Respuesta :
Answer:
a) 0.064 = 6.4% probability that the men's team experiences 2 or more concussions
b) 0.131 = 13.1% probability that the women's team experiences 2 or more concussions
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
In this question:
We have to find [tex]P(X \geq 2)[/tex] for both cases.
a. What is the probability that the men's team experiences 2 or more concussions?
For collegiate men soccer players the concussion rate is .41 per 1,000 participation hours. This means that [tex]\mu = 0.41[/tex]
Either less than 2 get concussed, or at least two do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.41}*(0.41)^{0}}{(0)!} = 0.664[/tex]
[tex]P(X = 1) = \frac{e^{-0.41}*(0.41)^{1}}{(1)!} = 0.272[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.664 + 0.272 = 0.936[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.936 = 0.064[/tex]
0.064 = 6.4% probability that the men's team experiences 2 or more concussions.
b. What is the probability that the women's team experiences 2 or more concussions?
For collegiate women soccer players the concussion rate is .63 per 1,000 participation hours. This means that [tex]\mu = 63[/tex]
Following the same logic as a.
[tex]P(X = 0) = \frac{e^{-0.63}*(0.63)^{0}}{(0)!} = 0.533[/tex]
[tex]P(X = 1) = \frac{e^{-0.63}*(0.63)^{1}}{(1)!} = 0.336[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.533 + 0.336 = 0.869[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.869 = 0.131[/tex]
0.131 = 13.1% probability that the women's team experiences 2 or more concussions
