Let's approach each question step-by-step.

1. Probability of an SRS of 100 first-time car buyers purchasing a car with a mean price greater than $13,000

Let:

• $\mu = 12,500$ (population mean price of first-time car buyers)
• $\sigma = 3000$ (population standard deviation)
• $n = 100$ (sample size)

Since the sample size is large, we can use the Central Limit Theorem to approximate the sampling distribution of the sample mean $\bar{X}$ as normally distributed with mean $\mu = 12,500$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$.

Step 1: Calculate the standard deviation of the sample mean:

$\sigma_{\bar{X}} = \frac{3000}{\sqrt{100}} = 300$

Step 2: Calculate the $Z$-score for $\bar{X} = 13,000$:

$Z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} = \frac{13,000 - 12,500}{300} = \frac{500}{300} \approx 1.67$

Step 3: Find the probability $P(\bar{X} > 13,000)$:

Using standard normal tables, $P(Z > 1.67) \approx 0.0475$.

Answer: The probability is approximately 0.0475, which does not match the given options exactly. However, let's select the closest option:

• Correct Answer: None of the given options is a close match for 0.0475.

---

2. Probability of an SRS of 50 people having a mean starting annual salary between $27,000 and $28,000

Let:

• $\mu = 26,000$ (population mean salary)
• $\sigma = 5200$ (population standard deviation)
• $n = 50$ (sample size)

The sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu = 26,000$ and standard deviation $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$.

Step 1: Calculate the standard deviation of the sample mean:

$\sigma_{\bar{X}} = \frac{5200}{\sqrt{50}} \approx 735.71$

Step 2: Calculate the $Z$-scores for $\bar{X} = 27,000$ and $\bar{X} = 28,000$:

For $\bar{X} = 27,000$: $Z = \frac{27,000 - 26,000}{735.71} \approx 1.36$

For $\bar{X} = 28,000$: $Z = \frac{28,000 - 26,000}{735.71} \approx 2.72$

Step 3: Find the probability $P(27,000 < \bar{X} < 28,000)$:

Using standard normal tables:

• $P(Z < 2.72) \approx 0.9967$
• $P(Z < 1.36) \approx 0.9131$

Thus, $P(27,000 < \bar{X} < 28,000) = 0.9967 - 0.9131 = 0.0836$.

Answer: The probability is approximately 0.0836, which is closest to option A.

• Correct Answer: A. 0.0837

---

If you have any questions or want further details, feel free to ask.

---

Additional Practice Questions

1. What is the probability of selecting a sample with a mean above $12,000?
2. If the sample size is increased to 200, how does it affect the probability calculation in question 1?
3. What if the standard deviation was 4000 instead of 3000 in question 1?
4. How would you calculate a confidence interval for the mean car price?
5. Why does increasing the sample size reduce the standard error?

---

Tip: For probability questions, calculating the $Z$-score is crucial. It standardizes values, allowing us to use the standard normal distribution for probabilities.