Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.