The problem provided shows a sampling distribution of sample means, and you're asked to find the population standard deviation $\sigma$. Here's how to approach the solution:

Given Information:

• Sample size $n = 81$
• The population distribution is normal.
• The graph shows the distribution of sample means with values approximately ranging from $77$ to $83$.

The standard deviation of the sample means (standard error) is related to the population standard deviation $\sigma$ by the formula:

$\text{Standard error} = \frac{\sigma}{\sqrt{n}}$

From the graph, the sample means range is around $6$ units (from about $77$ to $83$), and the sample mean is centered around $80$. Assuming the range is approximately $6$ units and represents about 6 standard errors (since this is a normal distribution, where approximately 99.7% of the data falls within 6 standard errors, i.e., 3 standard deviations in each direction):

$6 \cdot \text{Standard error} = 6$

Thus:

$\text{Standard error} \approx 1$

Now use the formula for the standard error:

$1 = \frac{\sigma}{\sqrt{81}}$

Simplifying:

$1 = \frac{\sigma}{9}$

Therefore, the population standard deviation $\sigma$ is:

$\sigma = 9$

So, the value of the population standard deviation $\sigma$ is 9.

Let me know if you need further details or have any questions.

Here are 5 questions to expand this concept:

1. What is the relationship between the sample size and the standard error?
2. How does the Central Limit Theorem help in approximating the distribution of sample means?
3. Why is the sample standard deviation divided by $\sqrt{n}$ when computing the standard error?
4. How would the population standard deviation change if the sample size were larger?
5. What does the 68-95-99.7 rule tell us about normal distributions?

Tip: When interpreting a sampling distribution graph, remember that the spread (standard error) decreases as the sample size increases.