Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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.
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
