Auto emissions The amount of nitrogen oxides (NOX) present in the exhaust of a particular model of old car varies from car to car according to a Normal distribution with mean 1.4 grams per mile $(\mathrm{g}/\mathrm{mi})$and standard deviation $0.3\mathrm{g}/\mathrm{mi}$. Two randomly selected cars of this model are tested. One has $1.1\mathrm{g}/\mathrm{mi}$of $\mathrm{NOX}$; the other has $1.9\mathrm{g}/\mathrm{mi}$. The test station attendant finds this difference in emissions between two similar cars surprising. If the NOX levels for two randomly chosen cars of this type are independent, find the probability that the difference is greater than $0.8$or less than $-0.8$.

The Mean and standard deviation,

For first car:

${\mu }_{X_1}=1.4\mathrm{g}/\mathrm{mi},{\sigma }_{X_1}=0.3\mathrm{g}/\mathrm{mi}$

Now For second car:

${\mu }_{X_2}=1.4\mathrm{g}/\mathrm{mi},{\sigma }_{X_2}=0.3\mathrm{g}/\mathrm{mi}$

Now the difference in emissions,

$-0.8<x<0.8$

Let

$x_1:$The first car's amount of nitrogen oxides

$X_2$: The second car's amount of nitrogen oxides

Then

$X_2-X_1$is the difference in nitrogen oxide emissions between two comparable cars

$X_2$and $X_1$are independent.

Since ${\mathrm{X}}_2$and ${\mathrm{X}}_1$have a Normal distribution, and the difference between them $\left({\mathrm{X}}_2-{\mathrm{X}}_1\right)$also has a Normal distribution.

Now,

If there are any two variables,

The difference in their means equals the difference in their means.

$\begin{array}{rcl}{\mu }_{X_1-X_2}& =& {\mu }_{X_1}-{\mu }_{X_2}\\ & =& 1.4-1.4\\ & =& 0\end{array}$

The variance of the difference is equal to the sum of the variances of the random variables when they are independent.

$\begin{array}{rcl}{\sigma }_{X_1-X_2}^2& =& {\sigma }_{X_1}^2+{\sigma }_{X_2}^2\\ & =& (0.3{)}^2+(0.3{)}^2\\ & =& 0.09+0.09\\ & =& 0.18\mathrm{g}/\mathrm{mi}\end{array}$

The square root of the variance is the standard deviation.

$\begin{array}{rcl}{\sigma }_{X_1-X_2}& =& \sqrt{{\sigma }_{X_1-X_2}^2}\\ & =& \sqrt{0.18}\\ & \approx & 0.4243\mathrm{g}/\mathrm{mi}\end{array}$

Thus,

The variation $\left(X_2-X_1\right)$follows a Normal distribution.

Now,

For Probability:

Calculate the z - score,

$z=\frac{-0.8-0}{0.4243}\approx -1.89$

Or

$z=\frac{0.8-0}{0.4243}\approx 1.89$

Use the normal probability table in the appendix to get the corresponding probability.

See the standard normal probability table's row beginning with -1.8 and column beginning with.09 for further information. $P(z<-1.89)$.

$P\left(X_1-X_2<-0.8\text{or}{X}_1-X_2>0.8\right)=P(z<-1.89\text{or}z>1.89)$

$$\begin{gathered} =2\left(P\right(z<-1.89) \\ =2(0.0294) \\ =0.0588 \end{gathered}$$

Thus,$0.0588$is there a chance that the difference in emissions between two automobiles that are similar is greater than 0.8 or less than $-0.8$.