Question

# Assume that the monthly worldwide average number of airplane crashes of commercial airlines is 2.2. What...

Assume that the monthly worldwide average number of airplane crashes of commercial airlines is 2.2. What is the probability that there will be

(a) more than 3 such accidents in the next month? 0.1807

(b) less than 2 such accidents in the next 3 months?

(c) exactly 6 such accidents in the next 4 months?

a)

Here, λ = 2.2 and x = 3
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X > 3) = 1 - P(X <= 3).
P(X > 3) = 1 - (2.2^0 * e^-2.2/0!) + (2.2^1 * e^-2.2/1!) + (2.2^2 * e^-2.2/2!) + (2.2^3 * e^-2.2/3!)
P(X > 3) = 1 - (0.1108 + 0.2438 + 0.2681 + 0.1966)
P(X > 3) = 1 - 0.8193 = 0.1807

b)

Here, λ = 6.6 and x = 2
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X < 2).
P(X < 2) = (6.6^0 * e^-6.6/0!) + (6.6^1 * e^-6.6/1!)
P(X < 2) = 0.0014 + 0.009
P(X <2) = 0.0104

c)

Here, λ = 8.8 and x = 6
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 6)
P(X = 6) = 8.8^6 * e^-8.8/6!
P(X = 6) = 0.0972
Ans: 0.0972

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