Final answer:
Whether the claim that the mean annual income is greater in county A than in county B is supported depends on the results of statistical hypothesis testing. Although we cannot provide a definitive answer without specific data, the analysis typically involves setting up hypotheses, choosing a test statistic, and calculating a p-value. The choice of mean or median as a better measure depends on the presence of outliers in the data.
Step-by-step explanation:
To assess the claim that the mean annual income is greater in county A than in county B, statistical analysis can be performed using the sample means and standard deviations. Since we do not have specific numerical values, we cannot directly affirm or refute the claim here. However, if one were to conduct a hypothesis test, they would typically set up null and alternative hypotheses, choose an appropriate test statistic (e.g., the t-test for means), and then calculate the probability (p-value) to determine if there is a statistically significant difference in the means of the two counties.
In general, it is indeed possible for the standard deviation to be greater than the average in distributions where there is a wide spread of data. The likelihood that the average salary falls within a certain range depends on the standard deviation, so if it's narrower (smaller standard deviation), it's more likely that the average would fall within a close range to the mean. Regarding the example of Company A and Company B, a hypothesis test analyzing the sample means and standard deviations might support the claim if the calculated p-value is less than the chosen significance level.
For the normal distribution example with a claim that the mean is greater than 12, a sample mean of 12.5 with a sample size of 36 could support this claim if the test statistic falls in the critical region of the chosen significance level.
The better measure of the center (mean or median) in a town with one high earner and many lower earners would likely be the median because it is not as sensitive to outliers as the mean is.