To determine the standard deviation of the population ($\sigma$), we can use the relationship between the standard deviation of the sampling distribution (denoted as $\sigma_{\bar{x}}$) and the population standard deviation $\sigma$.

The formula that relates them is:

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

Where:

• $\sigma_{\bar{x}}$ is the standard deviation of the sampling distribution (0.5 minutes in this case),
• $\sigma$ is the population standard deviation (what we're solving for),
• $n$ is the sample size (81 people).

Now, rearrange the formula to solve for $\sigma$:

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

Substitute the known values:

$\sigma = 0.5 \times \sqrt{81}$

Since $\sqrt{81} = 9$, we get:

$\sigma = 0.5 \times 9 = 4.5$

Thus, the population standard deviation ($\sigma$) is 4.5 minutes.

Would you like more details or have any other questions?

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Here are 5 related questions to expand on this:

1. How does increasing the sample size affect the standard deviation of the sampling distribution?
2. What is the significance of the standard deviation in a normal distribution?
3. If the sample size were 100 instead of 81, what would the new standard deviation of the sampling distribution be?
4. How would you calculate the margin of error using the standard deviation and sample size?
5. How does the Central Limit Theorem relate to this problem?

Tip: When dealing with sampling distributions, remember that larger sample sizes reduce the variability (standard deviation) of the sample means.