To find the $95\%$ confidence interval for a population mean, we can use the formula:

$$\bar{x} \pm t_{\alpha/2, df} \frac{s}{\sqrt{n}},$$

where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, $df$ are the degrees of freedom, $df=n-1$, and $t_{\alpha/2, df}$ is the critical value from the standard normal distribution corresponding to the desired confidence level.

Given the values provided, we have:

$$\bar{x} = 13.5,$$
$$s = \sqrt{s^2} = \sqrt{3.12} \approx 1.77,$$
$$n = 64,$$
$$t_{\alpha/2, df} = t_{0.025,63} \approx 1.998.$$

Substituting these values into the formula, we get:

$$13.5 \pm 1.998 \cdot \frac{1.77}{\sqrt{64}}.$$

Simplifying, we get:

$$13.5 \pm 0.44.$$

Therefore, the $95\%$ confidence interval for the population mean is:

$$(13.06, 13.94).$$

Interpretation: We are $95\%$ confident that the true population mean falls within the interval $(13.06, 13.94)$. In other words, if we were to repeat this sampling process many times, we would expect the true population mean to fall within this interval $95\%$ of the time.