Final answer:
Using the Central Limit Theorem, the sample mean is equal to the population mean, and the standard deviation of the sample mean is the population standard deviation divided by the square root of the sample size. The probabilities cannot be calculated without standard normal tables or calculator functions. Additional data or clarification may be needed to provide an accurate answer.
Step-by-step explanation:
The question pertains to the application of the Central Limit Theorem (CLT) to calculate the sample mean (\(\bar{x}\)), its standard deviation (\(\sigma_{\bar{x}}\)), and the probability of the sample mean being within a certain range for a normally distributed population. However, there appears to be a confusion in the data provided in the question, and, assuming that the population mean (\(\mu\)) and population standard deviation (\(\sigma\)) are correctly given as 28 and 20, respectively, I will proceed to calculate the requested values.
- For a sample size of n = 43, the sample mean is \(\bar{x} = \mu\), which is 28. The standard deviation of the sample mean is \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\), which calculates as \(\frac{20}{\sqrt{43}}\)
- For a sample size of n = 60, we again have \(\bar{x} = 28\), while \(\sigma_{\bar{x}}\) becomes \(\frac{20}{\sqrt{60}}\).
The probabilities P(28 \leq x \leq 30) for both cases need to be calculated using the z-score and then finding the cumulative probability from the standard normal distribution. This cannot be completed accurately without the actual z-scores and the standard normal cumulative distribution table or calculator function.
For the specific calculations and probabilities similar to the examples provided in the additional information, functions like 'normalcdf' are used. This indicates a need for technology like a graphing calculator or software capable of performing such calculations.