The population mean* (mu)

The standard error of the mean ($\sigma_{\bar{x}}$) is calculated as:

$$\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$$

For a given sample mean $\bar{x}$, the z-score is calculated as:

$$z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$$

a) For the mean between 66.8 and 68.3 in:

1. Calculate the z-score for $\bar{x}_1 = 66.8$: $$z_1 = \frac{66.8 - \mu}{\sigma_{\bar{x}}}$$
2. Calculate the z-score for $\bar{x}_2 = 68.3$: $$z_2 = \frac{68.3 - \mu}{\sigma_{\bar{x}}}$$
3. Find the probability $P(66.8 < \bar{x} < 68.3)$, which is equivalent to $P(z_1 < Z < z_2)$. This can be found using a standard normal distribution table or calculator: $$P(z_1 < Z < z_2) = P(Z < z_2) - P(Z < z_1)$$

b) For the mean less than 66.4 in:

1. Calculate the z-score for $\bar{x}_3 = 66.4$: $$z_3 = \frac{66.4 - \mu}{\sigma_{\bar{x}}}$$
2. Find the probability $P(\bar{x} < 66.4)$, which is equivalent to $P(Z < z_3)$. This can be found using a standard normal distribution table or calculator.

Multiply the probabilities found in Step 3 by the total number of samples ($N_{total}$).

a) Expected number of samples between 66.8 and 68.3 in:

$$N_a = P(66.8 < \bar{x} < 68.3) \times N_{total}$$

b) Expected number of samples less than 66.4 in:

$$N_b = P(\bar{x} < 66.4) \times N_{total}$$

Please provide the details from "Illustration 3" (population mean, population standard deviation, sample size, and total number of samples) to get the numerical answers.