Let's break this down step by step:

Part (a) - Condition for Using the Normal Model:

To use the normal model (i.e., the normal distribution) for computing probabilities involving the sample mean, one of the following must be true:

1. If the population distribution is normal, then the distribution of the sample mean $\overline{x}$ is normally distributed for any sample size $n$.
2. If the population distribution is not normal, the sample size must be sufficiently large (typically $n \geq 30$) for the Central Limit Theorem (CLT) to apply. This theorem states that the sampling distribution of $\overline{x}$ will approach normality as $n$ increases.

Since the sample size here is $n = 43$, which is greater than 30, we can safely use the normal model, assuming the population distribution is not extremely skewed or unusual.

Sampling distribution of $\overline{x}$:

• The sampling distribution of the sample mean $\overline{x}$ will be normally distributed with the following properties:
  • Mean of the sampling distribution: $\mu_{\overline{x}} = \mu = 67$
  • Standard deviation of the sampling distribution (Standard Error): $\sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{43}} \approx 2.287$

Part (b) - Find $P(\overline{x} < 70.6)$:

To find the probability that the sample mean is less than 70.6, we need to calculate the z-score: $z = \frac{\overline{x} - \mu_{\overline{x}}}{\sigma_{\overline{x}}}$

Substitute the values: $z = \frac{70.6 - 67}{2.287} \approx 1.575$

Now, we can use the standard normal distribution table (or a calculator) to find the probability corresponding to $z = 1.575$.

From the z-table, the probability $P(Z < 1.575)$ is approximately 0.9429. Therefore: $P(\overline{x} < 70.6) \approx 0.9429$

Part (c) - Find $P(\overline{x} \geq 68.1)$:

To find the probability that the sample mean is greater than or equal to 68.1, we first calculate the z-score: $z = \frac{\overline{x} - \mu_{\overline{x}}}{\sigma_{\overline{x}}}$

Substitute the values: $z = \frac{68.1 - 67}{2.287} \approx 0.481$

Using the standard normal distribution table or a calculator, we find that the probability corresponding to $z = 0.481$ is approximately 0.6843. However, since we are looking for $P(\overline{x} \geq 68.1)$, we need to find: $P(\overline{x} \geq 68.1) = 1 - P(Z < 0.481) \approx 1 - 0.6843 = 0.3157$

Final Answers:

• (a) To use the normal model, the population must either be normally distributed, or the sample size must be large enough ($n \geq 30$). The sampling distribution of $\overline{x}$ is normal with a mean of 67 and a standard error of approximately 2.287.
• (b) $P(\overline{x} < 70.6) \approx 0.9429$
• (c) $P(\overline{x} \geq 68.1) \approx 0.3157$

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Do you want more details on any part? Or have any questions?

5 Related Questions:

1. What is the Central Limit Theorem, and why is it important in statistics?
2. How do you calculate z-scores, and what do they represent?
3. Can the normal distribution model be used for any sample size?
4. How does increasing the sample size affect the standard error?
5. What other probability distributions can be used for small sample sizes?

Tip:

When dealing with sampling distributions, remember that as sample size increases, the distribution of the sample mean becomes more normally distributed, even if the population is not normal. This is the power of the Central Limit Theorem.