A contract runs for 300 days (independently) for a lorry hire to transport your goods. The lorry journey is 25km, and the random variable X is the time of the journey in minutes. $x = 60, 75, 90$, the probability function is given as $p(x) = 0.32, 0.43, 0.25$.
You are charged for the hire service at 0.70 cents per minute, and $2 per km.
What is the Expected total and Variance of the cost?
What I have tried:
Using the following
$$\
Y = \sum_{i=1}^n (a_i \cdot X_i+b) = \sum_{i=1}^n (a_i \cdot X_i) + n \cdot b\\
Var(Y) = \sum_{i=1}^n a^2_i var(X_i)$$
I have the following calculation: $$E(Y) =0.32 \cdot \left(0.7 \sum_{i=1}^{300}60 + 300 \cdot 50 \right) + 0.43 \cdot \left(0.7 \sum_{i=1}^{300}75 + 300 \cdot 50 \right) + 0.25 \cdot \left(0.7 \sum_{i=1}^{300}90 + 300 \cdot 50 \right)$$
$$E(Y^2) = 0.32 \cdot \left(0.7 \sum_{i=1}^{300}60 + 300 \cdot 50 \right)^2 + 0.43 \cdot \left(0.7 \sum_{i=1}^{300}75 + 300 \cdot 50 \right)^2 + 0.25 \cdot \left(0.7 \sum_{i=1}^{300}90 + 300 \cdot 50 \right)^2$$
$Var(Y) = E(X^2) - E(X)^2 = 9.37658×10^8 - 30529.5^2 = 5.60762975 × 10^6$
But these numbers seem far too large to be true. I'd really appreciate some support on approaching this!
First determine the mean and variance of the daily cost.
If $Y = 0.70X + 50$, then $\operatorname{E}(Y) = 0.70\operatorname{E}(X)+50$ and $\operatorname{Var}(Y) = 0.70^2 \operatorname{Var}(X)$, and we have
$$\begin{align} \operatorname{E}(X^2) &= \sum_{i=1}^3 {x_i}^2 P(X=x_i) = 60^2 \cdot 0.32 + 75^2 \cdot 0.43 + 90^2 \cdot 0.25 = 5595.75 \\ \operatorname{E}(X) &= \sum_{i=1}^3 x_i P(X=x_i) = 60 \cdot 0.32 + 75 \cdot 0.43 + 90 \cdot 0.25 = 73.95 \end{align}$$
which means $\operatorname{E}(Y) \approx 101.77$, and $\operatorname{Var}(X) \approx 62.30$ so that $\operatorname{Var}(Y) \approx 30.53$.
Now use this information to determine the mean and variance of the cost across $300$ days. Let $Z = \sum\limits_{i=1}^{300}Y_i$, where each of the $Y_i$ are identical and independently distributed with
$$P(Y = y) = \begin{cases}0.32 & \text{if }y=0.70\cdot60+50=92 \\ 0.43 & \text{if }y=0.70\cdot75+50=102.5 \\ 0.25 & \text{if }y=0.70\cdot90+50=113\end{cases}$$
Because the $Y_i$ are i.i.d with expectation $\operatorname{E}(Y)$ and variance $\operatorname{Var}(Y)$, we have
$$\begin{align} \operatorname{E}(Z) &= \operatorname{E}\left(\sum_{i=1}^{300} Y_i\right) = \sum_{i=1}^{300} \operatorname{E}(Y_i) = 300 \operatorname{E}(Y) \approx 30529.5 \\[1ex] \operatorname{Var}(Z) &= \operatorname{Var}\left(\sum_{i=1}^{300} Y_i\right) = \sum_{i=1}^{300} \operatorname{Var}(Y_i) = 300 \operatorname{Var}(Y) \approx 18690.7 \end{align}$$