What does “INFO (RANDOM_DELAY will be scaled with factor 84% if used)” actualy mean?

Question

What does the following log line mean?
INFO (RANDOM_DELAY will be scaled with factor 84% if used.)

Googling didn't find meaningful answers: is the answer only in the sources?
(OS is CentOS 7)
Edit
This line was received from logwatch, viz:
 ################### Logwatch 7.4.0 (03/01/11) ####################
        Processing Initiated: Wed Jul 15 09:38:02 2020
        Date Range Processed: yesterday
                              ( 2020-Jul-14 )
                              Period is day.
        Detail Level of Output: 0
        Type of Output/Format: mail / text
        Logfiles for Host: laptop.local
 ##################################################################

--------------------- Cron Begin ------------------------

**Unmatched Entries**
 INFO (RANDOM_DELAY will be scaled with factor 84% if used.)
 INFO (RANDOM_DELAY will be scaled with factor 47% if used.)
 INFO (RANDOM_DELAY will be scaled with factor 74% if used.)
 INFO (Shutting down)
 INFO (RANDOM_DELAY will be scaled with factor 23% if used.)

---------------------- Cron End -------------------------

The actual log, where these lines come from, was /var/log/cron.

Michael Hampton · Answer

The crontab(5) man page gives the answer:

The RANDOM_DELAY variable allows delaying job startups by random
   amount of minutes with upper limit specified by the variable. The
   random scaling factor is determined during the cron daemon startup so
   it remains constant for the whole run time of the daemon.

So if you put RANDOM_DELAY=60 in your crontab, the actual delay before the job starts will be the logged percentage, e.g. 84% of 60 minutes. Nothing happens if you don't use RANDOM_DELAY.

What does “INFO (RANDOM_DELAY will be scaled with factor 84% if used)” actualy mean?

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