Re: computing z-scores

On Thursday, May 24, 2018, Ron <ronljohnsonjr(at)gmail(dot)com> wrote:

> On 05/24/2018 10:15 AM, Martin Mueller wrote:
>
>> You construct a z-score for a set of values by subtracting the average
>> from the value and dividing the result by the standard deviation. I know
>> how to do this in a two-step procedure. First, I compute the average and
>> standard deviation. In a second run I use the formula and apply it to each
>> value.
>>
>> Is there a way of doing this in a single-step procedure or can you chain
>> the two parts together in one query? This goes beyond my SQL competence.
>>
>
> What about this?
>
> SELECT value, (value - AVG(value))/stddev(value) as zvalue
> FROM sometable
> WHERE some conditions
> GROUP by value
>

That is syntactically correct but doesn't provide the correct answer.

David J.