EXAMPLE 5.32 (Example 5.30 continued)

The total revenue from the sale of the three grades of gasoline on a particular day was $Y=3.0X_1+3.2X_2+3.4X_3$ , and we calculated ${\mu }_Y=5620$ and (assuming independence) ${\sigma }_Y=429.46$ . If the $X_i$ ’s are normally distributed, the probability that revenue exceeds 4500 is

$$\begin{array}{rl}P\left(Y>4500\right)& =P\left(Z>\frac{4500-5620}{429.46}\right)\\ & =P\left(Z>-2.61\right)\\ & =1-\Phi \left(-2.61\right)=0.9955\end{array}$$