To solve for $N$ using the formula with $n = 181$ and $e = 4\%$, the context of the problem is necessary, but a common formula where these variables might appear is the sample size determination formula for proportions in statistics:

$N = \frac{n}{1 - e^2}$

Where:

• $N$ is the adjusted sample size,
• $n$ is the initial sample size,
• $e$ is the margin of error (in decimal form).

Step-by-Step Solution:

1. Identify the formula: We'll use the formula for adjusting the sample size for a given margin of error: $N = \frac{n}{1 - e^2}$

2. Substitute the values: Given $n = 181$ and $e = 4\% = 0.04$, substitute these into the formula: $N = \frac{181}{1 - (0.04)^2}$

3. Square the margin of error: Calculate $e^2$: $0.04^2 = 0.0016$

4. Subtract from 1: Now subtract $0.0016$ from 1: $1 - 0.0016 = 0.9984$

5. Divide: Finally, divide the initial sample size $n$ by the result: $N = \frac{181}{0.9984} \approx 181.29$

Final Answer:

$N \approx 181.29$ So the adjusted sample size $N$ is approximately 181.29.

---

Would you like any further details or have any questions on this solution?

Here are some related questions:

1. What is the meaning of sample size adjustment in statistics?
2. How do you interpret the margin of error in this formula?
3. Why is the margin of error squared in the formula?
4. Can this formula be applied to any percentage of margin of error?
5. What happens to the sample size if the margin of error increases?

Tip: When solving problems with percentages, always convert percentages to decimals (e.g., $4\%$ becomes $0.04$) before applying them to formulas!