Assignment Previewer 2018 Quasion 1 2345678910 DevoreStat9 3.E.080.(40611331 1.

Question

Assignment Previewer 2018 Quasion 1 2345678910 DevoreStat9 3.E.080.(40611331 1. Question Details number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article for Probabilistic Life Prediction of Multiple-Anomaly Materials proposes a Poisson distribution for X. Suppose Methodology for Probabilistic Life that -4. (Round your answers to three decimal places.) (a) Compute both PX s 4) and P(X 4) (b) Compute Pi4 s x s 6). (c) Compute P6 s x). (d) What is the probability that the number of anomalies does not exceed the mean value by more than ane standard deviation? You may need to use the appropriate table in the Appendix of Tables to answer this question. DevoreStat9 3.E.050. [4061066 Question Detals A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls involve fax messages, and consider a sample of 25 incoming calls. (Round your answers to three decimal places.) (a) What is the probability that at most 5 of the calls involve a fax message? (b) What is the probability that exactly 5 of the calls involve a fax message? (c) What is the probability that at least 5 of the calls involve a fax message? (d) What is the probability that more than 5 of the calls involve a fax message? You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

1) = 4

P(X = x) = e- * x / x!

a) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

                   = e-4 * 40 / 0! + e-4 * 41 / 1! + e-4 * 42 / 2! + e-4 * 43 / 3! + e-4 * 44 / 4!

                   = 0.629

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                   = e-4 * 40 / 0! + e-4 * 41 / 1! + e-4 * 42 / 2! + e-4 * 43 / 3!

                   = 0.433

b) P(4 < X < 6) = P(X = 4) + P(X = 5) + P(X = 6)

                         = e-4 * 44 / 4! + e-4 * 45 / 5! +e-4 * 46 / 6!

                         = 0.456

c) P(X > 6) = 1 - P(X < 6)

                  = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5))

                  = 1 - (e-4 * 40 / 0! + e-4 * 41 / 1! + e-4 * 42 / 2! + e-4 * 43 / 3! + e-4 * 44 / 4! + e-4 * 45 / 5! )

                  = 1 - 0.785

                  = 0.215

d) mean = 4

standard deiviation = sqrt(4) = 2

mean + sd = 4 + 2 = 6

P(X < 6) = e-4 * 40 / 0! + e-4 * 41 / 1! + e-4 * 42 / 2! + e-4 * 43 / 3! + e-4 * 44 / 4! + e-4 * 45 / 5! + e-4 * 46 / 6!

              = 0.889

2) p = 0.25

n = 25

P(X = x) = 25Cx * 0.25x * (1 - 0.25)25-x

a) P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

                   = 25C0 * 0.250 * 0.7525 + 25C1 * 0.251 * 0.7524 + 25C2 * 0.252 * 0.7523 + 25C3 * 0.253 * 0.7522 + 25C4 * 0.254 * 0.7521 + 25C5 * 0.255 * 0.7520

                   = 0.378

b) P(X = 5) = 25C5 * 0.255 * 0.7520 = 0.165

c) P(X > 5) = 1 - (P(X < 5) - P(X = 5)) = 1 - (0.378 - 0.165) = 0.787

d) P(X > 5) = 1 - P(X < 5) = 1 - 0.378 = 0.622

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