A circle has a circumference whose length is 25 pi find the length of an arc whose central angle is 90 dregree

Orders arrive at a Web site according to a Poisson process with a mean of 11 per hour. Determine the following: a) Probability o

Nataliya [291]

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ 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 $\mu = 11h$ , in which h is the number of hours.

a) Probability of no orders in five minutes.

Five minutes means that $h = \frac{5}{60} = \frac{1}{12}$ , so $\mu = \frac{11}{12} = 0.9167$

This probability is P(X = 0). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.9167}*(0.9167)^{0}}{(0)!} = 0.39984$

0.39984 = 39.984% probability of no orders in five minutes.

b) Probability of 3 or more orders in five minutes.

This is:

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-0.9167}*(0.9167)^{0}}{(0)!} = 0.39984$

$P(X = 1) = \frac{e^{-0.9167}*(0.9167)^{1}}{(1)!} = 0.36653$

$P(X = 2) = \frac{e^{-0.9167}*(0.9167)^{2}}{(2)!} = 0.168$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.39984 + 0.36653 + 0.168 = 0.93437$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.93437 = 0.06563$

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:

$P(X = 0) = 0.001$

We have that:

$P(X = 0) = e^{-\mu}$

And

$\mu = 11h$

So

$P(X = 0) = 0.001$

$e^{-11h} = 0.001$

$\ln{e^{-11h}} = \ln{0.001}$

$-11h = \ln{0.001}$

$h =-\frac{\ln{0.001}}{11}$

$h = 0.63$

The length of time is 0.63 hours

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