When do we add variance and when do we use $Var(cx) = c^2Var(x)$? This is a problem that I'm working on:

A yoga studio is trying to estimate total class sales for next year.
They assume that:

*

*Between 10 and 14 people attend each class (uniformly distributed)

*Each person pays 15 dollars per class.

*Total attendance between any two class is independent.

*They host 2 classes a day, 5 days a week for the 50 weeks in the year (500 total classes).

We’ll estimate the total sales using a normal distribution.
Let $x$ be a random variable describing the number of people who attend
a single class, $E[x] = 12$ and $Var(x) = 2$.
i. Compute the expected value and variance of the revenue, $r = 15x$,
raised in a single class.
ii. Estimate the expected value and
variance of the revenue generated over the whole year of classes.

for i, I'm sure that the expected value is $12 \times 15 = 180$. For the variance, I think it's $Var(15x)=15^2\times2=450$ but I can't really explain why. Then for ii, I'm sure that the expected value is $180\times500=90000$. But for the variance, someone told me it's $500Var(r)=500\times450=225000$. Are these correct? Could someone please tell me when should I add variances and when should I use the linearity of expectation formula? My instructor said the following which I don't fully understand,

the sum is a sum of separate, random events, while
multiplication is an operation on the output of a single event.

 A: A sum of separate random events is a situation where we "roll the dice" again for every class. That's the actual scenario in this problem. Some random number of people between $10$ and $14$ show up for the first class. Later that day, some possibly-different number of people between $10$ and $14$ are in class. This continues for all $500$ classes that year.
Here, if $\mathbf R_i$ is the revenue from the $i^{\text{th}}$ class, the total revenue is $$\mathbf R_1 + \mathbf R_2 + \dots + \mathbf R_{500}.$$
We multiply a single random variable by $500$ if we only "roll the dice" once for all classes. This is a hypothetical alternative to the problem. Here, imagine there's a $500$-class program you sign up for at the beginning of the year. Some random number of people between $10$ and $14$ show up for the first class, and then those people will keep showing up twice a day for the rest of the year, no variation.
Here, if $\mathbf R$ is the revenue from one class, the total revenue is $500\mathbf R$. If you prefer,  you can think of this as $$\underbrace{\mathbf R + \mathbf R + \dots \mathbf R}_{\text{500 terms}}$$ which makes it easier to see that the only change is that the revenue is guaranteed to be the same every day.

Your expected value and variance for a single day are correct.
Expected values don't care about the difference between these two. We have $$\mathbb E[\mathbf R_1 + \dots + \mathbf R_{500}] = \mathbb E[\mathbf R_1] + \dots + \mathbb E[\mathbf R_{500}] = 500 \cdot \mathbb E[\mathbf R_1]$$where what I did in the last step was notice that even though the revenue might be different from day to day, they have the same distributions and therefore the same average amounts. We also have $\mathbb E[500\mathbf R] = 500 \cdot \mathbb E[\mathbf R]$.
With variances, the two situations are different:

*

*In the first scenario (the one we're actually in), we are adding $500$ independent random variables, so the variances also add. We get $\mathrm{Var}[\mathbf R_1] + \dots + \mathrm{Var}[\mathbf R_{500}]$ which simplifies to $500 \cdot \mathrm{Var}[\mathbf R_1]$ because all the variances are the same.

*In the second scenario, we would use the rule for scaling a variance: $\mathrm{Var}[500\mathbf R] = 500^2 \cdot \mathrm{Var}[\mathbf R]$.

