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soldier1979 [14.2K]
3 years ago
15

Suppose that the log-ons to a computer network follow a Poisson process with an average of 3 counts per minute. (a) What is the

mean time between counts (in minutes)? (Round yours answers to 3 decimal places.) (b) What is the standard deviation between counts (in minutes)? (Round yours answers to 3 decimal places.) (c) If it is an average of 3 counts per minute, find the value of such that . (Round yours answers to 4 decimal places.)
Mathematics
1 answer:
vampirchik [111]3 years ago
6 0

Answer: a) 0.333, b) 0.333 and c) 0.9986.

Step-by-step explanation:

Since we have given that

Average = 3 counts per minute = λ

a) What is the mean time between counts ?

Since it follows a Poisson Process, So,

E[x]=\dfrac{1}{\lambda}=\dfrac{1}{3}=0.333

(b) What is the standard deviation between counts ?

\sigma=\dfrac{1}{3}=0.333

(c) If it is an average of 3 counts per minute, find the value of such that .

If the average = 3 counts per minute.

Then, P(X

Hence, a) 0.333, b) 0.333 and c) 0.9986.

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Find a closed form expression for the nth right Riemann sum of this integral?
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Partitioning the interval [6,11] into n equally-spaced subintervals gives n rectangles of width \Delta x=\dfrac{11-6}n=\dfrac5n and of heights determined by the right endpoints of each subinterval.

If a=x_1=6, then x_2=6+\dfrac5n, x_3=6+2\dfrac5n, and so on, up to b=x_{n+1}=6+n\dfrac5n=11. Because we're using the right endpoints, the approximation will consider x_2,\ldots,x_{n+1}

The definite integral is then approximated by

\displaystyle\int_6^{11}(1-5x)\,\mathrm dx\approx\sum_{i=2}^{n+1}f(x_i)\Delta x=\sum_{i=2}^{n+1}\left(1-\left(6+(i-1)\dfrac5n\right)\dfrac5n

You have

\displaystyle\sum_{i=2}^{n+1}\left(1-5\left(6+(i-1)\dfrac5n\right)\dfrac5n=\sum_{i=2}^{n+1}\left(-29-\dfrac{25}n(i-1)\right)\dfrac5n
=\displaystyle-{145}n\sum_{i=2}^{n+1}1-\dfrac{125}{n^2}\sum_{i=2}^{n+1}(i-1)
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To check that this is correct, let's make sure the sum converges to the exact value of the definite integral. As n\to\infty, you have the sum converging to -\dfrac{145}2.

Meanwhile,

\int_6^{11}(1-5x)\,\mathrm dx=\left[x-\dfrac52x^2\right]_{x=6}^{x=11}=-\dfrac{415}2

so we're done.
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