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Let's break down and solve the problem step-by-step.

### Given:
- The average number of miles before tire replacement, \( \mu = 68 \) (in thousands).
- The standard deviation \( \sigma = 19 \) (in thousands).
- 10 randomly selected tires are tested.

We'll assume a normal distribution throughout this problem.

---

### Part a. 
**What is the distribution of \( X \)?**

\( X \) represents the number of miles a tire will last before replacement, and it follows a normal distribution \( X \sim N(\mu, \sigma^2) \).

Thus, the distribution is:
\[
X \sim N(68, 19^2) = N(68, 361)
\]

---

### Part b. 
**What is the distribution of \( \bar{X} \)?**

Since \( \bar{X} \) is the sample mean of 10 tires, the distribution of the sample mean is also normal with the same mean but with a reduced standard deviation.

\[
\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right) = N\left(68, \frac{19^2}{10}\right) = N(68, 36.1)
\]

---

### Part c. 
**Find the probability that the number of miles for a randomly selected tire is between 69.3 and 77.7.**

We need to find \( P(69.3 \leq X \leq 77.7) \).

First, calculate the z-scores for the bounds:
\[
z_1 = \frac{69.3 - 68}{19} = \frac{1.3}{19} \approx 0.0684
\]
\[
z_2 = \frac{77.7 - 68}{19} = \frac{9.7}{19} \approx 0.5105
\]

Next, using standard normal distribution tables or a calculator, find the probabilities corresponding to these z-scores:
- \( P(Z \leq 0.0684) \approx 0.5273 \)
- \( P(Z \leq 0.5105) \approx 0.6959 \)

So the probability is:
\[
P(69.3 \leq X \leq 77.7) = 0.6959 - 0.5273 = 0.1686
\]

---

### Part d. 
**Find the probability that the average miles for 10 tires is between 69.3 and 77.7.**

Now, we use the distribution of the sample mean from part b.

Find the z-scores for the bounds using the standard error \( \sigma_{\bar{X}} = \frac{19}{\sqrt{10}} \approx 6.01 \):
\[
z_1 = \frac{69.3 - 68}{6.01} \approx 0.2162
\]
\[
z_2 = \frac{77.7 - 68}{6.01} \approx 1.6113
\]

Using the standard normal table:
- \( P(Z \leq 0.2162) \approx 0.5856 \)
- \( P(Z \leq 1.6113) \approx 0.9467 \)

So the probability is:
\[
P(69.3 \leq \bar{X} \leq 77.7) = 0.9467 - 0.5856 = 0.3611
\]

---

### Part e. 
**Is the assumption that the distribution is normal necessary?**

Yes, the assumption of normality is necessary to use the normal distribution model to calculate probabilities and z-scores. Since the sample size is relatively small (n=10), we rely on the assumption of a normal distribution to use the properties of the sampling distribution of the mean.

---

### Final Answers:
- **a)** \( X \sim N(68, 361) \)
- **b)** \( \bar{X} \sim N(68, 36.1) \)
- **c)** Probability = 0.1686
- **d)** Probability = 0.3611
- **e)** Yes

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:
1. How would the probabilities change if the standard deviation was different?
2. What happens to the distribution of \( \bar{X} \) if the sample size increases to 30 tires?
3. How can the Central Limit Theorem be applied in this problem if the population was not normal?
4. What would happen to the z-scores if the mean was 70 instead of 68?
5. Can we compute probabilities using a different distribution for small sample sizes?

**Tip**: The larger the sample size, the more the sample mean \( \bar{X} \) approximates a normal distribution even if the population distribution is not normal (Central Limit Theorem).

The average number of miles (in thousands) that a car's tire will function before needing replacement is 68 and the standard deviation is 19. Suppose that 10 randomly selected tires are tested. Assume a normal distribution. Solve for parts a-e.

Solve problems related to normal distribution, sample mean distribution, and probability calculations for tire mileage. Learn how to calculate the probability that a tire lasts between specific mileage limits, and understand the importance of assuming a normal distribution.

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