[AccessD] Math problem??

Hi Jim

That's a rounding issue. To avoid that you would need to use ratings
like 4.3, 3.2, 3.9 etc. Or round 89.9 to 90 and 91.5 to 91.

A simplified system generates a simplified output.

/gustav


> Here are two examples of 4.7 average on a 1-5 scale.  Both have different outcomes on a 100 scale.  Averages are on the bottom row.

> Q 1     92      5               90      5
> Q 2     90      5               93.5    5
> Q 3     95      5               93      5
> Q 4     92      5               97.5    5
> Q 5     95      5               85      4
> Q 6     90      5               89.5    4
> Q 7     95      5               85.5    4
> Q 8     90      5               98.5    5
> Q 9     89      4               87.5    4
> Q 10    93.5    5               93      5
> Q 11    80      4               95      5
> Q 12    82      4               91      5
> Q 13    85      4               90      5

> Avg     89.9    4.7             91.5    4.7

> -----Original Message-----
> From: accessd-bounces at databaseadvisors.com
> [mailto:accessd-bounces at databaseadvisors.com]On Behalf Of Jim DeMarco
> Sent: Wednesday, August 18, 2004 11:09 AM
> To: Access Developers discussion and problem solving
> Subject: RE: [AccessD] Math problem??


> There are only 13 categories not 20.  89.9 is the average on the 100 scale.  When I convert each grade to a 1-5 value the average is 4.7.  Make sense?

> Jim D.

> -----Original Message-----
> From: accessd-bounces at databaseadvisors.com
> [mailto:accessd-bounces at databaseadvisors.com]On Behalf Of Scott Marcus
> Sent: Wednesday, August 18, 2004 10:40 AM
> To: Access Developers discussion and problem solving
> Subject: RE: [AccessD] Math problem??


> How does someone with a score of 89.9 get an average of 4.7? In my book, 89.9/20 = 4.495


> -----Original Message-----
> From: accessd-bounces at databaseadvisors.com
> [mailto:accessd-bounces at databaseadvisors.com]On Behalf Of Jim DeMarco
> Sent: Wednesday, August 18, 2004 10:13 AM
> To: Access Developers discussion and problem solving
> Subject: RE: [AccessD] Math problem??


> Interesting thought but I'm not sure that that's the answer either Gustav for a couple of reasons.  

> This would mean that every person with a 4.5 or greater would be in the "O" category when they probably belong in "V".  I've also got a person whose old average was 89.9 with a 1-5 average of 4.7
> (although maybe that 1/10 of a point is moot in this case I've got to lower one of the 13 grades by over 6 points to see a change in grouping level).

> I can always send you my test numbers if you feel like playing with them.

> Thanks,

> Jim D.

> -----Original Message-----
> From: accessd-bounces at databaseadvisors.com
> [mailto:accessd-bounces at databaseadvisors.com]On Behalf Of Gustav Brock
> Sent: Wednesday, August 18, 2004 9:53 AM
> To: Access Developers discussion and problem solving
> Subject: Re: [AccessD] Math problem??


> Hi Jim

> Isn't this a normal rounding issue?
> If you calculate the final grading as:

>   FinalGrading = Int(AverageGrading + 0.5)

> you will get an "O" for an average grading of 4.7.

> /gustav

>> We have an automated employee evaluation form that allows supervisors/managers to grade their staff in 13 categories on the following scale:

>> O               100 - 90
>> V               89 - 80
>> G               79- 70
>> I               69 - 60
>> U               Below 60

>> There's been much discussion among management to change this scale to a 1 to 5 rating but given a sample of 10 evaluations changing each grade above to a 1-5 value does not end up with the same
>> average over the 13 categories (in other words, on certain employees who averaged in the "O" or outstanding group the 1-5 average was 4.7 where I need it to be 5 to get them in the Outstanding
>> rating).

>> Is there a way for us to change our grading system without dropping certain grades from the (previously) correct bucket?