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Sagot :
Answer:
0.9898 = 98.98% probability that there will not be more than one failure during a particular week.
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.
3 failures every twenty weeks
This means that for 1 week, [tex]\mu = \frac{3}{20} = 0.15[/tex]
Calculate the probability that there will not be more than one failure during a particular week.
Probability of at most one failure, so:
[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^{-0.15}*0.15^{0}}{(0)!} = 0.8607[/tex]
[tex]P(X = 1) = \frac{e^{-0.15}*0.15^{1}}{(1)!} = 0.1291[/tex]
Then
[tex]P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898[/tex]
0.9898 = 98.98% probability that there will not be more than one failure during a particular week.
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