Standard deviation with multiple means and deviations

The amounts of a certain mineral that can be produced in a day from mines $1$, $2$, and $3$ are independent normal random variables with means equal to $80$, $90$, and $75$ pounds, respectively, and with standard deviations equal to $12$, $14$, and $10$ pounds, respectively. On a given day, what is the probability that the combined amount of mineral produced from all three mines exceeds $283$ pounds?

I know this might be a little advanced for me since I am teaching myself this topic. I am going ahead a little to see what certain problems might be like. I think it might be a good idea to see what the working of these certain problems might look like.

From my understanding, we could work with the sum random variable $Y=X_1+X_2+X_3$, where random variable $X_i$ stands for the daily amount of mineral that can be produced from mine $i$

I know we want to find $P(X>283)$.

Can someone help me with this? I am trying this problem for fun and it seems like something good to know for later. I think seeing this problem will help my understanding of this topic when learning it later by myself.

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You need $P(Y > 283)$ – Alex May 14 '14 at 8:07
en.wikipedia.org/wiki/… – Alex May 14 '14 at 8:07
I am learning this topic by myself. – earl May 14 '14 at 13:31
I did not get to learning about this yet. I am just trying it for fun. I think seeing this problem will help my understanding of what I will learn later on by myself. – earl May 14 '14 at 13:32

If you're learning this on your own — and I'm not sure in what order you're learning everything — the main takeaway would be that the sum of random variables (or any function of random variables for that matter) doesn't necessarily behave in a straightforward way. You probably know that, using your above numbers, $P(X_1 > 80) = 1/2$, $P(X_2 > 90) = 1/2$, and $P(X_3 > 75) = 1/2$; however, it's clearly not the case that $P(Y > 80 + 90 + 75) = 3/2$. These are separate mines, so should we treat them as independent variables?—i.e., maybe $P(Y>80+90+75) = (1/2)^3$. Actually, that's not true either. A better way to go about this is either using the properties of the distributions you're dealing with (which I'll show first) or taking advantage of some general method for working with multiple distributions (which I'll show second).

So. The sum of normal random variables is itself a normal random variable, with mean equal to the sum of the means and variance equal to the sum of the variances. In your case,

$$\mu_Y = \mu_{X_1} + \mu_{X_2} + \mu_{X_3} = 80 + 90 + 75 = 245$$

and

$$\sigma_Y^2 = \sigma_{X_1}^2 + \sigma_{X_2}^2 + \sigma_{X_3}^2 = 12^2 + 14^2 + 10^2 = 440$$

Therefore, $Y \sim N(245, 440)$. If we normalize $Y$, letting $Z = \frac{Y - 245}{\sqrt{440}}$, then the probability becomes

$$P\left(Z > \frac{283-245}{\sqrt{440}}\right) \approx P(Z > 1.812)$$

which you can find on a $Z$-chart.

As for why the sum of normal random variables is also normal: You can use moment generating functions to show this. If you've seen them already, you know that $M(t) = E\left[e^{Xt}\right]$. Since $Y=X_1 + x_2 + X_3$, we know that

\begin{aligned} \ E\left[e^{Yt}\right] &= E\left[e^{(X_1+X_2+X_3)t}\right] \\ \ &= E\left[e^{X_1 t}\right]E\left[e^{X_2 t}\right]E\left[e^{X_3 t}\right] \\ \end{aligned}

That is, the moment generating function for $Y$ is the product of the moment generating functions for $X_i$. Using the properties of the exponential function, you can simplify that product to obtain a single expression, which also has the shape of a normal distribution's MGF.

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