Discover a wealth of knowledge and get your questions answered on IDNLearn.com. Find in-depth and trustworthy answers to all your questions from our experienced community members.
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 value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
