The volume of a randomly selected bottle $X$ ~ $N(1000ml, 10^2ml)$. Measures, $Y$ are poured from a random bottle with $Y$ ~ $N(35ml, 1^2ml)$. 10 measures are poured.
What is the probability less than $700ml$ is left in the bottle after 10 measures?
Normally I would work on $p(X < 1050)$ if it was a single variable $X$ question and each measure was a fixed $35ml$ but I am not sure how to factor in $Y$ here? I need help with how to begin the approach to the answer please? After that I think I can work out the math OK.
Thanks
If I understood the question correctly, a bottle is picked whose volume $X$ has distribution $N(1000,10^2)$. Then, 10 volumes $Y_{1},\ldots,Y_{10}$ each with distribution $N(35,1^2)$ are poured out of the bottle. The random variable $Z=X-(Y_{1}+\cdots+Y_{10})$ corresponds to the amount of liquid left in the bottle after the 10 pours.
Assuming the picking of the bottle and pours are all independent, $Z$ is a sum of independent normal random variables and therefore also a normal random variable. That is, $Z\sim N(\mu_{Z},\sigma_{Z}^{2})$ with $\mu_{Z}=\mathbb{E}\left[Z\right]$ and $\sigma_{Z}^{2}=\text{Var}(Z)$. Once you have computed $\mu_{Z}$ and $\sigma_{Z}^{2}$, you can use these to compute $$\mathbb{P}(Z<700) = F\left(\frac{700 - \mu_Z}{\sigma_Z}\right)$$ where $F$ is the CDF of a standard normal random variable.
Here's a quick way to remember the sum of independent normals rule:
Let $X$ and $Y$ be independent normals. Then, the characteristic function of $Z=X+Y$ is $$ \mathbb{E}\left[e^{itZ}\right]=\mathbb{E}\left[e^{itX}\right]\mathbb{E}\left[e^{itY}\right]=e^{i\mu_{X}t-\sigma_{X}^{2}t/2}e^{i\mu_{Y}t-\sigma_{Y}^{2}t/2}=e^{i(\mu_{X}+\mu_{Y})t-(\sigma_{X}^{2}+\sigma_{Y}^{2})t/2} $$ which is the characteristic function of a normal variable with mean $\mu_X+\mu_Y$ and variance $\sigma_X^2+\sigma_Y^2$.