Review Assessment
0.5 of 1 Point
Part 3 of 4

Suppose a simple random sample of size $n=42$ is obtained from a population with $\mu=61$ and $\sigma=14$.
(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities regarding the sample mean? Assuming the normal model can be used, describe the sampling distribution $\bar{x}$.
(b) Assuming the normal model can be used, determine $P(\bar{x}< 64.6)$.
(c) Assuming the normal model can be used, determine $P(\bar{x} \geq 62.2)$.

Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2).
A. Approximately normal, with $\mu_{\bar{x}}=61$ and $\sigma_{\bar{x}}=14$
B. Approximately normal, with $\mu_{\bar{x}}=61$ and $\sigma_{\bar{x}}=\frac{14}{\sqrt{42}}$
C. Approximately normal, with $\mu_{\bar{x}}=61$ and $\sigma_{\bar{x}}=\frac{42}{\sqrt{14}}$
(b) $P(\bar{x}< 64.6)=\square$ (Round to four decimal places as needed.)

Step 1 ：The distribution of the population must be approximately normal in order to use the normal model to compute probabilities regarding the sample mean. The sampling distribution of the sample mean is also approximately normal, with a mean equal to the population mean, \( \mu_{\bar{x}} = 61 \), and a standard deviation equal to the population standard deviation divided by the square root of the sample size, \( \sigma_{\bar{x}} = \frac{14}{\sqrt{42}} \).

Step 2 ：To find the probability that the sample mean is less than 64.6, we first standardize the value of 64.6 by subtracting the mean of the sampling distribution and dividing by the standard deviation of the sampling distribution. This gives us a z-score of \( z = \frac{64.6 - 61}{\frac{14}{\sqrt{42}}} \approx 1.67 \).

Step 3 ：We then use the standard normal distribution to find the probability that a standard normal random variable is less than this z-score. This gives us \( P(\bar{x}<64.6) \approx 0.9522 \).

Step 4 ：Similarly, to find the probability that the sample mean is greater than or equal to 62.2, we standardize the value of 62.2 to get a z-score of \( z = \frac{62.2 - 61}{\frac{14}{\sqrt{42}}} \).

Step 5 ：We then use the standard normal distribution to find the probability that a standard normal random variable is greater than or equal to this z-score. This gives us \( P(\bar{x} \geq 62.2) \).

Step 6 ：Final Answer: The sampling distribution of the sample mean is approximately normal with \( \mu_{\bar{x}} = 61 \) and \( \sigma_{\bar{x}} = \frac{14}{\sqrt{42}} \). The probability that the sample mean is less than 64.6 is \(\boxed{0.9522}\).