Suppose that the average number of airline crashes in a country is 1.9 per month.(a) What is the probability that there will be at least 2 accidents in the next month?Probability =(b) What is the probability that there will be at least 4 accidents in the next two months?Probability =(c) What is the probability that there will be at most 3 accidents in the next four months?Probability =
Accepted Solution
A:
Answer:a) P [ x ≥ 2 ] = 0.5656 or 56.56 %b) P [ x ≥ 4 ] = 0.1258 or 12.58 % c) P [ x ≤ 3 ] = .8742 or 87.42 %Step-by-step explanation:We are going to solve a Poisson distribution problemλ = 1.9 crashes (per month)Poisson table we are going to use shows P [ X ≤ x]a) P [ x ≥ 2 ] P [ x ≥ 2 ] = 1 - P [ x ≤ 2-1 ] We can get he probability P [ x ≤ 2-1 ] from Poisson tableλ = 1.9 and x = 1In table we find λ value of 1,8 for x = 1 0.4628and λ value of 2 for x = 1 0.4060We need to interpolate: 0,2 ⇒ 0.0568 0,1 ⇒ ?? Δ = (0.0568)*(0,1)/0,2Δ = 0.0284P [ x ≤ 1 ] = 0.4344 then P [ x ≥ 2 ] = 1 - P [ x ≤ 2-1 ] = 1 - 0.4344 P [ x ≥ 2 ] = 0.5656 or 56.56 %b) P [ x ≥ 4 ]The same procedure P [ x ≥ 4 ] = 1 - P [ x ≤ (4-1) ] P [ x ≥ 4 ] = 1 - P [ x ≤ 3 ] From tables: λ value of 1,8 for x = 3 0.8913and λ value of 2 for x = 3 0.8571We need to interpolate: 0,2 ⇒ 0.0342 0,1 ⇒ ?? Δ = (0.0342)*(0,1)/0,2Δ = 0,0171Then P [ x ≤ 3 ] = 0.8742 andP [ x ≥ 4 ] = 1 - P [ x ≤ 3 ] ⇒ P [ x ≥ 4 ] = 1 - 0.8742P [ x ≥ 4 ] = 0.1258 or 12.58 %c) P [ x ≤ 3 ]In this case, the value is obtained interpolating from tableλ = 1.9 x = 3 1.8 0.8913 2.0 0.8571 Δ = 0,2 0.0342 0.1 ?? Δ = (0.0342)*(0.1)/0.2Δ = 0,0171Then P [ x ≤ 3 ] = 0.8913 -0.0171 = .8742 or 87.42 % P [ x ≤ 3 ] = .8742 or 87.42 %