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 =

Respuesta :

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 %