Rental car fleet evaluation. National Car Rental Systems, Inc., commissioned the U.S. Automobile Club (USAC) to conduct a survey of the general condition of the cars rented to the public by Hertz, Avis, National, and Budget Rent-a-Car.* USAC officials evaluate each company’s cars using a demerit point system. Each car starts with a perfect score of 0 points and incurs demerit points for each discrepancy noted by the inspectors. One measure of the overall condition of a company’s cars is the mean of all scores received by the company (i.e., the company’s fleet mean score). To estimate the fleet mean score of each rental car company, 10 major airports were randomly selected, and 10 cars from each company were randomly rented for inspection from each airport by USAC officials (i.e., a sample of size n = 100 cars from each company’s fleet was drawn and inspected).

a. Describe the sampling distribution of x, the mean score of a sample of n = 100 rental cars.

b. Interpret the mean of x in the context of this problem.

c. Assume$\mu =30$ and $\sigma =60$for one rental car company. For this company, find$P\left(\overline{x}⩾45\right)$ .

d. Refer to part c. The company claims that their true fleet mean score “couldn’t possibly be as high as 30.” The sample mean score tabulated by USAC for this company was 45. Does this result tend to support or refute the claim? Explain.

10 major airports were randomly selected and 10 cars from each company were randomly rented for inspection from each airport by USAC officials.

a.

Let the random variable x is defined as the scores of the company, and the sample mean is defined as the mean score of the company. The mean and standard deviation of the population are and respectively.

Then, the sampling distribution of the sample mean

As we know that the probability distribution of $\overline{x}$is normal with mean${\mu }_{\overline{x}}=\mu$

The size of the sample$n=100$

Since,

$\begin{array}{rcl}{\sigma }_{\overline{x}}& =& \frac{\sigma }{\sqrt{n}}\\ & =& \frac{\sigma }{\sqrt{100}}\\ & & \end{array}$

Hence, the mean of the sampling distribution of x is and standard deviation$\frac{\sigma }{\sqrt{100}}$ .

b.

The sampling distribution of the sample mean is nothing but the company’s fleet mean score distribution of rented cars.If we consider the mean fleet time of all the 10 rented cars selected from the 10 airports is the sample mean.The mean of all the population is considered as $\mu$for all the rental car company and standard deviation as$\sigma$ . According to the central limit theorem as sample size is increased then the sampling distribution is approximated to be normal.Therefore, the sampling distribution is approximately normal having mean$\mu$ and standard deviation of$\frac{\sigma }{\sqrt{100}}$

c.

This sample is drawn from the population having mean and standard deviation as $\mu$and$\sigma$respectively, hence,

localid="1658241367107" $\mu =30$, localid="1658241362349" $\sigma =60$and the size of the sample localid="1658241358142" $n=100$

As we know that the probability distribution of $\overline{x}$is normal with mean

localid="1658241371916" $\begin{array}{rcl}{\mu }_{\overline{x}}& =& \mu \\ & =& 30\\ & & \end{array}$

localid="1658241376261" $\begin{array}{rcl}{\sigma }_{\overline{x}}& =& \frac{\sigma }{\sqrt{n}}\\ & =& \frac{60}{\sqrt{100}}\\ & =& 6\\ & & \end{array}$

Herer, the probability for the event that the company’s fleet mean score greater than or equal to 45, which is same as that of finding the probability for the event localid="1658241381184" $\left(\overline{x}⩾45\right)$for one rental cars of the company.

Now, the standard normal random variable z is defined as:

localid="1658241387915" $z=\frac{\left(\overline{x}-{\mu }_{\overline{x}}\right)}{{\sigma }_{\overline{x}}}$

Therefore, for ,localid="1658241407753" $\overline{x}=19.6$

localid="1658241400589" $\begin{array}{rcl}z& =& \frac{\left(45-30\right)}{6}\\ & =& 2.5\\ & & \end{array}$

Now the probability for the eventlocalid="1658241395406" $\left(\overline{x}>45\right)$is calculated as follows:

The normal curve associated with this event is shown below:

The areas A is obtained from the statistical tables,

$\begin{array}{rcl}P\left(\overline{x}⩾45\right)& =& P\left(z⩾2.5\right)\\ & =& 0.5-A\\ & =& 0.5-0.4938\\ & =& 0.0062\\ & & \end{array}$

Hence, the probability for the event $\left(\overline{x}⩾45\right)$for one rental cars of the company is 0.0062.

d.

The probability that the mean fleet score takes greater than 45 is 0.0062. The probability is not negligible.The company’s claim is true when the probability is not negligible.From the normal distribution properties, we can say that the probability is 0.5 for the average fleet is high.