Chapter 18, Problem 47E

(a)

To determine

To explain what is the approximate model for the distribution of $\overline{y}$ .

Answer to Problem 47E

It follows Normal with the mean $2.9$ g/mi and the standard deviation $0.045$ g/mi.

Explanation of Solution

It is given in the question that CO emissions for a certain kind of car vary with mean $2.9$ g/mi and standard deviation $0.4$ g/mi. Let $\overline{y}$ represents the mean CO level for the company’s fleet. Thus, the conditions are as:

Random condition: It is reasonable to think as we are assuming that $80$ cars can be considered a representative sample of all cars of this type.

Independent condition: It is reasonable to think that the CO emissions for these cars are mutually independent.

$10\%$ condition: The sample size of $80$ cars in the fleet represents less than $10\%$ of all cars of this type.

Large enough sample condition: A sample of $80$ cars is large enough thus, it is satisfied.

Thus, all the conditions are met. And the mean and standard deviation is as follows:

  $\begin{array}{l}{\mu }_{\overline{y}}=2.9\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{0.4}{\sqrt{80}}\\ =0.045\end{array}$

(b)

To determine

To estimate the probability that $\overline{y}$ is between $3.0\text{ and 3}\text{.1}$ g/mi.

Answer to Problem 47E

  $0.0131$ .

Explanation of Solution

It is given in the question that CO emissions for a certain kind of car vary with mean $2.9$ g/mi and standard deviation $0.4$ g/mi. Let $\overline{y}$ represents the mean CO level for the company’s fleet. And the mean is $2.9$ g/mi and the standard deviation is $0.045$ g/mi. Then, let us first find the z -score as:

  $\begin{array}{l}{z}_{3.0}=\frac{3.0-2.9}{0.045}\\ =2.222\\ z_{3.1}=\frac{3.1-2.9}{0.045}\\ =4.444\end{array}$

Thus, the probability that $\overline{y}$ is between $3.0\text{ and 3}\text{.1}$ g/mi can be calculated as:

  $\begin{array}{l}P(2.222<z<4.444)=normalcdf(2.222,4.444,0,1)\\ =0.0131\end{array}$

Thus, the probability that $\overline{y}$ is between $3.0\text{ and 3}\text{.1}$ g/mi is approximately $0.0131$ .

(c)

To determine

To find out the fleet’s mean CO level is greater than what value.

Answer to Problem 47E

There is only a $5\%$ chance that the fleets mean CO level is greater than approximately $2.974$ g/mi.

Explanation of Solution

It is given in the question that CO emissions for a certain kind of car vary with mean $2.9$ g/mi and standard deviation $0.4$ g/mi. Let $\overline{y}$ represents the mean CO level for the company’s fleet. And the mean is $2.9$ g/mi and the standard deviation is $0.045$ g/mi. Thus, the fleet’s mean CO level is greater than what value can be calculated using the graphical utility as:

  $invNorm(0.95,0,1)=1.645$

Now put in $1.645$ in z as:

  $\begin{array}{l}1.645=\frac{y-2.9}{0.045}\\ \Rightarrow 0.074=y-2.9\\ \Rightarrow y=2.974\end{array}$

Thus, there is only a $5\%$ chance that the fleets mean CO level is greater than approximately $2.974$ g/mi.