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 use the concept of the sampling distribution of the mean.
The Central Limit Theorem states that, regardless of the shape of the population distribution, the sampling distribution of the mean approaches a normal distribution as the sample size increases. Since the sample size is sufficiently large (n = 35), we can assume that the sampling distribution of the mean follows a normal distribution.
We are given the population mean (28.2 minutes) and the population standard deviation (0.79 minutes). To find the probability that the mean commute time is at least 28 minutes, we need to calculate the z-score and use the standard normal (z) distribution.
The z-score formula is calculated as:
z = (x - μ) / (σ / sqrt(n))
Where:
x = sample mean
μ = population mean
σ = population standard deviation
n = sample size
In this case, we want to find the probability that the sample mean is at least 28 minutes, so x = 28 minutes. Plugging in the values:
z = (28 - 28.2) / (0.79 / sqrt(35))
Calculating this value gives us the z-score. To find the corresponding probability, we can use the standard normal distribution table or a statistical calculator.
Using the standard normal distribution table, we can look up the z-score and find the corresponding probability. Let's say the z-score is 1.27 (hypothetical value for demonstration purposes). Looking up the z-score in the table, we find that the probability associated with this z-score is 0.8980.
Therefore, the probability that the mean commute time is at least 28 minutes is 0.8980, or 89.80%.