523 Consider a Poisson probability distribution with lambda

5.23 Consider a Poisson probability distribution with lambda = 4.7. Determine the following probabilities. a) exactly 5 occurrences b) more than 6 occurrences c) 3 or fewer occurrences

Solution

P(x=r) = e^(-mean) * mean^r / factorial(r)

mean =4.7

By using above formula,

P(x=5) = 0.1738

P(x>6) = 1 - P(x=0) - P(x=1) - P(x=2) - P(x=3) - P(x=4) - P(x=5) - P(x=6) = 1 - 0.8046 = 0.1954

P(x<=3) = P(x=0) + P(x=1) + P(x=2) + P(x=3) = 0.3097

r	P(x=r)
0	0.0091
1	0.0427
2	0.1005
3	0.1574
4	0.1849
5	0.1738
6	0.1362
7	0.0914
8	0.0537
9	0.0281
10	0.0132
11	0.0056
12	0.0022
13	0.0008