Chapter 3.2, Problem 79E

(a)

To determine

To find out what percent of the $1120$ vehicles have worse gas mileage than the Malibu.

Answer to Problem 79E

About $92.92\%$ of the $1120$ vehicles have worse gas mileage than the Malibu.

Explanation of Solution

It is given that:

  $\begin{array}{l}\mu =18.7\\ \sigma =4.3\\ x=25\end{array}$

Thus the z -score will be as:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{25-18.7}{4.3}\\ =1.47\end{array}$

Thus the corresponding probability using the normal probability table will be as:

  $\begin{array}{l}P(X<25)\\ =P(Z<1.47)\\ =0.9292\\ =92.92\%\end{array}$

Thus, about $92.92\%$ of the $1120$ vehicles have worse gas mileage than the Malibu.

(b)

To determine

To explain how high must a vehicle’s gas mileage be in order to fall in the top $10\%$ of the $1120$ vehicles.

Answer to Problem 79E

The top $10\%$ of the $1120$ vehicles have a gas mileage above $24.204$ mpg.

Explanation of Solution

It is given that:

  $\begin{array}{l}\mu =18.7\\ \sigma =4.3\end{array}$

The cutoff value of the top $10\%$ is the same as the cutoff value of the bottom $90\%$ .

Thus the z -score with the probability of $90\%$ will be as:

  $z=1.28$

The z -score is as:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{x-18.7}{4.3}\end{array}$

Thus, we will find the value of x by computing as:

  $\begin{array}{l}1.28=\frac{x-18.7}{4.3}\\ \Rightarrow x-18.7=1.28(4.3)\\ \Rightarrow x=18.7+1.28(4.3)\\ \Rightarrow x=24.204\end{array}$

Thus the top $10\%$ of the $1120$ vehicles have a gas mileage above $24.204$ mpg.