probability - Find the waiting time at a copy machine with 2 classes and non-preemptive priority

People arrive at a copy machine according to a Poisson process with rate one per minute. The number of copies to be made by each person is uniformly distributed between 1 and 10. Each copy requires 3 seconds. Find the average waiting time in queue when:

People with no more than 2 copies to make are given non-preemptive priority over other
people.

Let $class \; 1$ be people with no more than 2 copies and $class \; 2$ others. I'm having trouble calculating $E[X_1^2]$ and $E[X_2^2]$. I'm pretty sure $\displaystyle E[X_1] = 3 \frac{2+1}{2} = 4.5$ and $\displaystyle E[X_2] = 3 \frac{10+3}{2} = 19.5$ . After this we have a waiting time formula which we plug these values into.

I tried looking at $\displaystyle E[X^2] = \frac{b^2 + ab + a^2}{3}$ so get $E[X_1^2] = 7$ and $E[X_2^2] = 133$ but these don't give a reasonable answer.

Answer

Ok so as requested I retype this into an answer.

In the original post the equation

$$E[X^2] = \frac{b^2 + ab + a^2}{3}$$

correspond to a continuous uniform distribution, $X \sim \text{Uniform}(a, b)$

However, as the question is modelling discrete counts, it should be referring to a discrete uniform distribution instead. For a discrete uniform random variable with integral support $\{a, a+1, \ldots, b\}$, we have

$$\begin{align} E[X^2] &= Var[X] + E[X]^2 \\ &= \frac {(b -a - 1)^2 + 1} {12} + \left(\frac {a + b} {2}\right)^2 \\ &= \frac {b^2 + a^2+1 - 2ab - 2b + 2a + 1 + 3a^2 + 6ab + 3b^2} {12} \\ &= \frac {2a^2 + 2b^2 + 2ab + a - b+ 1} {6} \end{align}$$

where the variance formula is quoted from
https://en.wikipedia.org/wiki/Uniform_distribution_(discrete)

One can also try to sum $\displaystyle \frac {1} {b-a+1} \sum_{x=a}^b x^2$ directly to obtain the answer.