Final answer:
To determine the probability that the sample mean will be greater than 22, one must calculate the z-score using the formula z = (X - μ) / SE, where SE is the standard error. Finding the z-score allows us to use the standard normal distribution to determine the probability.
Step-by-step explanation:
To answer the question, we need to calculate the probability that the sample mean is greater than 22 when selecting a sample of 89 measurements from a population with a mean (μ) of 21 and a standard deviation (σ) of 8. We use the central limit theorem, which states that the distribution of sample means will be approximately normally distributed if the sample size is large enough (n > 30 is a common rule of thumb). Since our sample size is 89, we can use the normal distribution. The standard error (SE) is the standard deviation of the sampling distribution of the sample mean, which is calculated as σ divided by the square root of the sample size (n).
The formula for the standard error (SE) is: SE = σ / √n
For this problem: SE = 8 / √89. Next, we calculate the z-score, which is the number of standard errors the sample mean is away from the population mean. The z-score formula is: z = (X - μ) / SE, where X is the sample mean.
Here, z = (22 - 21) / SE. After calculating the z-score, we can use the standard normal distribution table or a calculator to find the probability that the z-score is greater than the calculated value. This probability will answer the question about the likelihood of the sample mean being larger than 22.