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# Re: [libreoffice-users] VAR, STDEV, CHISQ.INV

```Guess what 1f*5/5==1f (Java; 1 AS float multiplied by 5 divided by 5 is equals to 1 as a float)
Everyone not studying computer science won't understand this at first glance (and Google has the
best explanations) but with Math.abs((1f/5*5)-1)<0.0001 you do not have a workaround, but the way
it has to be done.... (Die not read the thread, just want to explain that you cannot = on floats)

```
```Sounds to me like normal floating number behavior. I know someone is
trying
to document this and explain why it happens (it's a technical reason
that
goes beyond my skills) but this seems entirely normal behavior given
the
constraints of programming languages generally and hardware in
particular.

Best,
Joel

P.S. You can see similar issues in other spreadsheet software. This
seems
to give some indication as to the reasons why:
https://support.microsoft.com/en-us/kb/78113 (yes I know it's MSO
link...just explaining the reason why floating number isn't entirely
accurate all the time)

On Wed, Jul 8, 2015 at 11:33 AM, Rui Pedro Caldeira
<rpcaldeira@outlook.com>
wrote:

```
```Hey and thanks for your fast answer. This is the data set:

0.000001  0.000001  0.000001  0.000001  0.000001  0.000001
```
```0.000001
```
```0.000001  0.000001  0.000001  0.000001  0.000001  0.000001  0.000001
0.000002  0.000003  0.999997  0.999997  0.999997  0.999997  0.999998
1.000001  1.999997

Using the formula in the spreadsheet it gives the value: 0.328061419
```
```(VAR)
```
```and 0.5727664611 (STDEV)

And yes, i do mean CHISQ.INV :)

Thanks, Rui.

Cumprimentos,
Rui Pedro Caldeira

On Wed, Jul 8, 2015 at 7:24 PM, Brian Barker
```
```<b.m.barker@btinternet.com>
```
```wrote:

```
```At 18:44 08/07/2015 +0100, Rui Pedro Caldeira wrote:

```
```I'm having some problems with the VAR and STDEV function, when
```
```executing
```
```these lines of code. The average is well calculated but the
```
```remaining
```
```two
```
```return very low, unrealistic and equal values
```
```(6.93018471335974E-310).
```
```
```
```
That value is 2^-1027 - which is probably the smallest
```
```floating-point
```
```value your system can represent without underflow.

Have you compared the results produced from your data by equivalent
```
```normal
```
```formulae in a spreadsheet cell?

Do you mean CHISQ.INV and not CHISQINV?

Brian Barker

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