Final answer:
To determine the probability that the average years of education for a sample of 42 self-employed individuals is less than 9.5 years, we calculate the standard error, find the z-score for a sample mean of 9.5 years, and then consult the standard normal distribution to find the corresponding probability.
Step-by-step explanation:
To find the probability that the average years of education for a sample of 42 self-employed individuals is less than 9.5 years, we use the concept of sampling distributions. Because individual years of education are normally distributed with a mean (µ) of 10 years and a standard deviation (σ) of 3.3 years, the sampling distribution of the sample mean for a sample size (n) of 42 will also be normally distributed with a mean equal to the population mean (µ) and a standard deviation equal to the population standard deviation (σ) divided by the square root of the sample size (n), which is known as the standard error (SE).
The first step is to calculate the SE:
SE = σ / √n = 3.3 / √42
Next, we calculate the z-score that corresponds to a sample mean of 9.5 years:
z = (Xbar - µ) / SE
where Xbar is the sample mean. After calculating the z-score, we can use the standard normal distribution table or a calculator with statistical functions to find the probability corresponding to this z-score, which represents the probability that the sample mean is less than 9.5 years.