Final answer:
The probability that a sample mean of size 156 from a normal distribution with mean 97.4 and standard deviation 41 is greater than 92.5 can be found using the Central Limit Theorem, calculating the z-score, and then looking up the corresponding probability in a standard normal distribution table or calculator.
Step-by-step explanation:
To find the probability that the sample mean is greater than 92.5, we can use the Central Limit Theorem (CLT), which tells us that the distribution of the sample means will be normally distributed around the population mean μ with a standard deviation equal to the population standard deviation σ divided by the square root of the sample size n. The standard error (SE) of the sample mean is given by σ/√n.
The z-score for a sample mean x is calculated as (x - μ)/SE. Using μ=97.4 and σ=41 given in the original example, and a sample size of n=156, the SE would be 41/√156. We want the z-score for a sample mean of 92.5.
Once we have the z-score, we would look up the corresponding probability in a standard normal distribution table or use a statistical software or calculator. The probability we find is the area to the left of that z-score; since we want the area to the right (probability for a mean greater than 92.5), we subtract this value from 1.