The Poisson distribution defines the probability of k discrete and independent events occurring in a given time interval.
If λ = the average number of event occurring within the given interval, then

For the given problem,
λ = 6.5, average number of tickets per day.
k = 6, the required number of tickets per day
The Poisson distribution is

The distribution is graphed as shown below.
Answer:
The mean is λ = 6.5 tickets per day, and it represents the expected number of tickets written per day.
The required value of k = 6 is less than the expected value, therefore the department's revenue target is met on an average basis.
Answer:
d) (179.20, 212.716)
Step-by-step explanation:
1000 : 345
569 : 195.96
Midpoint of (179.20, 212.716) is 195.958 which is the closest to 195.96
1.
√(16+20)=√36=6
C
2.
√(21-9)=√(12)=(√4)(√3)=2√3
B
3. 2*√3=2*1.73205=3.464
B
1.C
2.B
3.B
It would equal 0.75 and as a fraction it is 3/4