In bash how do I divide two variables and output the answer rounded upto 5 decimal digits? [duplicate]

The problem here is that you missed the echo (or printf or any other thing) to provide the data to bc:

$ echo "scale=5; 12/7" | bc1.71428

Also, as noted by cnicutar in comments, you need to use $ to refer to the variables. sum is a string, $sum is the value of the variable sum.
All together, your snippet should be like:

sum=12n=7output=$(echo "scale=5;$sum/$n" | bc)echo "$output"

This returns 1.71428.

Otherwise, with "scale=5;sum/n"|bc you are just piping an assignment and makes bc fail:

$ "scale=5;sum/n"|bcbash: scale=5;sum/n: No such file or directory

You then say that you want to have the result rounded, which does not happen right now:

$ sum=3345699$ n=1000000$ echo "scale=5;($sum/$n)" | bc3.34569

This needs a different approach, since bc does not round. You can use printf together with %.Xf to round to X decimal numbers, which does:

$ printf "%.5f" "$(echo "scale=10;$sum/$n" | bc)"3.34570

See I give it a big scale, so that then printf has decimals numbers enough to round properly.

sum and n, these are bash variables. you should add $ to get their values. So, the solution should be:

echo "scale=5;($sum/$n)"|bc1.71428

awk 'BEGIN{sum=12;n=7;printf "%0.5f\n", sum/n}'1.71429

In this solution , awk uses printf to round up decimal to 5 places. If you wish to pass bash variables then use following :

 awk -v sum=12 -v n=7 'BEGIN{printf "%0.5f\n", sum/n}' 1.71429

On side notes, awk seem to be good in arithmetic :

sh-4.1$  time echo "scale=5; 12/7" | bc ;  time echo "scale=5;($sum/$n)"|bc;time awk 'BEGIN{sum=12;n=7;printf "%0.5f\n", sum/n}'1.71428real    0m0.004suser    0m0.001ssys     0m0.002s1.71428real    0m0.004suser    0m0.001ssys     0m0.001s1.71429real    0m0.002suser    0m0.001ssys     0m0.000ssh-4.1$