Answer: The Poisson probability that at least 2 will miss the flight P(x ≥ 2) is 0.22753
Explanation:
Given that;
probability that a ticket holder will miss a flight = 0.005
(number of passengers) sample size n = 180
so avg. number of passengers missing the flight will be λ = 0.005 × 180 = 0.9
so X: number of passengers will miss the flight
where x follows Poisson distribution with λ as 0.9
Probability mass function of Poisson is expressed as;
P(x) = (e^-λ)λˣ / x!
so we substitute
P(x) = (e^-0.9)0.9ˣ / x!
Now Probability that at least 2 will miss the flight;
P(x ≥ 2) = 1 - P(x < 2) = 1 - (P(x=0) + P(x=1) )
so
for P(x=0); P(0) = (e^-0.9)0.9⁰ / 0! = 0.40656
for P(x=1); P(1) = (e^-0.9)0.9¹ / 1! = 0.36591
hence;
(P(x=0) + P(x=1) ) = 0.40656 + 0.36591 = 0.77247
P(x ≥ 2) = 1 - P(x < 2) = 1 - (P(x=0) + P(x=1) )
P(x ≥ 2) = 1 - 0.77247 = 0.22753
Therefore, The Poisson probability that at least 2 will miss the flight P(x ≥ 2) is 0.22753