We observe the height of the population of a city as identically distributed random variables and Variance $\sigma^2=7.3$ cm and the mean of the height of 200 randomly chosen inhabitants is 176.2 cm. I need to determine the 99% confidence interval for the average $\mu$. And there is a small note in brackets "(Normal Approximation)".
So my approach so:
$$ 176.2 \pm Z \frac{\sqrt(7.3)}{\sqrt(200)}=176.2 \pm 0.4921 $$
But I am not sure if this was meant by this problem because it seems too simple? I would appreciate any help.