SOLUTION: 11 is both the median and the mode of a set of five positive integers. What is the least possible value of the average (arithmetic mean) of the set?

Question 1086906: 11 is both the median and the mode of a set of five positive integers.
What is the least possible value of the average (arithmetic mean) of
the set?
Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
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: 7
Answer by Edwin McCravy(20060) (Show Source): You can put this solution on YOUR website!
...a set of five positive integers.

Suppose the five positive integers are a,b,c,d,e where a ≤ b ≤ c ≤ d ≤ e, ascending order.

11 is both the median and the mode...

Since 11 is the median and the number of positive integers is 5, an odd number, the middle integer, c = 11. So, a ≤ b ≤ 11 ≤ d ≤ e

What is the least possible value of the average (arithmetic mean) of
the set?

We want the arithmetic mean (average) to be as small as possible, so we want to use the smallest positive integers as possible. The smallest we can take d and e to be is 11 each. Then 11 will also be the mode, which is what we want. Then the smallest that a and b can be is 1 each, So we have 1 ≤ 1 ≤ 11 ≤ 11 ≤ 11 The least possible average is Edwin

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