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20 May, 19:29

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 =

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  1. 20 May, 22:38
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    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 %
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