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 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 %
