To solve this problem, we use the concept of the sampling distribution of the sample mean. The sampling distribution of the sample mean follows a normal distribution with the following parameters:

1. Mean (μ): Equal to the population mean, $\mu = 62.35$ inches.
2. Standard deviation (σ): Equal to the population standard deviation divided by the square root of the sample size, $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{5.69}{\sqrt{50}}$.

Let's calculate $\sigma_{\bar{x}}$:

$\sigma_{\bar{x}} = \frac{5.69}{\sqrt{50}} \approx 0.8046$

Thus, the sampling distribution of the sample mean is:

$\bar{X} \sim N(62.35, 0.8046)$

We will calculate the required probabilities using the standard normal distribution $Z$, where:

$Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$

---

(a) Probability that the sample mean is less than 61 inches

Find $P(\bar{X} < 61)$:

1. Compute the $Z$-score: $Z = \frac{61 - 62.35}{0.8046} = \frac{-1.35}{0.8046} \approx -1.678$

2. Use the standard normal table or a calculator to find $P(Z < -1.678)$: $P(Z < -1.678) \approx 0.0468$

Thus, the probability is:

$P(\bar{X} < 61) \approx 0.0468 \, (4.68\%)$

---

(b) Probability that the sample mean is between 61.5 inches and 63 inches

Find $P(61.5 < \bar{X} < 63)$:

1. Compute the $Z$-scores: $Z_1 = \frac{61.5 - 62.35}{0.8046} = \frac{-0.85}{0.8046} \approx -1.056$ $Z_2 = \frac{63 - 62.35}{0.8046} = \frac{0.65}{0.8046} \approx 0.808$

2. Use the standard normal table or a calculator to find the probabilities: $P(Z < -1.056) \approx 0.1455, \quad P(Z < 0.808) \approx 0.7891$

3. Subtract the probabilities: $P(61.5 < \bar{X} < 63) = P(Z < 0.808) - P(Z < -1.056)$ $P(61.5 < \bar{X} < 63) \approx 0.7891 - 0.1455 = 0.6436 \, (64.36\%)$

---

(c) Probability that the sample mean is taller than 62 inches

Find $P(\bar{X} > 62)$:

1. Compute the $Z$-score: $Z = \frac{62 - 62.35}{0.8046} = \frac{-0.35}{0.8046} \approx -0.435$

2. Use the standard normal table or a calculator to find $P(Z > -0.435)$: $P(Z > -0.435) = 1 - P(Z < -0.435)$ $P(Z < -0.435) \approx 0.3327 \quad \text{(from the table)}$ $P(Z > -0.435) \approx 1 - 0.3327 = 0.6673 \, (66.73\%)$

---

Final Answers:

a. $P(\bar{X} < 61) \approx 0.0468 \, (4.68\%)$
b. $P(61.5 < \bar{X} < 63) \approx 0.6436 \, (64.36\%)$
c. $P(\bar{X} > 62) \approx 0.6673 \, (66.73\%)$

---

Would you like further details on these calculations?

---

Relative Questions:

1. How does the Central Limit Theorem justify the use of a normal distribution here?
2. What would happen to $\sigma_{\bar{x}}$ if the sample size were increased?
3. Why do we use the standard normal distribution for probabilities in this problem?
4. How can we interpret a Z-score in this context?
5. What if the population distribution were not normal—would the results change?

Tip: Always check whether the sample size is large enough for the Central Limit Theorem to apply, especially if the population distribution is not normal!