Question

Auto emissions 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 (g/mi) and standard deviation $0.3g/mi$. Two randomly selected cars of this type are tested. One has $1.1g/mi$of NOX; the other has $1.9g/mi$. 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 at least as large as the value the attendant observed.follow the four-step process.

Short answer

The difference might be as much as $0.8$grammes every mile, according to $2.94$percent of the time.

Step-by-step solution

Given Information

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

 Explanation

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 $P\left(X_1-X_2\ge 0.80\right)$

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

Therefore mean of $D$is

$$\begin{gathered} {\mu }_D={\mu }_{X_1}+{\mu }_{X_2} \\ =1.4-1.4 \\ =0 \end{gathered}$$

The standard deviation of $D$is:

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

The probability that the entire process will take less than 30 seconds. 

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

$P(D\ge 0.80)$

First we standardize $D=0.80$

$$\begin{gathered} z=\frac{x-{\mu }_D}{{\sigma }_D} \\ =\frac{0.80-0}{0.424} \\ =1.89 \end{gathered}$$

Using a table of typical normal probabilities as a guide,

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