To solve these problems, we assume that $X$, the diameter, follows a normal distribution. However, since you referred to "Exercise 46," I need details on the population mean $\mu$ and standard deviation $\sigma$. Assuming you have them, I will guide you through the calculations.

(a) Compute $P(69 \leq X \leq 71)$ when $n = 16$

Since the sample mean follows a normal distribution: $\bar{X} \sim N\left( \mu, \frac{\sigma}{\sqrt{n}} \right)$ we standardize it as: $Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}}$

Then, we compute: $P(69 \leq \bar{X} \leq 71) = P\left( \frac{69 - \mu}{\sigma / \sqrt{16}} \leq Z \leq \frac{71 - \mu}{\sigma / \sqrt{16}} \right)$

Let me know the values of $\mu$ and $\sigma$, or if you want me to proceed with symbolic representation.

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(b) Probability that sample mean exceeds 71 when $n = 25$

We compute: $P(\bar{X} > 71) = P\left( Z > \frac{71 - \mu}{\sigma / \sqrt{25}} \right)$

Again, I need the values of $\mu$ and $\sigma$ to proceed with calculations. If you have them, please provide them, and I will compute the final probabilities.

Would you like me to assume values or provide further clarification?

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Related Questions:

1. How does the Central Limit Theorem affect these probability calculations?
2. How would the probabilities change if the sample size were larger?
3. What happens if the population distribution were not normal?
4. How does increasing the standard deviation affect the probability?
5. Can we use a normal approximation for small sample sizes?

Tip: The standard error of the mean decreases as the sample size increases, making the distribution of the sample mean more concentrated around $\mu$.