Final answer:
To find the required sample size so that the sample mean is at least 2 units higher than the population mean with probability ≤ 0.02, calculate the Z-score for the 98th percentile of a standard normal distribution and rearrange the Z-score formula to solve for sample size. The calculated sample size is 10 when rounded up to the nearest whole number.
Step-by-step explanation:
To find the smallest sample size so that the sample mean exceeds the population mean by at least 2 with a probability not more than 0.02 when the population variance is 9, we can use the concept of the standard normal distribution and Z-scores.
First, we need to determine the Z-score that corresponds to the top 2% of a standard normal distribution. Looking up this value in a Z-table, we find that the Z-score is approximately 2.05 since we're dealing with the right tail of the distribution.
The formula to calculate the required sample size (n) using the Z-score is:
Z = (X‑bar - μ) / ( σ / √n )
Now, rearrange to solve for the sample size (n):
n = (Z * σ / (X‑bar - μ))^2
Plug in the known values:
n = (2.05 * 3 / 2)^2
n = (6.15 / 2)^2
n = 9.4225
To ensure our sample mean is more than 2 units above the population mean with a probability of at most 0.02, we would round up to the nearest whole number, so the smallest sample size needed is 10.