values, we get z = (6 - 5) / (2 / sqrt(5)). Finally, we look up this z-score in a standard normal distribution table or use a calculator to find the proportion of the area under the curve that is greater than this z-score, which corresponds to the proportion of sample means greater than 6.0. This calculation yields the result corresponding to option D.

Data Set 8-6 :A Population Mean (µ) of 5 and a Standard