Question 1121571
This is a Poisson distribution with variables random, independent, proportional to time, theoretically independent number in a time interval.

per minute there are 2.5 print jobs, and that is lambda, the Poisson parameter.
fewer than 5 print jobs is 0,1,2,3,4
can do by hand or table
by hand e^(-2.5)2.5^0/0! is for 0=0.082 from e^(-lambda)*lambda^x/x!
e^(-2.5)*2.5/1!=0.205 for 1
e^(-2.5)*2.5^2/2=0.257
e^(-2.5)*2.5^3/6=0.214
e^(-2.5)*2.5^4/24=0.134 for 4
0.891 is the probability if rounding occurs at the end (0.892 if the rounding occurs to 3 decimal places and those are added).
0.891 ia the probability from the table