A news report stated that the mean distance that commuters in the United States travel each way to work is 18 miles. Assume the standard deviation is 6 miles. A sample of 70 commuters is chosen.

Question

(b) What is the probability that the sample mean commute distance is between 19 and 20 miles? Round the answer to at least four decimal places.

use a T-84 calculator

GPT-4o mini · Answer

To find the probability that the sample mean commute distance is between 19 and 20 miles, we can use the properties of the sampling distribution of the sample mean.

Given data:

Population mean (μ) = 18 miles
Population standard deviation (σ) = 6 miles
Sample size (n) = 70

Step 1: Calculate the standard error of the mean (SEM):

The standard error of the mean (SEM) is calculated using the formula: \[ \text{SEM} = \frac{\sigma}{\sqrt{n}} \] Substituting the values we have: \[ \text{SEM} = \frac{6}{\sqrt{70}} \approx \frac{6}{8.3666} \approx 0.7161 \]

Step 2: Convert the sample mean values to z-scores:

We need to find the z-scores for the sample mean commute distances of 19 miles and 20 miles:

For 19 miles, the z-score is calculated as: \[ z = \frac{X - \mu}{\text{SEM}} = \frac{19 - 18}{0.7161} \approx \frac{1}{0.7161} \approx 1.3962 \]
For 20 miles, the z-score is calculated as: \[ z = \frac{X - \mu}{\text{SEM}} = \frac{20 - 18}{0.7161} \approx \frac{2}{0.7161} \approx 2.7930 \]

Step 3: Use the T-84 calculator to find the probabilities:

Now we can find the probability for the z-scores using the normal distribution:

Find the area to the left of z = 1.3962:
- On the T-84, you can calculate this using the normalCDF function. The command will be:
```
normalcdf(-1E99, 1.3962)
```
Find the area to the left of z = 2.7930:
- Use:
```
normalcdf(-1E99, 2.7930)
```

Step 4: Calculate the probabilities:

Let’s compute each of these:

Probability for z = 1.3962 (let’s call this P1)
Probability for z = 2.7930 (let’s call this P2)

Finally, the probability that the sample mean is between 19 and 20 miles can be computed as: \[ P(19 < \bar{X} < 20) = P2 - P1 \]

Example Calculation:

After performing the calculations on the T-84, you might typically find that:

\( P1 \approx 0.9177 \) (probability for z = 1.3962)
\( P2 \approx 0.9974 \) (probability for z = 2.7930)

Thus, the probability that the sample mean commute distance is between 19 and 20 miles is: \[ P(19 < \bar{X} < 20) = 0.9974 - 0.9177 = 0.0797 \]

Round the answer:

Rounding this to at least four decimal places gives: \[ P(19 < \bar{X} < 20) \approx 0.0797 \]

Final answer: The probability that the sample mean commute distance is between 19 and 20 miles is approximately 0.0797.