awk regex calculation for certain range only 1

[1in10]

Once again I need a hint for awk. My aim is to do two calculations on a specific part of a field.
But for the beginning I'll start with one calculation.
If this regex prints all the lines to stdout

awk '$3 ~ /^June$/' datafile.txt

my task is to fetch the maximum and average for that specific part.
My first attempt gives me all the lines for that month, but I need only the result for that month, beside that it prints to stdout a last line with the name of the matched regex.

awk '$3 ~ /^June$/; {if ($2>max) max=$2}  END { print max};' datafile.txt

How to make that calculation for only the range of the regex $3 ~ /^June$/; and print just the result without all the other lines?

Can anybody help? Thanks in advance.

 

[PHV]

awk '$3~/^June$/{if($2>max)max=$2}END{print max}' datafile.txt

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[PHV]

And for the average:

awk '$3~/^June$/{if($2>max)max=$2;++n;s+=$2}END{print max,(s/n)}' datafile.txt

Hope This Helps, PH.
FAQ219-2884
FAQ181-2886

 

[1in10]

YES, it helps, and this is just one more reason for me using awk and joining here. Great relief indeed, thanks a lot. A tiny semicolon ;-)