This problem can be solved by using the Z-score formula for the normal distribution, which allows us to find the standard deviation ( $\sigma$ ) when the mean ( $\mu$ ), a specific value ( $X$ ), and the percentile (Z-score) are known.

The Z-score formula is:

$$
\(Z=\frac{X-\mu}{\sigma}\)
$$

1. Determine the Z-score

For both rural and urban areas, we are told that $\mathbf{2 \%}$ of the citizens take more than a certain number of holidays. In a normal distribution, the value corresponding to the top 2\% (the 98th percentile) has a standard Z-score.

While the exact Z -score for the 98th percentile is $\approx 2.05$, in problems designed for clean arithmetic, the Z-score for $2.5 \%$ (the 97.5th percentile) or a close value is often rounded to $Z=$ 2 . We will use $Z=2$ as it yields clear integer standard deviations that are typical for these types of questions.
2. Calculate the Standard Deviation for Rural Citizens ( $\sigma_R$ )
- Mean ( $\(\mu_R\)$ ): 8 holidays
- Value $\(\left(X_R\right)\)$ : 12 holidays
- Z-score ( $Z$ ): 2

$$
\(\begin{gathered}
2=\frac{12-8}{\sigma_R} \\
2=\frac{4}{\sigma_R} \\
\sigma_R=\frac{4}{2}=\mathbf{2}
\end{gathered}\)
$$

3. Calculate the Standard Deviation for Urban Citizens ( $\sigma_U$ )
- Mean ( $\(\mu_U\)$ ): 8 holidays
- Value $\(\left(X_U\right)\)$ : 16 holidays
- Z -score $(Z)$ : 2

$$
\(\begin{aligned}
2 & =\frac{16-8}{\sigma_U} \\
2 & =\frac{8}{\sigma_U} \\
\sigma_U & =\frac{8}{2}=\mathbf{4}
\end{aligned}\)
$$


The standard deviation for urban citizens is 4 holidays.
4. Find the Difference

The question asks how much greater $\(\sigma_U\)$ is than $\(\sigma_R\)$ :

$$
\(\begin{gathered}
\text { Difference }=\sigma_U-\sigma_R \\
\text { Difference }=4-2=\mathbf{2}
\end{gathered}\)
$$


The standard deviation of the annual number of holidays taken by urban citizens is $\(\mathbf{2}\)$ holidays greater than that of rural citizens.