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## Comment on

Truth About Sets A & B## I'm having trouble

I'm stuck on the evaluation of condition I. The example you used that makes this untrue, (I have the video paused at 1:18) has the numbers for set B as {1,2,6}. While this does make the combined set have a median of 5, those numbers don't meet the criteria for set B of having an average of 3. (1,2,6) has an average of 4.5, so we can't use those numbers as an example.

Any clarification would be appreciated.

## Average of {1, 2, 6} = (1+2+6

Average of {1, 2, 6} = (1+2+6)/3 = 9/3 = 3 (not 4.5)

## Hello. Why couldn't the

## Set A can, indeed, be {1, 9},

Set A can, indeed, be {1, 9}, since the median would be 5.

I'm not sure why you feel that couldn't be the case.

## Set A can have 5-5, 4-6, 3-7,

## Hello, I had same question as

## In order for the values in a

In order for the values in a set to be equally spaced (aka evenly spaced), the difference between any pair of consecutive values must always be the same.

For example, in the set {6, 10, 14, 18, 22} the difference between any pair of consecutive values is always 4. So, the values in the set are equally spaced.

Likewise, in the set {3, 8, 13} the difference between any pair of consecutive values is always 5. So, the values in the set are equally spaced.

Likewise, in the set {2, 5} the difference between any pair of consecutive values is always 3. So, the values in the set are equally spaced. Yes, there are only 2 values in this last set, but the principle remains the same.

As such, we can say that ANY set of two values can be considered equally spaced.

As such, the median of any 2 values will equal the mean of those 2 values.

Does that help?

Cheers,

Brent

## I understood perfectly