To solve this problem, we need to calculate the probability that the sample mean will be within plus or minus 6 of the population mean. This means we want to find the probability that the sample mean, $\bar{x}$, is between 194 and 206, given that the population mean is 200 and the population standard deviation is 0.50.

Steps:

1. Determine the standard error of the mean ($\text{SE}$): The standard error (SE) is given by: $\text{SE} = \frac{\sigma}{\sqrt{n}}$ where $\sigma$ is the population standard deviation and $n$ is the sample size (which was not provided in the problem; I'll leave it as $n$ for now).

2. Find the z-scores for 194 and 206: The z-score is calculated as: $z = \frac{x - \mu}{\text{SE}}$ where $x$ is the value for which you are calculating the z-score, and $\mu$ is the population mean.

   So, for $x = 194$ and $x = 206$, the z-scores are: $z_{194} = \frac{194 - 200}{\text{SE}} = \frac{-6}{\text{SE}}$ $z_{206} = \frac{206 - 200}{\text{SE}} = \frac{6}{\text{SE}}$

3. Use the z-table to find the probabilities: Once we have the z-scores, we can use the z-table to find the probabilities corresponding to these z-scores.

If you can provide the sample size $n$, I can continue with the calculations and give you the final probability.

Would you like to provide $n$ or have more details on any of these steps?

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Here are 5 related questions to expand on the problem:

1. How does the sample size $n$ affect the standard error and the overall probability?
2. How would the probability change if the standard deviation was higher or lower?
3. What does it mean if the z-score is positive or negative?
4. How do you interpret a z-score in the context of a probability distribution?
5. What is the significance of using a z-table in statistical calculations?

Tip: The larger the sample size, the smaller the standard error, which leads to a more accurate estimate of the population mean.