For any sample that comes from a normally distributed population, we know that:
$\frac{(\bar{x} - \mu)}{(\sigma / \sqrt{n})} \sim N(0, 1)$ when $\sigma$ is known

Thus, a 95% CI for the mean, $\mu = \bar{x} \pm \frac{z_\alpha}{2} \times \frac{\sigma}{\sqrt{n}}$ where $\alpha = 0.05$

$$= 130 \pm 1.96 \times \frac{20}{\sqrt{50}} \\ = 130 \pm 5.5437$$

which can be expressed as $[124.5; 135.5]$