To determine the minimum sample size required for a 95% confidence level, we can use the following formula for sample size $n$:

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

Where:

• $Z$ is the z-score corresponding to the desired confidence level (95% confidence level has a z-score of 1.96).
• $\sigma$ is the population standard deviation (given as 12.6).
• $E$ is the margin of error or the desired accuracy (given as 1 unit).

Let's compute the sample size:

$n = \left( \frac{1.96 \cdot 12.6}{1} \right)^2$

Now I'll calculate the value.The minimum required sample size is approximately 610.

Would you like further details or have any questions? Here are 5 related questions that can expand your understanding:

1. What would the sample size be if the margin of error was 0.5 instead of 1?
2. How would the sample size change if the confidence level increased to 99%?
3. What is the significance of the z-score in sample size determination?
4. How does a larger standard deviation affect the required sample size?
5. Why is it important to assume a normal distribution for this calculation?

Tip: Reducing the margin of error or increasing the confidence level will significantly increase the required sample size.