Chapter 6.2, Problem 71E

To determine

Probability for the difference in emissions between two similar cars is greater than 0.8 or less than -0.8.

Answer to Problem 71E

Probability for the difference is greater than 0.8 or less than -0.8 is 0.0588.

Explanation of Solution

Given information:

Mean and standard deviation,

For first car:

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

For second car:

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

Difference in emissions,

  $-0.8<x<0.8$

Calculations:

Let

X₁ : Amount of Nitrogen oxides of the first car

X₂ : Amount of Nitrogen oxides of the second car

Then

X₂ - X₁ is the difference the amount of Nitrogen oxides for two similar cars

X₂and X₁ are independent.

Since X₂andX₁have a Normal distribution, their difference ( X₂ - X₁ ) also has a Normal distribution.

Now,

For any two variables,

The mean of the difference is equal to the difference of their means.

  ${\mu }_{X_1-X_2}={\mu }_{X_1}-{\mu }_{X_2}=1.4-1.4=0$

When the random variables are independent, the variance of the difference is equal to the sum of their variances.

  ${\sigma }_{X_1-X_2}^2={\sigma }_{X_1}^2+{\sigma }_{X_2}^2={\left(0.3\right)}^2+{\left(0.3\right)}^2=0.09+0.09=0.18\text{g}/\text{mi}$

We know that

The standard deviation is the square root of the variance.

  ${\sigma }_{X_1-X_2}=\sqrt{{\sigma }_{X_1-X_2}^2}=\sqrt{0.18}\approx 0.4243\text{g}/\text{mi}$

Thus,

The difference ( X₂ - X₁ )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 normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -1.8 and the column that starts with .09 of the standard normal probability table for $P\left(z<-1.89\right)$ .

  $\begin{array}{l}P\left(X_1-X_2<-0.8\text{or}{X}_1-X_2>0.8\right)=P\left(z<-1.89\text{or}z>1.89\right)\\ =2\left(P\left(z<-1.89\right)\right)\\ =2\left(0.0294\right)\\ =0.0588\end{array}$

Thus,

Probability for the difference in emissions between two similar cars is greater than 0.8 or less than -0.8 is 0.0588.