To solve this problem, we need to determine the required sample size $n$ to estimate a population mean with a specified confidence level and margin of error. Here’s the step-by-step solution:

Step-by-Step Solution

1. Given Data:

   • Population standard deviation ($\sigma$) = 45.1
   • Desired margin of error (E) = 0.2
   • Confidence level = 95%

2. Determine the Critical Value (Z-score): For a 95% confidence level, the critical Z-score (Z) is typically 1.960. This value corresponds to the 95% confidence interval in a standard normal distribution.

3. Use the Sample Size Formula: The formula to determine the sample size for estimating a population mean is: $n = \left( \frac{Z \cdot \sigma}{E} \right)^2$

4. Plug in the Values: Substituting the values we have: $n = \left( \frac{1.960 \times 45.1}{0.2} \right)^2$

5. Calculate the Result:

   • First, calculate the numerator: $1.960 \times 45.1 = 88.396$.
   • Then divide by the margin of error: $\frac{88.396}{0.2} = 441.98$.
   • Finally, square this value: $441.98^2 = 195344.0804$.

6. Round Up to the Nearest Whole Number: Since sample size must be a whole number, round up to ensure the desired confidence level and precision: $n = 195345$

Answer:

The required sample size $n$ is 195345.

Would you like more details on any of the steps, or have any other questions?

Related Questions:

1. How would the required sample size change if the margin of error were reduced to 0.1?
2. What is the effect of increasing the confidence level on the sample size?
3. How is the Z-score selected for different confidence levels?
4. Can this formula be used if the population standard deviation is unknown?
5. How would you calculate the sample size if the population distribution is not normal?

Tip:

Remember that increasing the desired confidence level or decreasing the margin of error will both increase the required sample size, as greater precision requires more data.