Suppose μ = 80 and σ = 20 for a population. In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall above 76.71.

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Quizplus · Accepted Answer

This statement is true because the sample mean follows a normal distribution with mean μ and standard deviation σ/sqrt(n), which in this case is 20/sqrt(100) = 2. Therefore, a 95% confidence interval for the sample mean is given by μ ± 1.96(σ/sqrt(n)), which is 80 ± 1.96(2) = (76.08, 83.92). Since we want to find the probability that the sample mean is above 76.71, we can find the z-score for this value using z = (76.71 - 80) / 2 = -1.645. The area to the right of this value under a standard normal curve is 0.95, which means that 95% of all possible sample means will fall above 76.71.

Suppose μ = 80 and σ = 20 for a Population