Final answer:
To find the value above which there is a 66% chance that the sample mean will fall, the Central Limit Theorem is applied to calculate the standard error, then use the Z-table or a calculator to find the desired percentile and finally calculate the value using the Z-score and standard error.
Step-by-step explanation:
The question concerns the determination of probabilities in relation to the sampling distribution of the mean for a normally distributed population. Given the population mean (μ) is 100, the standard deviation (σ) is 10, and the sample size (n) is 25, we aim to find the value above which there is a 66% chance that the sample mean will fall.
To determine this value, we apply the principles of the Central Limit Theorem which states that the sampling distribution of the sample mean for a sufficiently large sample size will be normally distributed with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size, known as the standard error.
Using this knowledge and the provided population parameters, we calculate the standard error as σ / sqrt(n) = 10 / sqrt(25) = 2. With this standard error, we can find the required value using Z-tables or a calculator that provides inverse normal probabilities.
Since the question asks for the 66th percentile of the distribution (the value above which 66% of the mean values will lie), we are seeking the Z-score that corresponds to the 34th percentile (since 100% - 66% = 34%). Once we have the Z-score, we can calculate the value by multiplying the Z-score by the standard error and adding it to the population mean.