Thequestion is in a 6th grade text book and says use the central tendency

Thequestion is in a 6th grade text book and says use the central tendency

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Kevin's definitions and solution are good. Another way to look at this: what are the numbers that would make the shortest list of numbers, and satisfy this situation (mean & median=10, mode=8)?

If we want to ensure the mode = 8, then we must use 8 the most often: let's use it twice, so we start with 8, 8. The median (middle number) needs to be 10, so now we have 8, 8, 10. To ensure that 10 ends up in the middle, we need two more numbers, for a total of 5 (8, 8, 10, x, y). The mean needs to be 10, so what two numbers averaged with 8, 8 and 10 would give us an average of 10? we know the mean, or average is adding all 5 together and dividing them by 5, so we have** (8+8+10+x+y) / 5 = 10**. If we multiply both sides by 5, then we have **8 + 8 + 10 + x + y = 50**.

There are probably many possible whole number (integer) combinations that would satisfy these conditions, but let's see if we can use numbers close to the ones we already have. Since we already have two 8's, We need numbers higher than 10 to give us an overall average (mean) of 10. Don't want to use 10 again, because we want to keep 8 as our mode.

We could try 11: 8+8+10+11+x = 50; 37 + x = 50, or x = 13

Now we have** 8, 8, 10, 11, 13**. Let's check these against what we need:

Most-used number = **8 = mode**, check

Middle number = **10 = median**, check

Average = 8+8+10+11+13 = 50, divided by 5 = **10 = mean**, check

We would need to know how many numbers there were to begin with, but this may help: The mode is the number that occurs most often, so 8 is the mode. The median is the number that is in the middle if the numbers are arranged in increasing order, for example, in the list 7, 8, 10, 12 and 13, the median is 10 The mean in the average. So add the numbers together and divide by the how may numbers there are; for example, in my list above, the average would be (7+8+10+12+13)/5 = 50/5 = 10