Professor Shankar has made the following observation regarding students\' perfor

Question

Professor Shankar has made the following observation regarding students' performance in ECE 361. 8. If a student reads the material ahead of the lecture, the student follows the lecture very well 95% of the time (student is ahead. If the student comes unprepared, the student is able to absorb the material only 10% of the time (student falls behind). There are three meetings (lectures or recitations) every week. If student A is well-prepared at the beginning of week # 1, what is the probability that the student is well prepared by the start of week # 2. '1's and -1's are transmitted in in a 0.6:0.4 ratio in a communication system. Because of the noise in the channel, is received as a '1, 50% of the time and ‘1' is received as a ‘-T 40% of the time and rest of the time '1, is received as a 'O. Similarly, .1, is received as a '-1, 50% of the time and .1, is received as a '1, 40% of the time and rest of the time '-1' is received as a '0', (i) Show the transition chart and the transition matrix i) What is the probability of receiving a 9. If a '1' is observed, what is the probability that it was transmitted as a 1? 10. The FLU season will be upon us shortly. CDC estimates that it is about 60% successful. On the other hand, the chance of contracting FLU among unvaccinated population is 75%, if only 30% of the people get vaccinated (not acceptable), what is the probability that a person selected at random will be afflicted with FLU this season? Show the transition chart and display the transition matrix

Explanation / Answer

Ans:

10)

P(not flu/vaccinated)=0.6

P(flu/vaccinated)=1-0.6=0.4

P(flu/unvaccinated)=0.75

P(no flu/unvaccinated)=1-0.75=0.25

P(vaccinated)=0.3

P(unvaccinated)=1-0.3=0.7

P(flu)=P(flu/vaccinated)*P(vaccinated)+P(flu/unvaccinated)*P(unvaccinated)

=0.4*0.3+0.75*0.7

=0.12+0.525=0.645

P(flu)=0.645

Transition matrix:

[P(no flu); P(flu)] = [P(no flu|vaccinated) P(no flu|unvaccinated); P(flu|vaccinated) P(flu|unvaccinated)] X [P(vaccinated); P(unvaccinated)]

[0.355 0.645]=[0.6 0.25; 0.4 0.75]*[0.3, 0.7]

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