Given $n=40$ (sample observations $y_i$) and sample mean $\overline{y}=195.47$, construct a $99\%$ confidence interval using the $G(\mu, 20)$ (normal) distribution and then interpret the interval.
My solution:
$$0.99=P\left( -2.58\le \frac{\overline{Y}-\mu}{\sqrt{20/40}} \le2.58 \right)$$ $$=P\left(\overline{Y}-1.7678\le \mu \le\overline{Y}+1.7678 \right)$$
Thus the confidence interval is $[193.7022,197.2378]$.
Interpretation: there is a 99% confidence that $\mu$ is in this interval.
I'd appreciate if someone would let me know if my construction is correct. Isn't my interval too narrow for this sample and mean? I know the concept should be pretty simple, but I got somewhat confused about some details from the text I read. Is my interpretation also correct?