A news report stated that the mean distance that commuters in the United States travel each way to work is 14 miles. Assume the standard deviation is 9 miles. A sample of 75 commuters is chosen.

Question

(a) What is the probability that the sample mean commute distance is greater than 12 miles? Round the answer to at least four decimal places.

Answer 1

To find the probability that the sample mean commute distance is greater than 12 miles, we can use the Central Limit Theorem. According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large (which it is in this case, n = 75).

Step 1: Determine the mean and standard error

The population mean (\( \mu \)) is 14 miles, and the population standard deviation (\( \sigma \)) is 9 miles.

The standard error of the mean (SEM) is calculated using the formula:

\[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \]

where:

• \( \sigma = 9 \) miles
• \( n = 75 \)

Calculating the SEM:

\[ \text{SEM} = \frac{9}{\sqrt{75}} \approx \frac{9}{8.6603} \approx 1.0392 \text{ miles} \]

Step 2: Calculate the Z-score

Next, we compute the Z-score to standardize the sample mean of 12 miles. The Z-score is computed as:

\[ Z = \frac{\bar{x} - \mu}{\text{SEM}} \]

where:

• \( \bar{x} = 12 \) miles
• \( \mu = 14 \) miles
• \( \text{SEM} \approx 1.0392 \text{ miles} \)

Now substituting the values:

\[ Z = \frac{12 - 14}{1.0392} \approx \frac{-2}{1.0392} \approx -1.926 \]

Step 3: Find the probability

To find the probability that the sample mean is greater than 12 miles, we need to find the corresponding probability for \( Z > -1.926 \).

Using the standard normal distribution table (or a calculator), we first find the probability for \( Z < -1.926 \):

\[ P(Z < -1.926) \approx 0.027 \]

Therefore, the probability that the Z score is greater than -1.926 is:

\[ P(Z > -1.926) = 1 - P(Z < -1.926) \approx 1 - 0.027 = 0.973 \]

Thus, the probability that the sample mean commute distance is greater than 12 miles is approximately:

\[ \boxed{0.9730} \]

This value is rounded to four decimal places as requested.

Answer 2

(b) What is the probability that the sample mean commute distance is between 15 and 16 miles? Round the answer to at least four decimal places.

Answer 3

To find the probability that the sample mean commute distance is between 15 and 16 miles, we will once again utilize the standard normal distribution.

Step 1: Find the Z-scores for 15 miles and 16 miles

We already calculated the standard error of the mean (SEM) in part (a):

\[ \text{SEM} \approx 1.0392 \text{ miles} \]

Now we find the Z-scores for both sample means.

1. Z-score for 15 miles:

\[ Z_{15} = \frac{15 - 14}{\text{SEM}} = \frac{15 - 14}{1.0392} \approx \frac{1}{1.0392} \approx 0.962 \]

2. Z-score for 16 miles:

\[ Z_{16} = \frac{16 - 14}{\text{SEM}} = \frac{16 - 14}{1.0392} \approx \frac{2}{1.0392} \approx 1.925 \]

Step 2: Find the probabilities

Next, we find the probabilities associated with these Z-scores.

1. Probability for Z = 0.962:

Using a standard normal distribution table or calculator:

\[ P(Z < 0.962) \approx 0.8315 \]

2. Probability for Z = 1.925:

Using a standard normal distribution table or calculator:

\[ P(Z < 1.925) \approx 0.9726 \]

Step 3: Calculate the probability between 15 and 16 miles

Now we find the probability that the sample mean is between 15 and 16 miles:

\[ P(15 < \bar{x} < 16) = P(Z < 1.925) - P(Z < 0.962) \]

Substituting the values we found:

\[ P(15 < \bar{x} < 16) = 0.9726 - 0.8315 = 0.1411 \]

Thus, the probability that the sample mean commute distance is between 15 and 16 miles is approximately:

\[ \boxed{0.1411} \]

This value is rounded to four decimal places as requested.