Orders arrive at a Web site according to a Poisson process with a mean of 11 per hour. Determine the following: a) Probability of no orders in five minutes. Round your answer to three decimal places (e.g. 98.765). Enter your answer in accordance to the item a) of the question statement b) Probability of 3 or more orders in five minutes. Round your answer to three decimal places (e.g. 98.765). Enter your answer in accordance to the item b) of the question statement c) Length of a time interval such that the probability of no orders in an interval of this length is 0.001. Round your answer to two decimal places (e.g. 98.76).

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

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