To solve this problem, we need to find the range around the population mean where 57% of the sample means will fall. Since the distribution is approximately normal, we will use the standard normal distribution and the properties of the sampling distribution of the mean.

Step 1: Calculate the standard error (SE) of the mean.

The standard error is given by:

$SE = \frac{\sigma}{\sqrt{n}}$

where $\sigma = 0.05$ inches (standard deviation of the population) and $n = 12$ (sample size).

$SE = \frac{0.05}{\sqrt{12}} \approx 0.0144$

Step 2: Determine the z-scores that correspond to the middle 57% of the normal distribution.

For a normal distribution, 57% in the middle leaves 43% in the tails, or 21.5% in each tail.

Using a z-table or calculator, the z-scores corresponding to the upper and lower 21.5% tail probabilities are approximately $\pm 0.78$.

Step 3: Find the range around the mean.

To find the boundaries, use the formula:

$\text{Boundary} = \mu \pm (z \times SE)$

where $\mu = 2.64$, $z = 0.78$, and $SE = 0.0144$.

Calculating the lower boundary:

$\text{Lower boundary} = 2.64 - (0.78 \times 0.0144) \approx 2.64 - 0.0112 = 2.63$

Thus, the lower boundary is approximately 2.63 inches.

Would you like a further explanation or have more questions?

Related Questions:

1. What is the upper boundary for this range?
2. How would the boundaries change if the sample size was increased to 20?
3. Why do we use the standard error in this calculation instead of the standard deviation?
4. How would you find the probability that a sample mean is below 2.63 inches?
5. What if the distribution was not normal? How would the approach differ?

Tip:

Always check whether the sample size is large enough (typically $n \geq 30$) to use the normal approximation for non-normal populations.