1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 6, 8, 10, 12
The mean is 4
The median is 2.5
The mode is 1
Mean = 14 Median = 12 Mode = 12
Range: 34 Mean: 21.71 Median: 15 Mode: 10
mean: 9 ordered sample: 7, 8, 8, 10, 12 median: 8 mode: 8 range: 12 - 7 = 5
Mean: 17 Median: 19 Mode: 20
There are infinitely many sets. One possible set is {10,10,12,13,15}
Mean = 14 Median = 12 Mode = 12
mean = 19 median = 12 mode = 12
Mean: 11.6 Median: 12 Mode: 10, 12, 15, 16, 5 Range: 11
Range: 34 Mean: 21.71 Median: 15 Mode: 10
Mean: 20.2222222 Median: 12 Mode: 11 Range: 79
what i have is mean is 22.2,median which is the middle is 23,range is 26, and mode is 12
mean: 9 ordered sample: 7, 8, 8, 10, 12 median: 8 mode: 8 range: 12 - 7 = 5
Mean: 37.2 Median: 20 Mode: 20, 12, 61, 74, 19
The average, median and mode of a list of numbers will be the same when the middle (or mean of the two middle numbers) is equal to the most common number in the list, and that number is also the mean. This assumes that the list has only a single mode. If you arrange such a list of numbers from least to greatest, the mode value will be grouped at the middle of the list, thus becoming the median as well. The average of the non-median values will be equal to the median/mode. Given a target mean/median/mode and a list length, you can construct an infinite number of lists that qualify. Here are some examples: 10 10 10 10 10 (or any list of only one number) 11 12 12 12 12 13 1 2 3 4 5 5 5 6 7 8 9
Mean: 17 Median: 19 Mode: 20
There are infinitely many sets. One possible set is {10,10,12,13,15}
Mode: 12 and 23Mean = 17.5Median = 17.5Mode: 12 and 23Mean = 17.5Median = 17.5Mode: 12 and 23Mean = 17.5Median = 17.5Mode: 12 and 23Mean = 17.5Median = 17.5