Calculating the independence sum of variance

Solution 1:

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

$$\begin{align} \operatorname{E}(Z) &= \operatorname{E}\left(\sum_{i=1}^{300} Y_i\right) = \sum_{i=1}^{300} \operatorname{E}(Y_i) = 300 \operatorname{E}(Y) \approx 30529.5 \\[1ex] \operatorname{Var}(Z) &= \operatorname{Var}\left(\sum_{i=1}^{300} Y_i\right) = \sum_{i=1}^{300} \operatorname{Var}(Y_i) = 300 \operatorname{Var}(Y) \approx 18690.7 \end{align}$$