Question

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 =

Answer 1

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 %

Explanation:

We are going to solve a Poisson distribution problem

λ = 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 table

λ = 1.9 and x = 1

In table we find λ value of 1,8 for x = 1 0.4628

and λ value of 2 for x = 1 0.4060

We need to interpolate:

0,2 ⇒ 0.0568

0,1 ⇒ ?? Δ = (0.0568)*(0,1)/0,2

Δ = 0.0284

P [ 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 0.8913

and λ value of 2 for x = 3 0.8571

We need to interpolate:

0,2 ⇒ 0.0342

0,1 ⇒ ?? Δ = (0.0342)*(0,1)/0,2

Δ = 0,0171

Then

P [ x ≤ 3 ] = 0.8742

and

P [ x ≥ 4 ] = 1 - P [ x ≤ 3 ] ⇒ P [ x ≥ 4 ] = 1 - 0.8742

P [ x ≥ 4 ] = 0.1258 or 12.58 %

c) P [ x ≤ 3 ]

In this case, the value is obtained interpolating from table

λ = 1.9 x = 3

1.8 0.8913

2.0 0.8571

Δ = 0,2 0.0342

0.1 ?? Δ = (0.0342)*(0.1)/0.2

Δ = 0,0171

Then

P [ x ≤ 3 ] = 0.8913 -0.0171 = .8742 or 87.42 %

P [ x ≤ 3 ] = .8742 or 87.42 %