Let's simplify the answer further by presenting the key points clearly:

1. **Given Information**:
   - Confidence level: 91.98%
   - Maximum confidence interval length: 2 cm
   - Population variance: \( \sigma^2 = 25 \, \text{cm}^2 \) (so \( \sigma = 5 \, \text{cm} \))

2. **Z-score**:
   - For a 91.98% confidence level, the Z-score is approximately \( Z_{\alpha/2} = 1.75 \).

3. **Formula for Minimum Sample Size \( n \)**:
   The required sample size \( n \) can be calculated using the formula:

   \[
   n = \left( \frac{5 \times Z_{\alpha/2}}{1} \right)^2
   \]

   Plugging in the Z-score:

   \[
   n = \left( \frac{5 \times 1.75}{1} \right)^2 = (8.75)^2 = 76.5625
   \]

4. **Smallest Sample Size**:
   Since the sample size must be a whole number, we round up to the next integer.

   \[
   n = 77
   \]

### Final Answer:
The smallest sample size needed is \( \boxed{77} \).

Would you like to explore any further aspects of this problem? Here are some related questions:
1. How would rounding down instead of up affect the confidence level?
2. What is the effect of increasing the variance on the sample size?
3. How can you calculate a confidence interval for a small sample size?
4. What assumptions are needed for the confidence interval formula?
5. How does the confidence interval change with different distributions?

**Tip:** When calculating sample sizes, always round up to ensure the desired confidence level is maintained.

Learn how to calculate the minimum sample size needed for a confidence interval with a detailed step-by-step solution. This problem involves statistics and confidence intervals, including the formula for minimum sample size and its application with specific data points.

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