Hi Philippe,

...... but =MEDIAN(A1:E6) wouldn't make sense unless you had an array ...... <<

Can you elaborate this statement, please? Can you give me an example that using CSE (Ctrl+Shift+Enter) will give the correct result whereas not using it will give the wrong result?

Using the data set suggested by you

A B C D E

1 1 1 2 2
2 2 2 3 3
3 4 4 4 4
4 4 5 5

and many other data sets that I created, I always came up with the same correct result without using CSE. I even used data sets that had values not in a sorted (ascending) sequence; no problem whatsoever.

I really want to know what you meant by way of an example. Thanks.

Epinn

"Philippe L. Balmanno" wrote in message ...
"Bruce" wrote in message
...
Greetings all: All I want to do is calculate the median of grouped data.
I
have five cols. of data corresponding to a 1:5 Likert scale. For example:
3
ones, 5 twos, 3 threes, 6 fours, 2 fives. I have 19 values, so the 10th
ranked value would be the median. The 10th value resides in the "threes"
column, so that's my median value. But how to do this via a function?

Do I expand these cells like this?:
=MEDIAN(1,1,1,2,2,2,2,2,3,3,3,4,4,4,4,4,4,5,5), which I don't know how to
do,
by the way, or is there a simpler way?

Many thanks in advance...
--
Bruce

Well if you have your values in a row columns A1:S1 then
if this was placed in T1: =MEDIAN(A1:S1) It would bring up your Median. If
your numbers were in a column like in A1:A19 then =MEDIAN(A1:A19) but
=MEDIAN(A1:E6) wouldn't make sense unless you had an array.