Access 2003: Math question....Calculating a percentage

Hi EE,

I got help last year from EE in how to calcuate percentages.

A student can have one of the following grades in a course:
------------------------------------------------------------------
H - honors
HP - high pass
P - pass

Using a cross tab query, calculated (number of student) totals for each grade for a course
for ex:

 course                 H       HP       P
=============================
MEDI 1107             26     41      102      <-- break down of how many students scored a particular grade

What the snippet below does is convert the total of students below to percentages.
            26+41+102 = 169

            26/169  * 100 = 15.38
            41/169  * 100 =  24.26
           102/169  * 100 =  60.35

      Notice how the decimals portion is less than .50,
              so the results of the calc are:
                         15%
                          24%
                          60%
              that totals 99%

         Is there any way to get the percentages to total 100%  or is this to be expected in math?

tx for your thoughts and ideas, sandra



lngHCount = CLng(Nz(Me.txtHCount, 0))
lngHPCount = CLng(Nz(Me.txtHPcount, 0))
lngPCount = CLng(Nz(Me.txtPcount, 0))


lngTotal = lngHCount + lngHPCount + lngPCount

If lngTotal > 0 Then
    lngHpct = (lngHCount / lngTotal) * 100
Else
    lngHpct = 0
End If

If lngTotal > 0 Then
   lngHPpct = (lngHPCount / lngTotal) * 100
Else
   lngHPpct = 0
End If

If lngTotal > 0 Then
   lngPpct = (lngPCount / lngTotal) * 100
Else
   lngPpct = 0
End If


Me.txtHpct = lngHpct
Me.txtHPpct = lngHPpct
Me.txtPpct = lngPpct
mytfeinAsked:
Who is Participating?
 
als315Connect With a Mentor Commented:
I think the only opportunity to get 100% is add this 1 to the value with maximal decimal part. In your example it will be 15.38. Change it to 16 and you will have minimal deviation from reality.
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RunriggerCommented:
when rounding to integer percentages, that is what you are going to get in this case, its a rareity for percentages to work out like that though!
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PHaddockConnect With a Mentor Commented:
You can't do it.  You also have to be careful what happens with the rounding  ie does .05 round up or down?

What you could do is make the last percentage the difference between 100 and the sum of the other two percentages.
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CluskittConnect With a Mentor Commented:
It's always a risk with rounding. And it's not that rare, Runrigger. Just imagine that you have 1/3 exactly distributed. Each would be 0.3. Rounding to integer would be 0, instead of 1. This applies in many scales. There is no way to avoid this, except programatically checking the sum, and if there's one missing, then check the highest decimal and add to that. In your case, it would add to H total.
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mytfeinAuthor Commented:
Hi RunRigger,

so the calcs are correct, even though it's odd
   (other class total to 100%, btw)

i googled this:
     http://msdn.microsoft.com/en-us/library/ck4c5842(VS.85).aspx

so the clng function is doing what the article above says, since decimal is less than .50,
does not round up the integer portion, just leaves as is....

pls confirm... tx, s
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RunriggerConnect With a Mentor Commented:
You could use a declaration of "double" instead of long and round the initial calculation to 2 decimal places.

assign the integer value to the forms text box (besure to round the insteger value from the double, that way, you may get a round up for the decimals greater than 0.5) and retain the decimal in an additional variable, do that for all three values.

At the end of the above bit of code above, add all three "integer" values and if not equal to 100, then simply add one/subtract to the form's text box where its respective decimal portion was the greatest (or least, if subtracting)!
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RunriggerConnect With a Mentor Commented:
Cluskit, I concede, I am not a statistician.

We are essentially advising Sandra the same thing, programmatically checking the total and making a corrective add/subtract, she will however need to change variable decalaration to double in order to check the decimal value.
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mytfeinAuthor Commented:
Hello als315
         Phaddock
         Cluskitt,

Thx for writing....

It's very clever of all of you with the suggestion to pgmatically adjust the percentage when the
total is less than 100.

What i am understanding is that i just adjust the number with the highest decimal portion.
Is it possible to explain why the number with the highest decimal portion, and not the number with
the highest INTEGER portion...

tx all very much, s
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PHaddockConnect With a Mentor Commented:
Personally I usually arbitrarily make it the third one that gets adjusted (in your example) because it's easy to write into a query because you don't need to devise a way to determine which of the three has the highest decimal before you adjust it.

The reason it is slightly more correct to adjust the highest decimal is because you are always making a smaller change to the numbers as a percentage of each number itself that gets changed.
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mytfeinAuthor Commented:
Hi RunRigger,

Thx for alerting me to work with DOUBLE variables....
Question, pls:
            ====>  HOW do you isolate  the decimal part of the double result
            to get  .38   of 15.38

In the code window is how would modify the code, but got stuck on the above question
tx, s


So would create another 3  variables:

dim dblHCount  as double
dim dblHPCount as double
dim dblPCount  as double

then i would execute the exising logic like this:

lngHCount = CLng(Nz(Me.txtHCount, 0))
lngHPCount = CLng(Nz(Me.txtHPcount, 0))
lngPCount = CLng(Nz(Me.txtPcount, 0))


lngTotal = lngHCount + lngHPCount + lngPCount

If lngTotal > 0 Then
    lngHpct = (lngHCount / lngTotal) * 100
Else
    lngHpct = 0
End If

If lngTotal > 0 Then
   lngHPpct = (lngHPCount / lngTotal) * 100
Else
   lngHPpct = 0
End If

If lngTotal > 0 Then
   lngPpct = (lngPCount / lngTotal) * 100
Else
   lngPpct = 0
End If

==== add this logic

lngTotalPercent = lngHpct + lngHPpct + lngPpct
If lngTotalPercent < 100 then
   dblHCount = (Nz(Me.txtHCount, 0))
   dblHPCount = (Nz(Me.txtHPcount, 0))
   dblPCount = (Nz(Me.txtPcount, 0))

====>  HOW do you isolate  the decimal part of the double result
            to get  .38   of 15.38
end if

Open in new window

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PHaddockConnect With a Mentor Commented:
This one originally came from http://www.freevbcode.com/ShowCode.Asp?ID=427

Public Function DecimalPortion(Number As Double) As Double
  Dim lPos As Long

  lPos = InStr(1, Number, ".")
  If lPos > 0 Then
    DecimalPortion = Val(Mid(Number, lPos))
  Else
    DecimalPortion = 0
  End If

End Function
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mytfeinAuthor Commented:
Thx sooo much LPaddock...

ok, will go work on it...

tx everyone, s

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als315Commented:
"Is it possible to explain why the number with the highest decimal portion, and not the number with
the highest INTEGER portion..."

You have good idea - take maximum from decimal part divided to value.
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CluskittCommented:
>> Is it possible to explain why the number with the highest decimal portion, and not the number with
the highest INTEGER portion

Because, statistically, that is the one closest to the next integer. Accounting for ratio, the highest integer would be the first, but with the current sample universe, the highest decimal is the closest to 1.
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