91Ó°ÊÓ

Given in the question that, the amount of nitrogen oxides {NOX}) present in the exhaust of a particular type of car varies from car to car according to a Normal distribution with mean $1.4$grams per mile and standard deviation $0.3$grams per mile

Consider $X_1$and $X_2$as the random variables, which showing the amount of nitrogen oxides present in the randomly selected cars.

Let's define a variable $D=X_1−X_2$

We have to find the $P\left(X_1−X_2â‰\yen 0.80\right)$

According to the information, the NOX levels of two cars of this type are independent.

Therefore mean of $D$is

$$\begin{gathered} {\mathrm{μ}}_D={\mathrm{μ}}_{X_1}+{\mathrm{μ}}_{X_2} \\ =1.4−1.4 \\ =0 \end{gathered}$$

The standard deviation of $D$is:

localid="1649864820999" $$\begin{gathered} {\mathrm{σ}}_D=\sqrt{{{\mathrm{σ}}_{X_1}}^2+{{\mathrm{σ}}_{X_2}}^2} \\ =\sqrt{{0.3}^2+{0.3}^2} \\ =0.424 \end{gathered}$$

We have to find the probability that the entire process will take less than $30$seconds. that is

$P(Dâ‰\yen 0.80)$

First we standardize $D=0.80$

$$\begin{gathered} z=\frac{x-{\mathrm{μ}}_D}{{\mathrm{σ}}_D} \\ =\frac{0.80−0}{0.424} \\ =1.89 \end{gathered}$$

Using a table of typical normal probabilities as a guide,

$$\begin{gathered} \mathrm{P}(\mathrm{Z}<1.89\text{or}\mathrm{Z}>1.89) \\ =2×\mathrm{P}(\mathrm{Z}<−1.89) \\ =2×0.0294 \\ =0.0588 \end{gathered}$$