# The content of bottles of water follows a normal distribution with average = 0.99 and standard deviation 0.02. Find the variance of sample

The content in litres of bottled water follows a normal distribution with average 0.99 and standard deviation of 0.02.

a) Suppose a sample of 16 bottles is chosen. Determine the probability that the average content of the sample is greater than 1 litre.

b) Determine the size of the sample so that the probability that the difference between the average of the sample and the average of the population doesn't exceed 0.01 is at least 0.95.

My professor solved a) this way: Why did he use $$\frac{0.02^2}{16}$$ for the variance and not just $$0.02^2$$?

He solved b) this way: I don't understand why he used the given mean 0.99 and for the variance he used $$\frac{\sigma}{\sqrt{n}}$$. Why do I have to estimate the variance if the variance is the stdev^2 and the stdev is given?

I assume the sample is supposed to be independent. I denote the content of the 16 sampled bottles by $$X_1,X_2,\ldots,X_{16}$$.
Regarding your first question: The variance of the sample mean is $$\operatorname{Var}(\bar{X})=\operatorname{Var}\left(\frac{1}{16}\sum_{i=1}^{16}{X_i}\right)=\frac{1}{16^2}\operatorname{Var}\left(\sum_{i=1}^{16} {X_i}\right)=\frac{1}{16^2}\sum_{i=1}^{16}\operatorname{Var}(X_i)=\frac{1}{16^2} 16\cdot 0.02^2 = \frac{0.02^2}{16},$$ where we have used independence and the rules for the variance.
The same reasoning applies to (b). By the same steps as above (just replace 16 by $$n$$) you can show that the variance of the sample mean is equal $$\frac{\sigma^2}{n}$$ and hence the standard deviation is $$\frac{\sigma}{\sqrt{n}}$$. Note that the variance of the sample mean does not equal to the variance of the underlying distribution (except if $$n=1$$ or the variance of the underlying distribution is $$0$$).
The expected value on the other hand is the same, which the following calculation shows: $$\mathbb{E}(\bar{X})=\mathbb{E}\left(\frac{1}{n}\sum_{i=1}^{n}{X_i}\right)=\frac{1}{n}\sum_{i=1}^{n}\mathbb{E}\left(X_i\right)=\frac{1}{n}n\cdot0.99=0.99,$$ where we have used the linearity of the expectation.
To think about this intuitively, imagine we collect a very small sample (so $$n$$ is small). Each observation carries much weight in the sample average (especially unusually small or large observations). If we draw repeated small samples, our sample average will hence vary a lot. If we have a very large sample ($$n$$ large), then we gonna observe all kinds of observation (most of them in a range that has much probability mass) and a few extreme observations won't move the sample average by much. If we draw repeated large samples, our sample average will not vary a lot. So, we would intuitively assume that the variance decreases as the sample size increases (indeed $$n$$ shows up in the denominator).