According to the Census Bureau's 2002 American Community Survey, the average travel time to work of workers 16 years and over living in Boston, MA, who did not work at home, was 28.2 minutes with a standard deviation of 0.79 minutes. A sample of 35 commute times is taken. What is the probability that the mean commute time is at least 28 minutes? We cannot assume that the data comes from a normal distribution.

To find the probability that the mean commute time is at least 28 minutes, we need to calculate the probability of obtaining a sample mean of at least 28 minutes from the given population.

Since the population standard deviation is known and the sample size is large (n=35), we can use the Central Limit Theorem to approximate the distribution of the sample mean as approximately normal.

The Central Limit Theorem states that as the sample size increases, the sampling distribution of the mean approaches a normal distribution, regardless of the shape of the population distribution.

To calculate the probability, we need to standardize the sample mean using the formula for the z-score:

z = (x - μ) / (σ / √n)

Where:
- x is the sample mean (28 minutes)
- μ is the population mean (28.2 minutes)
- σ is the population standard deviation (0.79 minutes)
- n is the sample size (35)

Let's calculate the z-score:

z = (28 - 28.2) / (0.79 / √35)
z = -0.2 / (0.79 / √35)
z = -0.2 / (0.79 / 5.92)
z = -0.2 / 0.133
z ≈ -1.504

Next, we need to find the probability of obtaining a z-score of -1.504 or less from the standard normal distribution. We can use a standard normal distribution table or a calculator to find this probability.

Using a standard normal distribution table, we can find that the probability corresponding to a z-score of -1.504 is approximately 0.065.

Therefore, the probability that the mean commute time is at least 28 minutes is approximately 0.065 or 6.5%.