On Sun, Jan 30, 2011 at 10:39 PM, Adam Richardson wrote:
> On Sun, Jan 30, 2011 at 8:44 PM, Marc Trudel wrote:
>
>> Just out of curiosity, wouldnt it be better to use floor($n) instead of
>> round($n-0.5)?
>>
>
> Hi Marc,
>
> I don't think so, although I could have missed something (I just quick
On Sun, Jan 30, 2011 at 8:44 PM, Marc Trudel wrote:
> Just out of curiosity, wouldnt it be better to use floor($n) instead of
> round($n-0.5)?
>
Hi Marc,
I don't think so, although I could have missed something (I just quick
cranked out the example.) The definition I linked to involves roundin
Just out of curiosity, wouldnt it be better to use floor($n) instead of
round($n-0.5)?
On Sun, Jan 30, 2011 at 5:48 AM, Paul Halliday wrote:
> On Sat, Jan 29, 2011 at 2:28 PM, Adam Richardson
> wrote:
> > For the nearest rank computation, you could use the following:
> >
> > $arr =
> >
> array(1
On Sat, Jan 29, 2011 at 2:28 PM, Adam Richardson wrote:
> For the nearest rank computation, you could use the following:
>
> $arr =
> array(12,89,65,23,90,99,9,15,56,67,3,52,78,12,10,88,77,77,77,77,77,77,77);
> sort($arr);
> $score_representing_95th_percentile = $arr[round((95/100) * count($arr) -
For the nearest rank computation, you could use the following:
$arr =
array(12,89,65,23,90,99,9,15,56,67,3,52,78,12,10,88,77,77,77,77,77,77,77);
sort($arr);
$score_representing_95th_percentile = $arr[round((95/100) * count($arr) -
.5)];
echo $score_representing_95th_percentile; // 90
That said, t
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