Sea Level Fluctuations

Since confidence intervals are constructed with the sample mean at its center, then $\bar X=\frac{1.75+2.13}{2}=1.94$ . Also, we must have $\bar X + z_{0.025} \cdot \frac{\sigma}{\sqrt{n}}=2.13$ , which implies $n=\left (\sigma\cdot \frac{z_{0.025}}{2.13-\bar X} \right )^2$ . Therefore the sample size is $n=\left (3.6478\cdot \frac{1.96}{2.13-1.94} \right )^2\approx 1416$ . Since the sample size is monthly, then the data covered 118 years.