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Sagot :
Answer:
0.8743 = 87.43% probability that more than one accident occurs per year
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 interval.
Buchtal, a manufacturer of ceramic tiles, reports on average 3.1 job-related accidents per year.
This means that [tex]\mu = 3.1[/tex]
What is the probability that more than one accident occurs per year?
This is:
[tex]P(X > 1) = 1 - P(X \leq 1)[/tex]
In which
[tex]P(X \leq 1) = P(X = 0) + P(X = 1)[/tex]
Then
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.6}*(3.6)^{0}}{(0)!} = 0.0273[/tex]
[tex]P(X = 1) = \frac{e^{-3.6}*(3.6)^{1}}{(1)!} = 0.0984[/tex]
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0273 + 0.0984 = 0.1257[/tex]
[tex]P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1257 = 0.8743[/tex]
0.8743 = 87.43% probability that more than one accident occurs per year
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