Try the following... a bit of a lengthy array formula, but should work.
(For the symbols =, <=, =, <, , <)

=MEDIAN(IF(A3:A6-$B$2=0,IF(ISERROR(FIND("=",$A$2)),"",B3:B6),IF(A3: A6-$B$20,IF(ISERROR(FIND("",$A$2)),"",B3:B6),IF(ISE RROR(FIND("<",$A$2)),"",B3:B6))))

You'll have to adjust the range. I tested in A3:A6.

Scott

kcbannon wrote:
I having no luck in substituting cell references for actual operators or
independent variables in a MEDIANIF array. It has worked in simpler
functions like COUNTIF. Am I missing something?

In the table below, I want to find the median of the returns in B4:B7
subject to the values in A4:A7 being less than, greater than, equal to
or not equal to (operator in cell A2) a certain variable (value in cell
B2).

Oper Variable
= 5

Value Return
2.5 15.1%

5.0 12.2%

6.2 9.8%

7.3 21.3%

The way I think the formula should be written is:
=median(if(A3:A7=5,B3:B7,""))

However, I want to be able to manipulate the operator (A2) and the
independent variable (B2) by referring to the cells. This is where is
am having problems; the formula returns #Value!
=median(if(A3:A7,$A$2&$B$2),B3:B7,"")))

Tracing through the second formula, it appears that the comma after A7
marks the end of the Logical Test and "$A$2&$B$2" become the Value if
True.

Any help would be greatly appreciated.




--
kcbannon