\(Explanation\)

In a list of 10 consecutive integers, the mean is the average of the 5th and 6th numbers.

Therefore, the 6th through 10th integers (five total integers) is greater than the mean. Since probability is determined by the number of desired items divided by the total number of choices, the probability that the number chosen is greater than the average of all 10 integers is \(\frac{5}{10} = \frac{1}{2}\).


Another approach to this problem is to create a set of 10 consecutive integers; the easiest such list contains the numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. The mean is one-half the sum of the first element plus the last element, or \(\frac{10+1}{2}= 5.5\).

Therefore, there are 5 numbers greater than the mean in the list: 6, 7, 8, 9 and 10. Again, the probability of choosing a number greater than the average of all 10 integers is \(\frac{5}{10} = \frac{1}{2}\).