the profit of pizza

Question

The number of pizzas delivered to university students each month is a random variable with the following probability distribution $x=\{0,1,2,3\}$ and $$P(X=x)=\{.1,.3,.4,.2\}$$ respectively. If the pizzeria makes a profit of $\$3$ per pizza, determine the mean and variance of the profits per student. Please give the answers to two decimal places.

$$E(\text{Profit}) = ??, \qquad V(\text{Profit}) = ??$$

I have been having trouble with this. I tried to find the answers by dividing each number by the total, then adding them together, which doesn't seem right.

Answer 1

$\text{Mean} = \sum p_i X_i $

Where $X_i=3x_i$

$\text{Variance} =(\sum p_iX_i^2) - mean^2$

Answer 2

Hint: The monthly profit per student $Y$ is a random variable which is given by $$Y=3X$$ Thus by the properties of mean and variance we have that $$E[Y]=E[3X]=3E[X]=3\sum_{k=0}^{4}kP(X=k)$$ and $$Var(Y)=Var(3X)=3^2Var(X)=9\left(E[X^2]-E[X]^2\right)$$ where $$E[X^2]=\sum_{k=0}^{4}k^2P(X=k)$$

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Added: $$E[Y]=3\sum_{k=0}^{3}kP(X=k)=3\left(0\cdot0.1+1\cdot0.3+2\cdot0.4+3\cdot0.2\right)=5.1$$ and $$\begin{align*}E[Y^2]&=E[(3X)^2]=9E[X^2]=\\&=9\sum_{k=0}^{3}kP(X=k)=9\left(0^2\cdot0.1+1^2\cdot0.3+2^2\cdot0.4+3^2\cdot0.2\right)=33.3\end{align*}$$ Therefore $$Var(Y)=33.3 - 5.1^2=7.29$$