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A news report stated that the mean distance that commuters in the United States travel each way to work is 16 miles. Assume the standard deviation is 8 miles. A sample of 80 commuters is chosen.

a) What is the probability that the sample mean commute distance is greater than 15 miles? Round the answer to at least four decimal places.

USE A T-84 CALCULATOR

3 answers

To find the probability that the sample mean commute distance is greater than 15 miles, we can use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normally distributed if the sample size is large enough (which it is in this case with n = 80).

Given:

  • Population mean (\( \mu \)) = 16 miles
  • Population standard deviation (\( \sigma \)) = 8 miles
  • Sample size (\( n \)) = 80

Step 1: Calculate the standard error (SE) of the sample mean

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{8}{\sqrt{80}} \approx 0.8944 \]

Step 2: Calculate the Z-score for the sample mean of 15 miles

The Z-score formula is:

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

Where:

  • \( \bar{x} \) is the sample mean (15 miles)
  • \( \mu \) is the population mean (16 miles)

Calculating the Z-score:

\[ Z = \frac{15 - 16}{0.8944} \approx \frac{-1}{0.8944} \approx -1.118 \]

Step 3: Find the probability associated with this Z-score

You want to find \( P(\bar{x} > 15) \).

Using the Z-table (or calculator) to find \( P(Z < -1.118) \):

  1. First, find \( P(Z < -1.118) \). This is the area to the left of \( Z = -1.118 \).
  2. Use a normal distribution calculator or Z-table to find this value. By consulting the Z-table, we find that \( P(Z < -1.118) \approx 0.1303 \).
  3. Therefore, the probability that the sample mean is greater than 15 miles is:

\[ P(Z > -1.118) = 1 - P(Z < -1.118) \approx 1 - 0.1303 \approx 0.8697 \]

Final Answer:

\[ P(\bar{x} > 15) \approx 0.8697 \]

So the probability that the sample mean commute distance is greater than 15 miles is approximately 0.8697 (rounded to four decimal places).

(b) What is the probability that the sample mean commute distance is between 17 and 18 miles? Round the answer to at least four decimal places.

To find the probability that the sample mean commute distance is between 17 and 18 miles, we'll again use the standard normal distribution, calculated as follows:

Step 1: Calculate the Z-scores for 17 and 18 miles

Using the information from the previous part:

  • Population mean (\(\mu\)) = 16 miles
  • Standard error (\(SE\)) = 0.8944 (calculated earlier)

Z-score for 17 miles

\[ Z_{17} = \frac{17 - 16}{0.8944} = \frac{1}{0.8944} \approx 1.1180 \]

Z-score for 18 miles

\[ Z_{18} = \frac{18 - 16}{0.8944} = \frac{2}{0.8944} \approx 2.2360 \]

Step 2: Find the probabilities associated with these Z-scores

Next, we will find \(P(17 < \bar{x} < 18)\), which is equivalent to finding \(P(Z_{17} < Z < Z_{18})\).

Finding the probabilities for each Z-score

  1. For \(Z = 1.1180\):

    • Using a Z-table or calculator, we find \(P(Z < 1.1180) \approx 0.8686\).

  2. For \(Z = 2.2360\):

    • Using a Z-table or calculator, we find \(P(Z < 2.2360) \approx 0.9878\).

Step 3: Calculate the probability between the two Z-scores

Now, we can find the probability that the sample mean is between 17 and 18 miles:

\[ P(17 < \bar{x} < 18) = P(Z < 2.2360) - P(Z < 1.1180) \] \[ P(17 < \bar{x} < 18) \approx 0.9878 - 0.8686 = 0.1192 \]

Final Answer

So, the probability that the sample mean commute distance is between 17 and 18 miles is approximately 0.1192 (rounded to four decimal places).