EX 4.25

In recent years the Weibull distribution has been used to model engine emissions of various pollutants. Let $X$ denote the amount of ${\mathrm{NO}}_x$ emission $\left(\mathrm{g}/\mathrm{gal}\right)$ from a randomly selected four-stroke engine of a certain type, and suppose that $X$ has a Weibull distribution with $\alpha =2$ and $\beta =10$

• suggested by information in the article “Quantification of Variability and Uncertainty in Lawn and Garden Equipment ${\mathrm{NO}}_x$ and Total Hydrocarbon Emission Factors,” J. of the Air and Waste Management Assoc., 2002: 435-448).

The corresponding density curve looks exactly like the one in Figure 4.28 for $\alpha =2,\beta =1$ except that now the values 50 and 100 replace 5 and 10 on the horizontal axis. Then$$ \begin{align} P\left( X \leq 10 \right) &= F\left( 10; 2, 10 \right) \ &= 1 - e^{-\left( \frac{10}{10} \right)^2} \ &= 1 - e^{-1} \ &= 0.632 \end{align}

Similarly, $P\left( {X \leq {25}}\right) = {.998}$, so the distribution is almost entirely concentrated on values between 0 and 25. The value $c$ which separates the $5\%$ of all engines having the largest amounts of ${\mathrm{{NO}}}_{x}$ emissions from the remaining ${95}\%$ satisfies

{.95} = 1 - {e}^{-{\left( c/{10}\right) }^{2}}

Isolating the exponential term on one side, taking logarithms, and solving the resulting equation gives $c \approx {17.3}$ as the 95th percentile of the emission distribution.