Isn't the zero result coming from all the zeros you are getting from the
exception side of your IF statement (i.e. when you don't meet the B and C
conditions)?

Will =MEDIAN(IF((B$2:B$1000=2)*(C$2:C$1000=2006),D$2:D$ 1000,"")) work for
you (array-entered)?
--
David Biddulph

"jhicsupt" wrote in message
...
I was using the sort of the same principle for MEDIAN. However, if value
is
true, it's still coming up with 0. What am I doing wrong?

Thanks again.

=MEDIAN(IF((B$2:B$1000=2)*(C$2:C$1000=2006),D$2:D$ 1000,0))

"Bob Phillips" wrote:

=QUARTILE(IF((B$2:B$1000=2)*(C$2:C$1000=2006),D$2: D$1000,0),1)

still array entered.

--
---
HTH

Bob

(there's no email, no snail mail, but somewhere should be gmail in my
addy)



"jhicsupt" wrote in message
...
If criteria is not met, it returns a #NUM. How do I get it to return
Nothing?

"Bob Phillips" wrote:

=QUARTILE(IF((B$2:B$1000=2)*(C$2:C$1000=2006),D$2: D$1000),1)

which is an array formula, it should be committed with
Ctrl-Shift-Enter,
not
just Enter.
Excel will automatically enclose the formula in braces (curly
brackets),
do
not try to do this manually.
When editing the formula, it must again be array-entered.

--
---
HTH

Bob

(there's no email, no snail mail, but somewhere should be gmail in my
addy)



"jhicsupt" wrote in message
...
I want to do median with multiple columns criteria.

If B2:B1000=2
AND
If C2:C1000=20006

QUARTILE(D2:D1000)

The below returns an error. What am I missing? Thanks in advance.

=QUARTILE(IF(AND(B$2:B$1000=2,C$2:C$1000=2006),D$2 :D$1000,1),0)