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).

Answers

Answer 1

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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Answers

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Step-by-step explanation:

Explanation:

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For this situation, the sample size is 65, and the standard error of the mean is 44.8. Substitute these values into the formula, and solve for .

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We have:

Melvin conducted a series of surveys asking political candidates about the number of constituents they mail literature to each sample contains 65 responses, and the standard error of the mean is 44.8.

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14x = -126 + 28

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Answer:

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Step-by-step explanation:

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The Equation Is Solved With:

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Step-by-step explanation:

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A Ferris wheel is 35 meters in diameter and boarded from a platform that is 3 meters above the ground. The six o'clock position of the Ferris wheel is level with the loading platform. The wheel completes one revolution in 8 minutes. The function, h(t), gives your height in meters above the ground t minutes after the wheel beings to turn. How high are you off the ground after 6 minutes

Answers

Answer:

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Step-by-step explanation:

Given the data in the question;

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Step-by-step explanation:

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