So I have a question about combining random variables. While it is given that a bottle filling machine fills with $\sigma=5$, the observed variable here is the weight of $n$ bottles, where the bottle itself has a normal distribution also.
Essentially I need to find the missing $\mu$ for the amount of wine in a filled bottle, but my intuition says that as I am observing a different variable (the weight) that itself has a distribution, I must also find a new $\sigma$.
The question is below:
My approach so far:
Let random variables $L$, $B$ and $W$ represent the liquid amount, bottle weight and total weight respectively.
$L \sim N(\mu,25)$
$B \sim N(250,225)$
$W=L+B$
$Var(W)=Var(L)+2Cov(L,B)+Var(B)=Var(L)+Var(B)=250$
$W \sim N(998,250)$ where $998$ comes from observation of 16 samples.
Now what? I think it would be wrong to calculate the confidence interval by using $\mu =998-250$ and use $\sigma^2=25$ as this would not use the results derived above. The error in bottle weighs would not be accounted for.
Is it correct to use the standard deviation of $W$ as this also accounts for the uncertainty in the observed weight?