How can I find greater than value of column 4 and print the whole line ? My output is a status report of a storage box. eg:

Source Destination State Lag Status host1:SystemState backup1:/vol/xxxxx/xxxxxx Snapvaulted 11:24:21 Idle host2:/vol/xxxxx/xxxxxx backup2:/vol/xxxxx/xxxxxx Snapvaulted 1898:58:16 Idle host3:/vol/xxxxx/xxxxxx backup3:/vol/xxxxx/xxxxxx Snapvaulted 18:58:02 Idle host4:/vol/xxxxx/xxxxxx backup4:/vol/xxxxx/xxxxxx Snapvaulted 19:46:24 Idle host5:/vol/xxxxx/xxxxxxr backup5:/vol/xxxxx/xxxxxxr Snapvaulted 5009:22:26 Idle

I have tried the following command :

ssh backupFiler snapvault status | awk '{print $4}'| cut -d ':' -f 1 | awk '{ if ( $1 > 24 ) print $0 }'

Problem is I get only the 4th column as output

ssh backupFiler snapvault status | awk '{ if ( $4 > 24 ) print $0 }'

Output is not accurate as the column is being considered as a string.

I need my output to show only the lines that have Lag greater than 24hrs.