Mathematics: What is the answer to the question?

What is the answer to the question?

jjimenez0276

23.12.2021

(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:

$P(X=x)=\frac{e^{-3}3^{x}}{x!};\ x=0,1,2,3...$

(1)

Compute the probability that a school-age child will sustain 2 injuries in a year as follows:

$P(X=2)=\frac{e^{-3}3^{2}}{2!}=\frac{0.049\times9}{2}=0.221$

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)

$=1-\frac{e^{-3}3^{0}}{0!}-\frac{e^{-3}3^{1}}{1!}\\=1-0.049-0.147\\=0.804$

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)

$=\frac{e^{-3}3^{0}}{0!}+\frac{e^{-3}3^{1}}{1!}+\frac{e^{-3}3^{2}}{2!}\\=0.049+0.147+0.221\\=0.417$

Thus, the probability that a school-age child will sustain at most 2 injuries in a year is 0.417.

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