Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab

Question

atroni

2 months ago

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Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab

ilities that the number of inquiries in a particular 50-millisecond stretch will be:

Mathematics

1 answer:

Svet_ta [12.7K] · Answer 1 · 2026-03-10T14:59:37+03:00

Response: a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Detailed explanation:

In Problem 8-4, the computer time-sharing system experiences teleport inquiries at an average rate of 0.1 per millisecond. We are tasked with determining the probabilities of the inquiries over a specific period of 50 milliseconds:

Given that

$\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5$

Applying the Poisson process, we find that

(a) at most 12

probability= $P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980$

(b) exactly 13

probability= $P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013$

(c) more than 12

probability= $P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020$

(d) exactly 20

probability= $P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026$

(e) within the range of 10 to 15, inclusive

probability= $P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318$

Thus, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318