Chapter 7, Problem 128CR

Suppose that fuel efficiency for a particular model car under specified conditions is normally distributed with a mean value of 30.0 mpg and a standard deviation of 1.2 mpg.

1. a. What is the probability that the fuel efficiency for a randomly selected car of this type is between 29 and 31 mpg?
2. b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than 25 mpg?
3. c. If three cars of this model are randomly selected, what is the probability that all three have efficiencies exceeding 32 mpg?
4. d. Find a number c* such that 95% of all cars of this model have efficiencies exceeding c* (that is. P(x > c*) = 0.95).

To determine

Obtain the probability that the fuel efficiency for a randomly selected car for a particular type is between 29 and 31 mpg.

Answer to Problem 128CR

The probability that the fuel efficiency for a randomly selected car for a particular type is between 29 and 31 mpg is 0.5934.

Explanation of Solution

Calculation:

It is given that the fuel efficiency for a particular type of car is normally distributed with mean and standard deviation as 30.0 mpg and 1.2 mpg, respectively.

That is, $\mu =30.0\text{ mpg and }\sigma =1.2\text{ mpg}$

Define the random variable x as fuel efficiency for a randomly selected car of a particular type.

The general formula to x value to z score is $z*=\frac{x*-\mu }{\sigma }$

Standardize $a=29$ using the above formula.

$\begin{array}{c}z*=\frac{a*-\mu }{\sigma }\\ =\frac{29-30}{1.2}\\ =\frac{-1}{1.2}\\ =-0.83\end{array}$

Standardize $b=31$ using the above formula.

$\begin{array}{c}z*=\frac{b*-\mu }{\sigma }\\ =\frac{31-30}{1.2}\\ =\frac{\text{1}}{1.2}\\ =0.83\end{array}$

The required probability is obtained as given below:

$P\left(-0.83<z<0.83\right)=P\left(z<0.83\right)-P\left(z<-0.83\right)$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<-0.83\right)$.

Procedure:

• In the $z*$ row locate −0.8
• In column locate .03
• The intersection of the row −0.8 and column .03 gives 0.2033

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<0.83\right)$.

Procedure:

• In the $z*$ row locate 0.8
• In column locate .03
• The intersection of the row 0.8 and column .03 gives 0.7967

The required probability is obtained as given below:

$\begin{array}{c}P\left(-0.83<z<0.83\right)=P\left(z<0.83\right)-P\left(z<-0.83\right)\\ =0.7967-0.2033\\ =\text{0}\text{.5934}\end{array}$

Thus, the probability that the fuel efficiency for a randomly selected car for a particular type is between 29 and 31 mpg is 0.5934.

To determine

Check if it the efficiency of a randomly selected car of this model is less than 25 mpg surprises or not.

Answer to Problem 128CR

Yes, the result that the efficiency of a randomly selected car of this model is less than 25 mpg surprises.

Explanation of Solution

Calculation:

Standardize $c=25$ using the above formula.

$\begin{array}{c}z*=\frac{c*-\mu }{\sigma }\\ =\frac{25-30}{1.2}\\ =\frac{-5}{1.2}\\ =-4.17\end{array}$

Since, the Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain does not contain −4.17 consider −3.89.

Procedure:

• In the $z*$ row locate −3.8
• In column locate .09
• The intersection of the row −3.8 and column .09 gives 0.000

The required probability is obtained as given below:

$P\left(z<-4.17\right)=\text{0}\text{.000}$

Since, the probability is 0, it is very unlikely to select a car with fuel efficiency less than 25 mpg.

Hence, it is surprised to get the result that the efficiency of a randomly selected car of this model is less than 25 mpg.

To determine

Obtain the probability that all three cars have efficiencies exceeding 32 mpg when they are randomly selected.

Answer to Problem 128CR

The probability that all three cars have efficiencies exceeding 32 mpg when they are randomly selected is 0.0001.

Explanation of Solution

Calculation:

Standardize $c=32$ using the above formula.

$\begin{array}{c}z*=\frac{c*-\mu }{\sigma }\\ =\frac{32-30}{1.2}\\ =\frac{2}{1.2}\\ =1.67\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<1.67\right)$.

Procedure:

• In the $z*$ row locate 1.6
• In column locate .07
• The intersection of the row 1.6 and column .07 gives 0.9525

The probability that a randomly selected car has fuel efficiency exceeding 32 mpg is obtained as given below:

$\begin{array}{c}P\left(z\ge 1.67\right)=1-P\left(z<1.67\right)\\ =1-0.9525\\ =\text{0}\text{.0475}\end{array}$

The required probability is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{three randomly selected cars}\\ \text{ all have at least 32 mpg}\end{array}\right)={\left(\text{0}\text{.0475}\right)}^3\\ =\text{0}\text{.0001}\end{array}$

Thus, the probability that all three cars have efficiencies exceeding 32 mpg when they are randomly selected is 0.0001.

To determine

Obtain the value of c* such that 95% of all cars of the model have efficiencies exceeding c*.

Answer to Problem 128CR

The value of c* is 28.026.

Explanation of Solution

Calculation:

It is given that 95% of all cars of the model have efficiencies exceeding c*.

$\begin{array}{c}P\left(x\ge c*\right)=0.95\\ P\left(x<c*\right)=1-0.95\\ =0.05\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain z*.

In the table, locate probability value of approximately 0.05. It lies between 0.0505 and 0.0495. The corresponding z* scores are −1.64 and −1.65. Hence, the z* corresponding to the probability 0.05 is the average of the two z* scores.

Thus, z* is −1.645.

The value of c* is obtained as given below:

$\begin{array}{c}c=\mu +z*\sigma \\ =30+\left(-1.645\right)\left(1.2\right)\\ =30-\text{1}\text{.974}\\ =\text{28}\text{.026}\end{array}$

Thus, the value of c* is 28.026.