Chapter 7.5, Problem 26E

The mean annual cost of automobile insurance is $939 (CNBC, February 23, 2006). Assume that the standard deviation is σ = $245.

1. a. What is the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for each of the following sample sizes: 30, 50, 100, and 400?
2. b. What is the advantage of a larger sample size when attempting to estimate the population mean?

To determine

Find the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for each of the following sample sizes: 30, 50, 100, and 400.

Answer to Problem 26E

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 30 is 0.4246.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 50 is 0.5284.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 100 is 0.6922.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 400 is 0.9586.

Explanation of Solution

Calculation:

The standard deviation is $\sigma =\$245$ and the mean annual cost of automobile insurance is $939.

Sampling distribution of $\overline{x}$:

• The expected value of $\overline{x}$ is, $E\left(\overline{x}\right)=\mu$.
• The standard deviation of $\overline{x}$ is

  For finite population, ${\sigma }_{\overline{x}}=\sqrt{\frac{N-n}{N-1}}\left(\frac{\sigma }{\sqrt{n}}\right)$

  For infinite population, ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

• When the population size is infinite or finite and $\frac{n}{N}\le 0.05$ then the standard deviation of $\overline{x}$ is ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$.

The expected value of $\overline{x}$ is obtained below:

$\begin{array}{c}E\left(\overline{x}\right)=\mu \\ =\$939\end{array}$

Thus, the expected value of $\overline{x}$ is $939.

For sample size 30:

The standard deviation is ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

Substitute $\sigma$ as $245 and n as 30 in the formula,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{245}{\sqrt{30}}\\ =\frac{245}{5.4772}\\ =44.73\end{array}$

Thus, the standard deviation of $\overline{x}$ is $44.73.

Therefore, the sampling distribution of the sample mean $\overline{x}$ is approximately normal with $\mu =\$939$ and ${\sigma }_{\overline{x}}=\$44.73$.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 30 is calculated as follows:

$\begin{array}{c}P\left(\left|\overline{x}\right|\le \mu \pm 25\right)=P\left(\mu -25\le \overline{x}\le \mu +25\right)\\ =P\left[\frac{-25}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{25}{{\sigma }_{\overline{x}}}\right]\\ =P\left(\frac{-25}{44.73}\le z\le \frac{25}{44.73}\right)\\ =P\left(-0.56\le z\le 0.56\right)\\ =P\left(z\le 0.56\right)-P\left(z\le -0.56\right)\end{array}$

For $z=0.56$:

From Table 1: Cumulative probabilities for the standard normal distribution,

• Locate the value 0.5 in the first column.
• Locate the value 0.06 in the first row corresponding to the value 0.5 in the first column.
• The intersecting value of row and column is 0.7123.

For $z=-0.56$:

From Table 1: Cumulative probabilities for the standard normal distribution,

• Locate the value –0.5 in the first column.
• Locate the value 0.06 in the first row corresponding to the value –0.5 in the first column.
• The intersecting value of row and column is 0.2877.

Therefore,

$\begin{array}{c}P\left(\mu -25\le \overline{x}\le \mu +25\right)=0.7123-0.2877\\ =0.4246\end{array}$

Thus, the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 30 is 0.4246.

For sample size 50:

The standard deviation is obtained below:

Substitute $\sigma$ as $245 and n as 50 in the formula.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{245}{\sqrt{50}}\\ =\frac{245}{7.0711}\\ =34.65\end{array}$

Thus, the standard deviation of $\overline{x}$ is 34.65.

Therefore, the sampling distribution of the sample mean $\overline{x}$ is approximately normal with $\mu =\$939$ and ${\sigma }_{\overline{x}}=\$34.65$.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 50 is calculated as follows:

$\begin{array}{c}P\left(\left|\overline{x}\right|\le \mu \pm 25\right)=P\left(\mu -25\le \overline{x}\le \mu +25\right)\\ =P\left[\frac{-25}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{25}{{\sigma }_{\overline{x}}}\right]\\ =P\left(\frac{-25}{34.65}\le z\le \frac{25}{34.65}\right)\\ =P\left(-0.72\le z\le 0.72\right)\\ =P\left(z\le 0.72\right)-P\left(z\le -0.72\right)\end{array}$

For $z=0.72$:

From Table 1: Cumulative probabilities for the standard normal distribution,

• Locate the value 0.7 in the first column.
• Locate the value 0.02 in the first row corresponding to the value 0.7 in the first column.
• The intersecting value of row and column is 0.7642.

