Question

Invent a data set with 7 values that has a mean of 10 and a median of 12. Draw a dot plot to represent your data set. Help!!!!

Answer 1

One possible answer: {1, 2, 5, 12, 14, 16, 20}  

See figure 1 below

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Another possible answer:  {1, 1, 5, 12, 15, 16, 20}

See figure 2 below

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Another possible answer:  {1, 1, 1, 12, 15, 20, 20}

See figure 3 below

Infinitely many answers are possible. You only need to pick one as what you'll show to your teacher.

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Explanation:

Let A,B,C,D,E,F,G be the 7 unknown numbers. We'll have the numbers go from smallest to largest. Though as you'll see later, through some shuffling around, we can have repeats. For now, we'll have A through G be distinct values.

The median is 12, and we have an odd number of values here. This tells us that D = 12 must be the case since D is right in the middle.

The set is now A,B,C, 12, E, F, G

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Recall the process to find the arithmetic mean is to

• step 1) add up the numbers
• step 2) divide the sum by the number of values

We're given a mean of 10 which tells us,

mean = (sum of the values)/(number of values)

10 = (A+B+C+12+E+F+G)/(7)

(A+B+C+12+E+F+G)/(7) = 10

A+B+C+12+E+F+G = 10*7

A+B+C+12+E+F+G = 70

A+B+C+E+F+G = 70-12

A+B+C+E+F+G = 58

Unfortunately at this point we don't have enough info to find the solutions to A,B,C,E,F,G. There are infinitely many ways to solve this.

But we do know that A,B,C are less than 12 while E,F,G are larger than 12.

Let's pick at random 5 values to assign to A,B,C,E, and F. The remaining value G will be solved for.

The random numbers I'll pick are:

• A = 1
• B = 2
• C = 5
• E = 14
• F = 16

So,

A+B+C+E+F+G = 58

1+2+5+14+16+G = 58

38+G = 58

G = 58-38

G = 20

So the set

{A,B,C,D,E,F,G} = {1, 2, 5, 12, 14, 16, 20}

has D = 12 in the middle as the median

Also, the values add up to 1+2+5+12+14+16+20 = 70 and it leads to the arithmetic mean of 70/7 = 10

This confirms the set {1, 2, 5, 12, 14, 16, 20} fits the description. It is one possible answer (out of infinitely many).

How do we make the dot plot? Well we draw out a number line, and then place a dot above each number mentioned. We'll have a dot above 1, then above 2, then 5, then 12, and so on. Refer to figure 1 below.

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If you want, you can stop here since the dot plot and data set are already done. Read on if you want to see how to generate other possible solutions.

Admittedly, the dot plot is rather boring. It's just a flat horizontal set of dots. But we can add a bit of variety.

If we had repeated numbers, then the dots would stack up.

We'll need to keep the 12 anchored where it is. That will not change.

The other values around it can change, but do so carefully. Let's change the "2" into a "1".

So we go from {1, 2, 5, 12, 14, 16, 20} to {1, 1, 5, 12, 14, 16, 20}. There's a problem though: the arithmetic mean is not 10 anymore.

The drop from 2 to 1 is -1. So we need to have a +1 to exactly one of the other values (not the 12) to balance things out. Let's say we add 1 to 14 to get 15

So the set {1, 1, 5, 12, 14, 16, 20} changes to {1, 1, 5, 12, 15, 16, 20} which is another possible answer.

Refer to figure 2 for the dot plot of this. We have exactly one stack of dots (2 units high) over the "1". Every other dot is on its own.

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Now let's say we wanted to form two stacks.

Starting from {1, 1, 5, 12, 15, 16, 20} we can subtract 4 from 5 to get 5-4 = 1

So we can change the "5" into "1" to get {1, 1, 1, 12, 15, 16, 20}

Then add 4 to another value, let's say the 16 to get 16+4 = 20. This is to balance out the -4 earlier.

Therefore we have {1, 1, 1, 12, 15, 20, 20} as another possible answer

Now we have two stacks. The left most stack has 3 items, and the right most has 2 items. The 12 and 15 are separate dots.

Refer to figure 3 for the dot plot.

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This process of shuffling around pebbles so to speak can be done an infinite number of ways. Feel free to get creative to try out different sets and different dot plots.

Make sure that whatever you end up going with, the 12 stays in the middle and the values add up to 70.