To solve this problem, we need to calculate the probability that the sample mean $M$ is greater than 91.1, given a normal distribution with parameters:

• Population mean $\mu = 86.5$
• Population standard deviation $\sigma = 64.9$
• Sample size $n = 242$

Since we’re dealing with the sample mean, we will use the sampling distribution of the sample mean, which has a mean $\mu_M = \mu$ and a standard deviation $\sigma_M = \frac{\sigma}{\sqrt{n}}$.

Step 1: Calculate the Standard Deviation of the Sample Mean

$\sigma_M = \frac{\sigma}{\sqrt{n}} = \frac{64.9}{\sqrt{242}}$

Calculating $\sigma_M$: $\sigma_M \approx \frac{64.9}{15.556} \approx 4.171$

Step 2: Calculate the Z-score for $M = 91.1$

The Z-score formula for a sample mean is: $Z = \frac{M - \mu_M}{\sigma_M} = \frac{91.1 - 86.5}{4.171}$

Calculating the Z-score: $Z \approx \frac{4.6}{4.171} \approx 1.1026$

Step 3: Find the Probability

Now, we look up the Z-score of 1.1026 in the standard normal distribution table or use a calculator to find the probability that $Z > 1.1026$.

The probability $P(Z > 1.1026)$ corresponds to the area to the right of 1.1026 on the standard normal curve. From the standard normal distribution table: $P(Z > 1.1026) \approx 0.1355$

Answer

$P(M > 91.1) \approx 0.1355$

Would you like further clarification on any step?

1. How is the standard error of the mean calculated, and why is it important?
2. How does sample size affect the probability in such problems?
3. How does this problem change if we want $P(M < 91.1)$?
4. What if the standard deviation were different? How would it affect the outcome?
5. How can we use technology, like a calculator or software, to automate such calculations?

Tip: Remember that increasing the sample size decreases the standard error, making the sample mean more closely clustered around the population mean.