count if function help

  • Thread starter Thread starter Matthew Dyer
  • Start date Start date
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Matthew Dyer

I have cells with values ranging from 40-59. I love the countif
function, but I want to do something a little different. I want to see
how many values fall within a certain range (ex. between 50 and 55).
How may I achieve this? I really hate to use the frequency function
becuase of how complicated and rigid it is.
 
Hi Matthew

=COUNTIF(A:A,">49")-COUNTIF(A:A,">55")

--
Regards
Roger Govier

Matthew Dyer said:
I have cells with values ranging from 40-59. I love the countif
function, but I want to do something a little different. I want to see
how many values fall within a certain range (ex. between 50 and 55).
How may I achieve this? I really hate to use the frequency function
becuase of how complicated and rigid it is.

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__________ Information from ESET Smart Security, version of virus signature database 4772 (20100114) __________

The message was checked by ESET Smart Security.

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Count the number bigger than (or equal to???) 50 and subtract the ones that are
too big (bigger than (or equal to????)) 55.



=countif(a1:a10,">"&50) - countif(a1:a10,">"&55)

You may want ">=" in either/both of these expressions
 
XL2000-2010
=COUNTIF($A2:B$25,"<=55")-COUNTIF(A2:$A$25,"<50")

XL2007-2010(Only)
=COUNTIFS($A$2:$A$25,"<=55",$A$2:$A$25,">=50")
 
I knew it was going to be something simple like that... I kept trying
to put one countif function inside of another countif function, and as
you can imagine that didnt work out at all. Thanks for all your help
guys!
 
Here's another question... I would like to do a sumif using two
different criteria.

ex: sum values of column E if column D is >=58 AND column H is >0

Help..?
 
Well, you can't use SUMIF for that. Use this instead:

=SUMPRODUCT((D2:D100>=58)*(H2:H100>0),E2:E100)

Adjust the ranges to suit.

Hope this helps.

Pete
 
Well, you can't use SUMIF for that. Use this instead:

=SUMPRODUCT((D2:D100>=58)*(H2:H100>0),E2:E100)

Adjust the ranges to suit.

Hope this helps.

Pete





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That works perfectly... but do you mind explaining how this particular
function... um... functions?
 
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