Business

Business, 02.11.2019 04:31, blazepixi

Calls arrive at lynn ann fish's hotel switchboard at a rate of 2.5 per minute. the average time to handle each is 20 seconds. there is only one switchboard operator at the current time. the poisson and negative exponential distributions appear to be relevant in this situation. a) the probability that the operator is busy

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Mathematics, 24.06.2019 05:50, ctyrector

Normal approximation of the sampling distribution of a sumsuppose that the expected number of phone calls that are handled by a switchboard in each second is 5.35. assume that the distribution of the number of phone calls per second follows the poisson distribution. tasks 10-12 refer to this information.question 10the number of phone calls that the switchboard is expected to handle in 1 minute is question 11the 0.80-percentile of the number of calls per minute is (approximately) equal to: (use the normal approximation. the answer may be rounded up to 3 decimal places of the actual value.) ? question 12the probability that the switchboard will need to handle no more than 300 call in a minute is: (use the normal approximation, without a continuity correction. the answer may be rounded up to 3 decimal places of the actual value.)

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Mathematics, 10.07.2019 03:10, sarah7484

Calls arrive at lynn ann fish's hotel switchboard at a rate of 2.0 per minute. the average time to handle each is 20 seconds. there is only one switchboard operator at the current time. the poisson and negative exponential distributions appear to be relevant in this situation. a) the probability that the operator is busy = nothing (round your response to two decimal places). b) the average time that a caller must wait before reaching the operator = nothing minutes (round your response to two decimal places). c) the average number of calls waiting to be answered = nothing calls (round your response to two decimal places).

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Mathematics, 07.08.2019 03:10, lizzie3545

Calls arrive at lynn ann fish's hotel switchboard at a rate of 2.0 per minute. the average time to handle each is 20 seconds. there is only one switchboard operator at the current time. the poisson and negative exponential distributions appear to be relevant in this situation. a) the probability that the operator is busy = nothing (round your response to two decimal places). b) the average time that a caller must wait before reaching the operator = nothing minutes (round your response to two decimal places). c) the average number of calls waiting to be answered = nothing calls (round your response to two decimal places).

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