Paper 2 · Poisson Distribution
A company receives on average 6 orders per day, following a Poisson distribution. What is the probability that no more than 2 orders will be received on a given day, to 3 decimal places?
A0.045
B0.089
C0.062
D0.120
Explanation: With X ~ Poisson(6), P(X <= 2) = P(0)+P(1)+P(2) = e^-6(1+6+18) = 25e^-6, which is approximately 0.062.
Derived from ZIMSEC Statistics Paper 2, November 2018, Q8
