A simple example is the set { 1, 2, 3,4, 5,6,41} The median is 4, the mean is 62/7 ~ 8.86 which is greater than 4 , the range is 40. There is no mode because all members are unique. The mean is ~ 8.86, the standard deviation is ~ 13.2, so the last entry (41) is (41 - 8.86)/ 13.2 = 2.43 standard deviations above the mean. With only 7 numbers, having one more than 2 standard deviations above the mean is highly unlikely. Under the assumption of normality, the fraction of values greater than 2.43 standard deviations above the mean is less than 0.01. Of course 0.01 x 7 is much less than 1, so the value 41 should be viewed as an outlier. This argument can be put on a firmer basis by using a hypothesis testing approach.
