Hi Jerry,

I tried your solution and it worked, but it's new for me that COUNT(B5:B20)
used as a criterium returns a Boolean value required by the IF function as
first argument, while using as a separate function returns an integer. It
seems that if COUNT(B5:B20) returns zero then its logical value is FALSE, it
it returns a positive integer then its logical value is TRUE. Is it so? How
do you find out such tricks, Help doesn't mention this possibility?

Regards,
Srefi


€žJerry W. Lewis€� ezt Ã*rta:

If in the workbook this will be repeated many times or involve a much
larger range, then it might recalculate faster to cut out the math
=IF(COUNT(B5:B20),AVERAGE(B5:B20),0)

Jerry

Stefi wrote:

=IF(ERROR.TYPE(AVERAGE(B5:B20))=2,0,AVERAGE(B5:B20 ))

Regards,
Stefi

€žPete Cumberland€� ezt Ã*rta:


I'm having a problem with a spreadsheet I'm developing which records average
marks of groups of students over several weeks. I want to display a
"Running Average" of those averages as I enter the data, however, because I
do this week by week there are inevitably weeks which have not had any data
entered and her I see the error #DIV/0! for the formula
=AVERAGE(IF(R5:R20<0, R5:R20,"")). I do understand why I'm getting this
(because I'm dividing by zero) but would like to return a zero so that the
average of the averages will return a number rather than an error.
Can anybody help me?


Pete