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More mean median mode (Posted on 2014-02-21) Difficulty: 3 of 5
For each part begin with the set {8, 10, 11}.

The goal is to add as few integers as possible to the set to create a new set having the given values. If there is more than one possibility then the range should be minimized.

a. mean=8, median=10, mode=11

b. mean=8, median=11, mode=10

c. mean=10, median=8, mode=11

d. mean=10, median=11, mode=8

e. mean=11, median=8, mode=10

f. mean=11, median=10, mode=8

Notes: Mean refers to the arithmetic mean. There should be a single mode (no ties).

No Solution Yet Submitted by Jer    
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Solution possible solution | Comment 1 of 3

a. There need be at least two 11's. So far the total is 40. We need 5*8 = 40, so the remaining integer is zero: {0, 8, 10, 11, 11}

b. To begin: {8, 10, 10, 11, ...} We'll need four numbers higher than 11, but no duplicates: {..., 8, 10, 10, 11, 12, 13, 14, 15}, so far a total of 93. Nine members averaging 8 would be 72; the first integer needs to be -21.

c. Start with {8,11,11}, then have two smaller numbers so the total is 5*10=50, but this requires numbers higher than 8, so no good. So add another 11: {8,11,11,11}. The three new numbers must add to 7*10-41 = 29. Still no luck getting below 8. We have to get some of the value by making the third 11 into a higher number. We make the lower three numbers 5,6 and 7: {5,6,7,8,11,11,70-48}, i.e.,  {5,6,7,8,11,11,22}.

d. {4, 8, 8, 11, 12, 13, 14}

e.  {5, 6, 7, 8, 10, 10, 31}

f. {8, 8, 10, 14, 15}


  Posted by Charlie on 2014-02-21 15:49:04
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