Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A 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.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
