Assume that the number of airplane crashes per month worldwide has a Poisson distribution. The monthly worldwide average number of airplane crashes is 1.2. What is the probability that there will be more than 2 such accidents in the next month

0.12

Step-by-step explanation:

Poisson distribution is used to approximate the probability of a given number of event with a known average occurrence. The Formula for this distribution is denoted by:

P(X=k) = e^-λ × (λ^k/k!)

Where k = variable probability value.

λ = average occurrence of event

e = exponential value = 2.71828

For determining on the number of accidents greater than 2,we determine for when k = 0,1,2 and subtract the answer from 1, that is P(X>2) = 1 - P(X≤2)

When k=0,

P(X=0) = e^-1.2 × (1.2^0/0!) = 0.3012

When k=1

P(X=1) = e^-1.2 × (1.2^1/1!) = 0.3614

When k=2

P(X=2) = e^-1.2 × (1.2²/2!) = 0.2169

When k=0,1,2, the sum of probability = 0.3012 + 0.3614 + 0.2169 = 0.8795.

Hence, P(X>2) = 1 - 0.8795 = 0.1205 ≈ 0.12

cause it equals more