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Averages Part II

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Posted · Report post

post-53485-0-04700400-1375111190_thumb.p

This one may be too easy for brainden but why not, sometimes it is the obvious that proves most challenging.

Find an average height of the bars (you cannot throw out any elements of the data). Use a strategy that best calculates the average which is most representative of the population. Show that your solution is the best.

The numbers underneath the bars represent the heights of each bar.

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Posted · Report post

you simply are talking "arithmetic mean"?

So, I'll take the first obvious answer!

35+48+35+40+50+35+35+40+150+35+40+35+45+45=668
668/14 = 47.714285...

I assume you are expecting us to consider the fact that the 150 is obviously an outlier of some sort and we should take that into account...but without throwing out any of the elements...

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Posted · Report post

Isn't the strategy for calculate the average always the same (add up all the values and then divide)?

I do know that the median is more representative of the population than the mean. Is that what you're looking for?

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Posted · Report post

There are many types of strategies for determining 'average'. Which is the best with this dataset?

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Posted · Report post

if you have outliers in your data, and you want to keep the data whole (not removing any of it to influence your results), then I typically go with either the MEDIAN or MODE for my "average"...

So, in the case of this data, we obviously have an outlier...

MEAN with whole data: 47.714285

MEAN excluding outlier: 39.846153

MEDIAN: 40

MODE: 35

As a result, I would simply go with the median as our "average" for this data set...since it keeps the data whole (I kept all data points), but still gives the most accurate representation of the mean without outliers influencing it too much.

The MODE is also pretty good when the majority of your data points are a single number. In this case, 6/14 are "35"...so, IMHO, that's right on the border of wanting to use that number as your "average".

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Posted · Report post

Is there no more precise calculation?

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Posted · Report post

Is there no more precise calculation?

Do you mean something like... Avg X with a standard deviation of y?

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Posted · Report post

...was to pick a number S (ideal for differences), calculate and average all the differences D, then calculate the average to be S+D.

Because 35 is the mode, it makes a good choice for S (6 differences will be 0). The respective differences are:

0,13,0,5,15,0,0,5,115,0,5,0,10,10

These sum to 158, so the average difference D is 178/14 = 12+10/14 = 12+5/7

So the average is 35+12+5/7 = 47+5/7 = 47.714285(repeating)

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Posted · Report post

...was to pick a number S (ideal for differences), calculate and average all the differences D, then calculate the average to be S+D.

Because 35 is the mode, it makes a good choice for S (6 differences will be 0). The respective differences are:

0,13,0,5,15,0,0,5,115,0,5,0,10,10

These sum to 158, so the average difference D is 178/14 = 12+10/14 = 12+5/7

So the average is 35+12+5/7 = 47+5/7 = 47.714285(repeating)

Is this easier than just computing the average directly?

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Posted · Report post

...was to pick a number S (ideal for differences), calculate and average all the differences D, then calculate the average to be S+D.

Because 35 is the mode, it makes a good choice for S (6 differences will be 0). The respective differences are:

0,13,0,5,15,0,0,5,115,0,5,0,10,10

These sum to 158, so the average difference D is 178/14 = 12+10/14 = 12+5/7

So the average is 35+12+5/7 = 47+5/7 = 47.714285(repeating)

Is this easier than just computing the average directly?

this method is very easy with small data sets or data sets with a significant mode subset

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