How Do You Round?

[The Hammer]

If given the list of numbers in this poll and asked to round them, how would you do that?

 

[Twilight Hue]

Easy.

 

[JustGeorge]

I guess I'd round up.

Probably because I don't want to be caught at the cash register short changed....

 

[vulcanlogician]

I mean, 5 and .5 are supposed to be rounded up. But there may be instances where you want to round down (ie. you seek a conservative estimate etc.)

 

[Heyo]

If given the list of numbers in this poll and asked to round them, how would you do that?

0, 0, 0, 0
(rounded to full 10s).

 

[exchemist]

I guess I'd round up.

Probably because I don't want to be caught at the cash register short changed....

Yes rounding up is the usual convention - except with the UK tax authority, which is generous enough to round down to the nearest whole number.

 

[The Hammer]

Why Rounding 5s Up Is Wrong

It's best to round to the even number, so as to reduce error.

"Let’s look at an example. Consider the numbers 0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, and 9.5. The average of these numbers is 5. If we were to round all of these numbers as taught, you get 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. The average of these is 55/10=5.5, which is 0.5 higher than the real average. This means we have an error of 0.5 through this rounding method.

A better rounding method for numbers with 5 as their last digit would be to round to the nearest even digit. This gets rid of systematic error as you would be rounding up and down equally often (at least in theory). We can see this clearer through an example. Using the same dataset as above, if we round each number to the nearest even digit, we get 0, 2, 2, 4, 4, 6, 6, 8, 8, 10. The average of this is 50/10=5, meaning we have an error of 0."


@JustGeorge , @vulcanlogician , @exchemist , @Heyo

 

[lewisnotmiller]

Why Rounding 5s Up Is Wrong

It's best to round to the even number, so as to reduce error.

"Let’s look at an example. Consider the numbers 0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, and 9.5. The average of these numbers is 5. If we were to round all of these numbers as taught, you get 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. The average of these is 55/10=5.5, which is 0.5 higher than the real average. This means we have an error of 0.5 through this rounding method.

A better rounding method for numbers with 5 as their last digit would be to round to the nearest even digit. This gets rid of systematic error as you would be rounding up and down equally often (at least in theory). We can see this clearer through an example. Using the same dataset as above, if we round each number to the nearest even digit, we get 0, 2, 2, 4, 4, 6, 6, 8, 8, 10. The average of this is 50/10=5, meaning we have an error of 0."


@JustGeorge , @vulcanlogician , @exchemist , @Heyo

How is that 'best'?

It's a way of reducing variance over a larger dataset, but much of our requirement for rounding comes from discrete instances.

Indeed, when dealing with a larger dataset, in my experience you don't round the individual values at all, but instead may round the end result.

Consider the numbers 0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, and 9.5. The average of these numbers is 5.


0.5 + 1.5 + 2.5 + 3.5 + 4.5 + 5.5 + 6.5 + 7.5 + 8.5 + 9.5 = 50

50/10 = 5.0

If we add another number to the dataset, the advantage of rounding the end result, but not each value in the array would be a little more obvious.

0.5 + 1.5 + 2.5 + 3.5 + 4.5 + 5.5 + 6.5 + 7.5 + 8.5 + 9.5 + 10.5 = average of (60.5/11) = 5.5

Rounding to the nearest even number gives 0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10 = average of (60/11) = 5.45 repeated.

 

[exchemist]

Why Rounding 5s Up Is Wrong

It's best to round to the even number, so as to reduce error.

"Let’s look at an example. Consider the numbers 0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, and 9.5. The average of these numbers is 5. If we were to round all of these numbers as taught, you get 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. The average of these is 55/10=5.5, which is 0.5 higher than the real average. This means we have an error of 0.5 through this rounding method.

A better rounding method for numbers with 5 as their last digit would be to round to the nearest even digit. This gets rid of systematic error as you would be rounding up and down equally often (at least in theory). We can see this clearer through an example. Using the same dataset as above, if we round each number to the nearest even digit, we get 0, 2, 2, 4, 4, 6, 6, 8, 8, 10. The average of this is 50/10=5, meaning we have an error of 0."


@JustGeorge , @vulcanlogician , @exchemist , @Heyo

Interesting. I have never come across this, though it seems to make sense. I wonder if @Polymath257 is familiar with it.

 

[Polymath257]

Yes. You can also decide to round to the odd number. Alternating between the two for large pieces of large data sets is even better.

This is a well known issue when dealing with data, which is generically known as 'binning'.

Another situation which can be even more complicated is latitude and longitude on a sphere (or RA and declination in the sky). How do you round these in a way so that the statistics don't get messed up (think about what happens at the poles)?

 

[Heyo]

A better rounding method for numbers with 5 as their last digit would be to round to the nearest even digit. This gets rid of systematic error as you would be rounding up and down equally often (at least in theory). We can see this clearer through an example. Using the same dataset as above, if we round each number to the nearest even digit, we get 0, 2, 2, 4, 4, 6, 6, 8, 8, 10. The average of this is 50/10=5, meaning we have an error of 0."

Assuming that the predecessors of the to round 5 are evenly distributed. Which, for special cases, we know they aren't.
As @Polymath257 pointed out, an even better method would be to alternate with rounding up and down.
Best is still to not round numbers on which later operations are performed. Only round end results.

 

[exchemist]

Assuming that the predecessors of the to round 5 are evenly distributed. Which, for special cases, we know they aren't.
As @Polymath257 pointed out, an even better method would be to alternate with rounding up and down.
Best is still to not round numbers on which later operations are performed. Only round end results.

That's sound advice, but a luxury only available since my time at university. We did it all with a slide rule. 3 sig figs was the best we could do.

 

[Polymath257]

Assuming that the predecessors of the to round 5 are evenly distributed. Which, for special cases, we know they aren't.
As @Polymath257 pointed out, an even better method would be to alternate with rounding up and down.
Best is still to not round numbers on which later operations are performed. Only round end results.

The problem also arises when doing measurements. So, when you look at a ruler to measure a length, and the actual length appears between two marks, which way do you round? The problem is that you may not know the 'exact' value, so some rounding is unavoidable even before the calculations are done.

Then comes the issue of tracking the possible errors, from measurement through calculation, to the final result.

This is why it is usually better to give a number with error bars than round. So, 3.5+-.3 makes it clear that rounding is going to be an issue no matter what, but allows for calculations to be done intelligently.

 

[Polymath257]

That's sound advice, but a luxury only available since my time at university. We did it all with a slide rule. 3 sig figs was the best we could do.

Unless you looked up the values in a book of tables. Then you might be able to get 4 or 5....until you actually started to calculate.

 

[The Hammer]

As @Polymath257 pointed out, an even better method would be to alternate with rounding up and down.

That's what this method is supposed to do: alternate rounding up and down to the nearest even number.

 

[Revoltingest]

2, 3, 4, 5 is the standard for adding machines.
It's called the "5/4" cut-off.