The question in the image asks for the required sample size needed to produce a standard deviation for the sample mean ($\sigma_{\bar{x}}$) of 0.18, given the following:

• Population standard deviation ($\sigma$) = 3.3 pounds
• Desired standard deviation for the sample mean ($\sigma_{\bar{x}}$) = 0.18 pounds

We can solve this using the formula for the standard error of the mean:

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

Where:

• $\sigma_{\bar{x}}$ is the standard deviation of the sample mean
• $\sigma$ is the population standard deviation
• $n$ is the sample size

Step-by-step solution:

1. Rearrange the formula to solve for $n$:

$n = \left( \frac{\sigma}{\sigma_{\bar{x}}} \right)^2$

2. Substitute the given values:

$n = \left( \frac{3.3}{0.18} \right)^2$

3. Calculate the division:

$\frac{3.3}{0.18} = 18.33$

4. Square the result:

$n = (18.33)^2 = 336.11$

Since the sample size must be a whole number, we round 336.11 to 336.

Final Answer:

The required sample size is 336.

Thus, the correct option is:

336

Would you like further details on this topic, or have any follow-up questions?

Here are five related questions for practice:

1. How does increasing the sample size affect the standard deviation of the sample mean?
2. If the population standard deviation were smaller, how would the required sample size change?
3. What is the difference between population standard deviation and standard error of the mean?
4. Why do we square the ratio of the standard deviations when calculating the sample size?
5. How would the calculation change if a 95% confidence interval were involved?

Tip: Always check whether rounding is allowed for statistical calculations to avoid small discrepancies in critical results.