What is a mean of 11 a median of 12 a mode of 8 and a range of 10

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Answer 1

If you have a certain number n numerical data values arranged in order from least to greatest, then the median is the central value if n is odd, or it's the average of the two central values if n is even. That tells us where the 12 is in the dataset. 

If you add all the data values in the set then divide by n the answer is the mean 11. Another way of looking at this is the total sum of all data values is 11n. 

The mode is 8, which means that the value 8 appears more than once in the dataset, and also means that if other data values are also repeated, then 8 is repeated the most.

A range of 10 means that the difference between the smallest and largest data values is 10.

Can we find examples of datasets meeting the requirements? To answer this, let's consider the range. If the lowest data value is X₁ then the highest is X_n=X₁+10. X₁<8 (otherwise X₁ could not be the lowest data value) and X_n>12, the median. So X₁+10>12, and X₁>2, therefore 2<X₁<8.

Now consider n: first consider n is odd, so that the median 12 is the central value. n≥7 because the minimum arrangement would be X₁,8,8,12,...,...,X₇.

Take n=7. If the mean is 11, then the sum of the data=77. This gives us a clue about X₅ and X₆, which can be represented by 12+x₅ and 12+x₆ because both data values must be greater than the median. However, these data values must also be less than X₇. So we have 12+x_i<X₇ where i is 6 or 7. The total sum of the data values is 2X₁+38+24+x₅+x₆=77. Therefore:

2X₁+x₅+x₆=15. But 2<X₁<8, so x₅+x₆=15-2X₁. The RHS has a range of values: (-1,11). Since x₅ and x₆ must be positive, this range is limited to (0,11) corresponding to (7.5,2) for X₁ and (17.5,12) for X_n. For example, if X₁=7, x₅+x₆=1, so, since x₆>x₅, x₆>0.5 and x₅<0.5. Suppose x₅=0.4 and x₆=0.6 then we would have 7,8,8,12,12.4,12.6,17, which has a mean of 11, a median of 12, a mode of 8, and a range of 10.

Another example: X₁=6, x₅+x₆=3, let x₅=1.4 and x₆=1.6. Dataset is 6,8,8,12,13.4,13.6,16. Note that the gap between X₆ and X₇ is decreasing.

When X₁=5, x₅+x₆=5. Let x₅=2.4 and x₆=2.6. Dataset is 5,8,8,12,14.4,14.6,15. Now the gap between X₆ and X₇ narrows further. 

The minimum value of X₁ is determined by the inequality: 12+½(15-2X₁)<10+X₁, 24+15-2X₁<20+2X₁, 19<4X₁, X₁>19/4 or X₁>4.75. The valid range for X₁ is therefore (4.75,7.5) when n=7.