In 1999, the average charge for tax preparation by H&R Block was $84.57. Assuming a normal distribution and a standard deviation of $10, what proportion of H&R Block’s tax preparation fees were more than $84.57?

The answer is 0.5 or 50%. This is because the question is asking for the proportion of fees that were more than the average fee of $84.57. In a normal distribution, the mean (or average) is at the center, and 50% of the data falls above the mean and 50% falls below. Since the standard deviation is given as $10, we can use a z-score formula to convert the given value of $84.57 into a standardized value (z-score) that we can use to calculate the proportion of fees that were more than $84.57. z = (X - ?) / ? z = ($84.57 - $84.57) / $10 z = 0 A z-score of 0 means that $84.57 is exactly at the mean of the distribution. To calculate the proportion of fees that were more than $84.57, we need to find the area under the normal curve to the right of the mean. This can be done using a standard normal distribution table, or a calculator with a normal distribution function. The area to the right of the mean (z-score of 0) is 0.5 or 50%. Therefore, we can conclude that 50% of H&R Block's tax preparation fees were more than $84.57.