Westonci.ca is the best place to get answers to your questions, provided by a community of experienced and knowledgeable experts. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
Answer:
a) 0.39984 = 39.984% probability of no orders in five minutes.
b) 0.06563 = 6.563% probability of 3 or more orders in five minutes.
c) The length of time is 0.63 hours
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.
Orders arrive at a Web site according to a Poisson process with a mean of 11 per hour.
This means that [tex]\mu = 11h[/tex], in which h is the number of hours.
a) Probability of no orders in five minutes.
Five minutes means that [tex]h = \frac{5}{60} = \frac{1}{12}[/tex], so [tex]\mu = \frac{11}{12} = 0.9167[/tex]
This probability is P(X = 0). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.9167}*(0.9167)^{0}}{(0)!} = 0.39984[/tex]
0.39984 = 39.984% probability of no orders in five minutes.
b) Probability of 3 or more orders in five minutes.
This is:
[tex]P(X \geq 3) = 1 - P(X < 3)[/tex]
In which
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.9167}*(0.9167)^{0}}{(0)!} = 0.39984[/tex]
[tex]P(X = 1) = \frac{e^{-0.9167}*(0.9167)^{1}}{(1)!} = 0.36653[/tex]
[tex]P(X = 2) = \frac{e^{-0.9167}*(0.9167)^{2}}{(2)!} = 0.168[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.39984 + 0.36653 + 0.168 = 0.93437[/tex]
[tex]P(X \geq 3) = 1 - P(X < 3) = 1 - 0.93437 = 0.06563[/tex]
0.06563 = 6.563% probability of 3 or more orders in five minutes.
c) Length of a time interval such that the probability of no orders in an interval of this length is 0.001.
This is h for which:
[tex]P(X = 0) = 0.001[/tex]
We have that:
[tex]P(X = 0) = e^{-\mu}[/tex]
And
[tex]\mu = 11h[/tex]
So
[tex]P(X = 0) = 0.001[/tex]
[tex]e^{-11h} = 0.001[/tex]
[tex]\ln{e^{-11h}} = \ln{0.001}[/tex]
[tex]-11h = \ln{0.001}[/tex]
[tex]h =-\frac{\ln{0.001}}{11}[/tex]
[tex]h = 0.63[/tex]
The length of time is 0.63 hours
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
