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.