To solve this problem, we need to calculate the smallest sample size $n$ required to achieve a confidence interval for the mean with a specified confidence level (91.98%) and a maximum allowable length of 2 cm. Given that the population is normally distributed with a variance of $\sigma^2 = 25 \, \text{cm}^2$, we will follow these steps:

Step-by-Step Solution

1. Confidence Level and Z-score: A 91.98% confidence level corresponds to a Z-score. Since the confidence level is not a standard value, we need to find the Z-score that leaves 4.01% in each tail (i.e., 100% - 91.98% = 8.02%, so 4.01% in each tail).

   We will denote this Z-score as $Z_{\alpha/2}$, where $\alpha = 1 - 0.9198 = 0.0802$.

2. Formula for the Confidence Interval: The formula for the confidence interval for a mean $\mu$ is:

   $\text{CI} = \bar{x} \pm Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right)$

   The length of the confidence interval is given by:

   $\text{Length of CI} = 2 \times Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right)$

   We are given that this length should be no more than 2 cm, so:

   $2 \times Z_{\alpha/2} \left( \frac{\sigma}{\sqrt{n}} \right) \leq 2$

3. Rearrange to Find $n$: Substituting the value for $\sigma$:

   $Z_{\alpha/2} \left( \frac{\sqrt{25}}{\sqrt{n}} \right) \leq 1$

   Since $\sqrt{25} = 5$:

   $Z_{\alpha/2} \left( \frac{5}{\sqrt{n}} \right) \leq 1$

   Solving for $n$:

   $\frac{5 Z_{\alpha/2}}{\sqrt{n}} \leq 1 \implies 5 Z_{\alpha/2} \leq \sqrt{n}$

   $n \geq (5 Z_{\alpha/2})^2$

4. Find the Z-score: We need the Z-score corresponding to the 91.98% confidence level, which leaves 4.01% in each tail.

Let's calculate the Z-score and the smallest sample size $n$.The Z-score corresponding to a 91.98% confidence level is approximately $1.75$.

The smallest sample size needed to ensure that the 91.98% confidence interval for the mean is no longer than 2 cm is $\boxed{77}$.

Would you like more details on this calculation, or do you have any questions?

Here are some related questions you might find useful:

1. How would the required sample size change with a different confidence level?
2. What if the variance were different, say 16 cm² or 36 cm²?
3. How can you interpret confidence intervals in the context of hypothesis testing?
4. Why do we use the Z-score for normally distributed populations?
5. How would this calculation differ if the population were not normally distributed?

Tip: Confidence intervals provide a range of values within which the true population parameter is expected to lie, given a specified level of confidence.