Question

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:

Answer 1

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

Step-by-step explanation:

Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probabilities that the number of inquiries in a particular 50-millisecond stretch will be:

Since we have given that

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

Using the poisson process, we get that

(a) less than or equal to 12

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

(b) equal to 13

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

(c) greater than 12

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

(d) equal to 20

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

(e) between 10 and 15, inclusively

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

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