Question 1086906
<font color="black" face="times" size="3">Let's consider five cases:
Case A: There is only one copy of 11
Case B: There are two copies of 11
Case C: There are three copies of 11
Case D: There are four copies of 11
Case E: There are five copies of 11
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Case A isn't possible because we're told that the mode is 11. There needs to be at least two copies of 11 to have the mode be 11. Of course, whatever the count is for 11, the other data values can't repeat as frequently.


The other cases are possible. Below are examples for each case
Case B example: {2, 3, 11, 11, 15}
Case C example: {2, 3, 11, 11, 11}
Case D example: {2, 11, 11, 11, 11}
Case E example: {11, 11, 11, 11, 11}
Each example above has '11' in the exact center


The mean will be smallest if we go with case C. This is to make the first two values as small as possible. We keep the '11's just where they are. So if we have {1, 1, 11, 11, 11} then the mean is (1+1+11+11+11)/5 = 7 which is the smallest possible mean. Keep in mind that 1 is the smallest positive number. We can have 1 repeat because it's not as frequent as the three copies of 11. 


Answer: <font color=red>7</font></font>