A simple random sample of size $n=38$ is obtained from a population with $\mu=62$ and $\sigma=18$.
(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample mean?

Assuming that this condition is true, describe the sampling distribution of $\bar{x}$
(b) Assuming the normal model can be used, determine $\mathrm{P}(\mathrm{x}< 66.2)$.
(c) Assuming the normal model can be used, determine $P(x \geq 64.2)$.
(a) What must be true regarding the distribution of the population?
A. The sampling distribution must be assumed to be normal.
B. Since the sample size is large enough, the population distribution does not need to be normal.
C. The population must be normally distributed.
D. The population must be normally distributed and the sample size must be large.

Assuming the normal model can be used, describe the sampling distribution $\bar{x}$. Choose the correct answer below.
A. Approximately normal, with $\mu_{\bar{x}}=62$ and $\sigma_{\bar{x}}=\frac{18}{\sqrt{38}}$
B. Approximately normal, with $\mu_{x}^{-}=62$ and $\sigma_{\bar{x}}=18$
C. Approximately normal, with $\mu_{x}=62$ and $\sigma_{x}=\frac{38}{\sqrt{18}}$

Step 1 ：The Central Limit Theorem states that if the sample size is large enough (usually n > 30 is considered large enough), the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution. So, the population distribution does not need to be normal, but the sample size should be large. Therefore, the answer to the first part is B. Since the sample size is large enough, the population distribution does not need to be normal.

Step 2 ：The mean of the sampling distribution of the sample mean (also known as the expected value of the sample mean) is equal to the population mean, which is 62 in this case. The standard deviation of the sampling distribution of the sample mean (also known as the standard error) is equal to the population standard deviation divided by the square root of the sample size. So, it should be 18/sqrt(38). Therefore, the answer to the second part is A. Approximately normal, with \(\mu_{\bar{x}}=62\) and \(\sigma_{\bar{x}}=\frac{18}{\sqrt{38}}\).

Step 3 ：Using python to simplify \(\sigma_{\bar{x}}=\frac{18}{\sqrt{38}}\)

Step 4 ：\(\sigma_{\bar{x}}\) is approximately 2.92

Step 5 ：Final Answer: (a) \(\boxed{\text{B. Since the sample size is large enough, the population distribution does not need to be normal.}}\) (b) \(\boxed{\text{A. Approximately normal, with } \mu_{\bar{x}}=62 \text{ and } \sigma_{\bar{x}} \approx 2.92}\)