Dr. Smith is a teacher in the education department who serves one student at a time...

  

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Dr. Smith is a teacher in the education department who serves one student at a time throughout the day. He can serve an average of 8 students per hour and the students arrive at his office at a rate of 3.6 per hour. a) What is the average time a student spends at Dr. Smith's office (either waiting or meeting with Dr. Smith)? (7 Points) b) What is the probability that 4 students will arrive at Dr. Smith's office in the next hour? (7 Points) c) What is the average number of students at Dr. Smith's office (either waiting or meeting with Dr. Smith)? (7 Points) Amos is going to Dr. Smith's office. Remember that a person approaching a waiting line system is not a part of that system. d) What is the probability Amos will find exactly 1 person waiting in line? (8 Points) e) What is the probability Amos will find more than 1 person waiting in line? (11 Points) This is another queueing theory problem. Dr. Smith's office can be modeled as an M/M/1 queue, where the arrival rate is λ = 3.6 students per hour and the service rate is µ = 8 students per hour. a) The average time a student spends at Dr. Smith's office (either waiting or meeting with Dr. Smith) is given by the formula W = 1 / (μ-X). Plugging in the values for λ and μ, we get W = 1 / (8-3.6) = 0.1923 hours, or about 11.54 minutes. b) The number of arrivals in a fixed period of time follows a Poisson distribution with parameter Xt, where t is the length of the time period. In this case, t = 1 hour, so the parameter is At = 3.6*1 = 3.6. The probability of k arrivals in this time period is given by the formula P(k) = (At)^k *exp(-λt) / k!. So, the probability of 4 arrivals in the next hour is P(4) = (3.6)^4 *exp(-3.6) / 4! = 0.1742. c) The average number of students at Dr. Smith's office (either waiting or meeting with Dr. Smith) is given by the formula L = λ/ (μ-X). Plugging in the values for λ and μ, we get L = 3.6 / (8-3.6) = 0.6923. d) The probability that there are n customers in the system (i.e., either waiting or meeting with Dr. Smith) is given by the formula P(n) = (^/μ)^n * (1 -/u). So, the probability that Amos will find exactly 1 person waiting in line when he arrives is P(1) = (3.6/8)^1 * (1-3.6/8)= 0.3375. e) The probability that Amos will find more than 1 person waiting in line when he arrives is given by the formula P(more than 1 waiting) = 1 - P(0 waiting) - P(1 waiting). Using the formula for P(n), we have P(more than 1 waiting) = 1 - P(0) - P(1) = 1 - [(3.6/8)^0 * (1-3.6/8)] - [(3.6/8)^1 *(1-3.6/8)] = 0.405. Dr. Smith is a teacher in the education department who serves one student at a time throughout the day. He can serve an average of 8 students per hour and the students arrive at his office at a rate of 3.6 per hour. a) What is the average time a student spends at Dr. Smith's office (either waiting or meeting with Dr. Smith)? (7 Points) b) What is the probability that 4 students will arrive at Dr. Smith's office in the next hour? (7 Points) c) What is the average number of students at Dr. Smith's office (either waiting or meeting with Dr. Smith)? (7 Points) Amos is going to Dr. Smith's office. Remember that a person approaching a waiting line system is not a part of that system. d) What is the probability Amos will find exactly 1 person waiting in line? (8 Points) e) What is the probability Amos will find more than 1 person waiting in line? (11 Points) This is another queueing theory problem. Dr. Smith's office can be modeled as an M/M/1 queue, where the arrival rate is λ = 3.6 students per hour and the service rate is µ = 8 students per hour. a) The average time a student spends at Dr. Smith's office (either waiting or meeting with Dr. Smith) is given by the formula W = 1 / (μ-X). Plugging in the values for λ and μ, we get W = 1 / (8-3.6) = 0.1923 hours, or about 11.54 minutes. b) The number of arrivals in a fixed period of time follows a Poisson distribution with parameter Xt, where t is the length of the time period. In this case, t = 1 hour, so the parameter is At = 3.6*1 = 3.6. The probability of k arrivals in this time period is given by the formula P(k) = (At)^k *exp(-λt) / k!. So, the probability of 4 arrivals in the next hour is P(4) = (3.6)^4 *exp(-3.6) / 4! = 0.1742. c) The average number of students at Dr. Smith's office (either waiting or meeting with Dr. Smith) is given by the formula L = λ/ (μ-X). Plugging in the values for λ and μ, we get L = 3.6 / (8-3.6) = 0.6923. d) The probability that there are n customers in the system (i.e., either waiting or meeting with Dr. Smith) is given by the formula P(n) = (^/μ)^n * (1 -/u). So, the probability that Amos will find exactly 1 person waiting in line when he arrives is P(1) = (3.6/8)^1 * (1-3.6/8)= 0.3375. e) The probability that Amos will find more than 1 person waiting in line when he arrives is given by the formula P(more than 1 waiting) = 1 - P(0 waiting) - P(1 waiting). Using the formula for P(n), we have P(more than 1 waiting) = 1 - P(0) - P(1) = 1 - [(3.6/8)^0 * (1-3.6/8)] - [(3.6/8)^1 *(1-3.6/8)] = 0.405.

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