To calculate the standard deviation of the mean of the sampling distribution of sample proportions, we use the following formula:

$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

Where:

• $p$: The true population proportion of defective pumpkins.
• $n$: The sample size.

Step 1: Identify the given values

• $p = \frac{1}{50} = 0.02$ (since 1 in 50 pumpkins are not fit).
• $n = 260$.

Step 2: Calculate $1 - p$

$1 - p = 1 - 0.02 = 0.98$

Step 3: Plug values into the formula

$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.02 \cdot 0.98}{260}}$

Step 4: Perform the calculation

First, calculate the numerator: $0.02 \cdot 0.98 = 0.0196$

Next, divide by $n$: $\frac{0.0196}{260} \approx 0.00007538$

Now, take the square root: $\sigma_{\hat{p}} = \sqrt{0.00007538} \approx 0.00868$

Final Answer:

The standard deviation of the sampling distribution of sample proportions is approximately 0.00868.

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Questions to Explore Further:

1. What is the interpretation of this standard deviation in the context of the problem?
2. How does increasing the sample size affect the standard deviation of the sampling distribution?
3. What would happen to $\sigma_{\hat{p}}$ if the defect rate $p$ was higher, e.g., $p = 0.05$?
4. How can this standard deviation help in constructing confidence intervals for the defect rate?
5. What is the probability that a sample proportion falls within one standard deviation of the mean?

Tip: Always ensure the sample size is large enough to approximate the sampling distribution of proportions using a normal distribution ($np \geq 10$ and $n(1-p) \geq 10$).