91影视

The Mean and standard deviation,

For first car:

${渭}_{X_1}=1.4\mathrm{g}/\mathrm{mi},{蟽}_{X_1}=0.3\mathrm{g}/\mathrm{mi}$

Now For second car:

${渭}_{X_2}=1.4\mathrm{g}/\mathrm{mi},{蟽}_{X_2}=0.3\mathrm{g}/\mathrm{mi}$

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.

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

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

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

The square root of the variance is the standard deviation.

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

Thus,

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

Now,

For Probability:

Calculate the z - score,

$z=\frac{-0.8-0}{0.4243}鈮�-1.89$

Or

$z=\frac{0.8-0}{0.4243}鈮�1.89$

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 $-0.8$.