To determine the sample size $n$ needed to estimate the mean $\mu$ with the given margin of error, we use the formula for the margin of error for a confidence interval:

$E = z \frac{\sigma}{\sqrt{n}}$

Where:

• $E$ = margin of error (0.10),
• $z$ = critical value for the confidence level (2.57 for 99% confidence level),
• $\sigma$ = standard deviation of the population (0.85),
• $n$ = required sample size.

Rearranging for $n$:

$n = \left( \frac{z \cdot \sigma}{E} \right)^2$

Step-by-Step Calculation:

1. Substitute the values: $n = \left( \frac{2.57 \cdot 0.85}{0.10} \right)^2$

2. Compute the numerator: $2.57 \cdot 0.85 = 2.1845$

3. Divide by $E$: $\frac{2.1845}{0.10} = 21.845$

4. Square the result: $n = (21.845)^2 = 477.46$

5. Round up to the next whole number: $n = 478$

Final Answer:

The required sample size is 478.

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Do you have any further questions or need additional clarifications? Here are some related questions:

1. How is the critical value $z$ determined for different confidence levels?
2. What would the sample size be if the standard deviation were different?
3. How do changes in margin of error affect the required sample size?
4. What is the impact of lowering the confidence level to 95%?
5. How would this formula adapt if estimating proportions instead of means?

Tip: Always round the sample size up to ensure the desired margin of error is met.