Respuesta :
Answer:
(1) The probability that a school-age child will sustain 2 injuries in a year is 0.221.
(2) The probability that a school-age child will sustain at least 2 injuries in a year is 0.0804.
(3) The probability that a school-age child will sustain at most 2 injuries in a year is 0.417.
Step-by-step explanation:
Let X = number of injuries per year for school-age children.
The average number of injuries per year for school-age children is, λ = 3.
The random variable X follows a Poisson distribution with parameter λ.
The probability mass function of X is:
[tex]P(X=x)=\frac{e^{-3}3^{x}}{x!};\ x=0,1,2,3...[/tex]
(1)
Compute the probability that a school-age child will sustain 2 injuries in a year as follows:
[tex]P(X=2)=\frac{e^{-3}3^{2}}{2!}=\frac{0.049\times9}{2}=0.221[/tex]
Thus, the probability that a school-age child will sustain 2 injuries in a year is 0.221.
(2)
Compute the probability that a school-age child will sustain at least 2 injuries in a year as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)
[tex]=1-\frac{e^{-3}3^{0}}{0!}-\frac{e^{-3}3^{1}}{1!}\\=1-0.049-0.147\\=0.804[/tex]
Thus, the probability that a school-age child will sustain at least 2 injuries in a year is 0.0804.
(3)
Compute the probability that a school-age child will sustain at most 2 injuries in a year as follows:
P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)
[tex]=\frac{e^{-3}3^{0}}{0!}+\frac{e^{-3}3^{1}}{1!}+\frac{e^{-3}3^{2}}{2!}\\=0.049+0.147+0.221\\=0.417[/tex]
Thus, the probability that a school-age child will sustain at most 2 injuries in a year is 0.417.
