A tutoring class consists of 5 students and meets for a total of $11$ times. After each lesson, students take a quiz. The score of each of the $5$ students is equally likely to be $1,2,3, 4$ independently of other students. Let $X$ be the sum of the quiz scores of the students in today’s lesson. What is the moment generating function of $X$?
I am very confused about what type of distribution $X$ follows. I figured I would assign a single distribution to each of the students, $X_k$, for the student $k$. Thus far, my guess is that $X$ follows a binomial distribution but the parameters are not clear. Any assistance is much appreciated.
One approach for this specific situation is to consider each student's score to be the result of two coin flips. So it is two sets of $5$ coin flips; the first set can be zero or one, the second set can be the "2's digit" and would add $2$ to the score if it is a $1$ and $0$ if it is a zero.
Suppose the first student gets a $1$. That would correspond to $X_1 = 0$ and $Y_1 = 0$ and the first score would be $2X_1 + Y_1 + 1$. If the second student got a $4$, that would correspond to $X_2 = 1$ and $Y_2 = 1$ for a score of $2X_2 + Y_2 + 1$. The total score would be $\sum_{i = 1}^{5} \big(2X_{i} + Y_{i} + 1\big)$.
So you would have a sum of $5$ Bernoulli random variables (a binomial with $n=5$ and $p = 1/2$), and $2$x a sum of $5$ other Bernoulli random variables. From there you can use theorems about moment generating functions.