For $z=-0.72$:

From Table 1: Cumulative probabilities for the standard normal distribution,

• Locate the value – 0.7 in the first column.
• Locate the value 0.02 in the first row corresponding to the value –0.7 in the first column.
• The intersecting value of row and column is 0.2358.

Therefore,

$\begin{array}{c}P\left(\mu -25\le \overline{x}\le \mu +25\right)=0.7642-0.2358\\ =0.5284\end{array}$

Thus, the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 50 is 0.5284.

For sample size 100:

The standard deviation is obtained below:

Substitute $\sigma$ as $245 and n as 100 in the formula.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{245}{\sqrt{100}}\\ =\frac{245}{10}\\ =24.5\end{array}$

Thus, the standard deviation of $\overline{x}$ is $24.5.

The sampling distribution of the sample mean $\overline{x}$ is approximately normal with $\mu =\$939$ and ${\sigma }_{\overline{x}}=\$24.5$.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 100 is calculated as follows:

$\begin{array}{c}P\left(\left|\overline{x}\right|\le \mu \pm 25\right)=P\left(\mu -25\le \overline{x}\le \mu +25\right)\\ =P\left[\frac{-25}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{25}{{\sigma }_{\overline{x}}}\right]\\ =P\left(\frac{-25}{24.5}\le z\le \frac{25}{24.5}\right)\\ =P\left(-1.02\le z\le 1.02\right)\\ =P\left(z\le 1.02\right)-P\left(z\le -1.02\right)\end{array}$

For $z=1.02$:

From Table 1: Cumulative probabilities for the standard normal distribution,

• Locate the value 1.0 in the first column.
• Locate the value 0.02 in the first row corresponding to the value 1.0 in the first column.
• The intersecting value of row and column is 0.8461.

For $z=-1.02$:

From Table 1: Cumulative probabilities for the standard normal distribution,

• Locate the value –1.0 in the first column.
• Locate the value 0.02 in the first row corresponding to the value –1.0 in the first column.
• The intersecting value of row and column is 0.1539.

Therefore,

$\begin{array}{c}P\left(\mu -25\le \overline{x}\le \mu +25\right)=0.8461-0.1539\\ =0.6922\end{array}$

Thus, the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 100 is 0.6922.

For sample size 400:

The standard deviation is obtained below:

Substitute $\sigma$ as $245 and n as 400 in the formula.

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{245}{\sqrt{400}}\\ =\frac{245}{20}\\ =12.25\end{array}$

Thus, the standard deviation of $\overline{x}$ is $12.25.

The sampling distribution of the sample mean $\overline{x}$ is approximately normal with $\mu =\$939$ and ${\sigma }_{\overline{x}}=\$12.25$.

The probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 400 is calculated as follows:

$\begin{array}{c}P\left(\left|\overline{x}\right|\le \mu \pm 25\right)=P\left(\mu -25\le \overline{x}\le \mu +25\right)\\ =P\left[\frac{-25}{{\sigma }_{\overline{x}}}\le \frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{25}{{\sigma }_{\overline{x}}}\right]\\ =P\left(\frac{-25}{12.25}\le z\le \frac{25}{12.25}\right)\\ =P\left(-2.04\le z\le 2.04\right)\\ =P\left(z\le 2.04\right)-P\left(z\le -2.04\right)\end{array}$

For $z=2.04$:

From Table 1: Cumulative probabilities for the standard normal distribution,

• Locate the value 2.0 in the first column.
• Locate the value 0.04 in the first row corresponding to the value 2.0 in the first column.
• The intersecting value of row and column is 0.9793.

For $z=-2.04$:

From Table 1: Cumulative probabilities for the standard normal distribution,

• Locate the value –2.0 in the first column.
• Locate the value 0.04 in the first row corresponding to the value –2.0 in the first column.
• The intersecting value of row and column is 0.0207.

Therefore,

$\begin{array}{c}P\left(\mu -25\le \overline{x}\le \mu +25\right)=0.9793-0.0207\\ =0.9586\end{array}$

Thus, the probability that a sample of automobile insurance policies will have a sample mean within $25 of the population mean for the sample size 400 is 0.9586.

To determine

Explain the advantage of a larger sample size to estimate the population mean.

Explanation of Solution

From part a, it was observed that the probability of being with $\pm 25$ of $\mu$ is from 0.4246 for sample size 30 to 0.9586 for sample size 400.

Hence, it can be concluded that as the sample size increases, the probability becomes closer to 1.