Final answer:
To find the probability that the sample mean commute distance is greater than 15 miles, we can use the z-score formula. By calculating the z-score and using a standard normal distribution table, we find that the probability is practically 0. To find the probability that the sample mean commute distance is between 18 and 19 miles, we can calculate the z-scores for both values and find the probability in the same way.
Step-by-step explanation:
To answer part A, we need to find the probability that the sample mean commute distance is greater than 15 miles. We can use the z-score formula to calculate this probability. The z-score can be calculated as:
z = (x - μ) / (σ / √n)
Where x is the sample mean commute distance (15 miles), μ is the population mean commute distance (70 miles), σ is the standard deviation (10 miles), and n is the sample size (75).
By plugging in the values, we get z = (15 - 70) / (10 / √75) = -44.1421
Now, we can use a standard normal distribution table or a calculator to find the probability of a z-score less than -44.1421. This probability will be extremely close to 0, indicating that the probability of the sample mean commute distance being greater than 15 miles is practically 0.
For part B, we need to find the probability that the sample mean commute distance is between 18 and 19 miles. We can use the same z-score formula to calculate the z-scores for both 18 and 19 miles. Let's calculate the z-score for 18 miles:
z = (x - μ) / (σ / √n) = (18 - 70) / (10 / √75) = -33.8656
Similarly, let's calculate the z-score for 19 miles:
z = (x - μ) / (σ / √n) = (19 - 70) / (10 / √75) = -32.5493
By using the standard normal distribution table or a calculator, we can find the probability that the z-score is between -33.8656 and -32.5493, indicating the probability that the sample mean commute distance is between 18 and 19 miles.