![]() |
using percentages in an if statement
Is there a way I can incorporate the percentage of values that meet a
critiera into an If statement. For instance, say I've got 1 2 3 4 5 6 7 8 9 10 Is there any way that I can return TRUE if 90% of the values are larger than 1, but FALSE if less than 90% are greater than 1? |
Hi!
Really don't need an IF: =COUNTIF(A1:A10,"1")/COUNT(A1:A10)=0.9 But, IF you really want one: =IF(COUNTIF(A1:A10,"1")/COUNT(A1:A10)=0.9,TRUE) You might also want to make sure there are enough values in the range so that you don't get an error return: =IF(COUNT(A1:A10)<1,"",IF(COUNTIF(A1:A10,"1")/COUNT(A1:A10)=0.9,TRUE)) Biff "matt" wrote in message ... Is there a way I can incorporate the percentage of values that meet a critiera into an If statement. For instance, say I've got 1 2 3 4 5 6 7 8 9 10 Is there any way that I can return TRUE if 90% of the values are larger than 1, but FALSE if less than 90% are greater than 1? |
Is this what you mean:
=COUNTIF(A1:A10,"1")=9 ? -- HTH, RD --------------------------------------------------------------------------- Please keep all correspondence within the NewsGroup, so all may benefit ! --------------------------------------------------------------------------- "matt" wrote in message ... Is there a way I can incorporate the percentage of values that meet a critiera into an If statement. For instance, say I've got 1 2 3 4 5 6 7 8 9 10 Is there any way that I can return TRUE if 90% of the values are larger than 1, but FALSE if less than 90% are greater than 1? |
All times are GMT +1. The time now is 07:56 AM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
ExcelBanter.com