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## Comment on

Mean, Median and Mode## I'm at cloud nine now! I'm a

## Thanks! Glad you like the

Thanks! Glad you like the videos,

## Hi, I just want to make sure

## In the case that all of the

In the case that all of the values are different, then we can say that there is a multi-way tie for the most common value.

So, for example, the set {2, 3, 6, 9} has four modes: 2, 3, 6 and 9

## http://greprepclub.com/forum

plz explain this sum

## You bet.

You bet.

I've posted my solution here: http://greprepclub.com/forum/course-of-an-experiment-95-measurements-wer...

Cheers,

Brent

## Thankyou.

## http://greprepclub.com/forum

Sometimes i get confused to put answers in 0.1, 0.01 similar like accuracies. Can you please help me in understanding those concepts well.

## We cover rounding here: https

We cover rounding here: https://www.greenlighttestprep.com/module/gre-arithmetic/video/1062

## Hi

Can you pl provide the soln to this - A set of 7 integers has a range of 2, an average of 3, and a mode of 3.

## Set A can be EITHER {2, 2, 3,

Set A can be EITHER {2, 2, 3, 3, 3, 4, 4} OR {2, 3, 3, 3, 3, 3, 4}

Both of these sets meet the given conditions.

Can you tell me how the question is worded?

## This is the link to the exact

I was able to come to the answer. But I wasn't sure till the end. Would be nice if you could help me think through.

## You bet.

You bet.

Here's my step-by-step solution: http://greprepclub.com/forum/a-set-of-7-integers-has-a-range-of-2-an-ave...

Cheers,

Brent

## 2 3 3 3 3 3 4 is one possible

Since they told mode 3 expect 1st and 7th, all member can be 3, we met mode restriction. Range has to be 2 so let 1st member be 2 and certainly 7th member will be 4.

## Hi Brent,

For the following question, can we say using the logic of weighted average that the qty B is greater ? Qty A has equal weight for x and y. So average has to be in the middle of x and y. However, Qty B has 2 weights of y and 1 weight of x. So average has to be closer to y than x. Since y > x, the average from qty B has to be greater.

http://greprepclub.com/forum/x-y-3608.html

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Also for this question, http://greprepclub.com/forum/graph-above-shows-the-frequency-distribution-of-50-integer-1805.html, 23 values <5 (median), 16 values =5, 11 values >5. More weightage in < median. So average has to be < 5.

Thanks :)

## Hi Vinodhini,

Hi Vinodhini,

Question link: http://greprepclub.com/forum/x-y-3608.html

Yes, the weighted approach you describe is perfect. GREAT WORK!!

------------------------------------------

Question link: http://greprepclub.com/forum/graph-above-shows-the-frequency-distributio...

Your approach says that, since there are more values BELOW the median than there are values ABOVE the median, the average must be LESS THAN the median.

This approach can get you in trouble, because the mean (average) depends on the MAGNITUDE of the values above and below the median.

For example, if we replace one of the eleven 6's with 1,000,000 then the average of the 50 values will be greater than the median.

Cheers,

Brent

## Hi Brent,

Thank you. Got my mistake in 2nd approach..

## Hi Brent ! I wanted to ask if

## Sorry, but I don't have a

Sorry, but I don't have a list of all reinforcement problems in the course. Perhaps this is something I will add in the future.

## Ok np sir! Thanks for all the

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