excans:9dde9115ca: Difference between revisions
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(Created page with "The number of calls received after <math>t</math> seconds equals <math>N_t</math> has a Poisson distribution with mean <math>\lambda t </math> where <math>\lambda </math> is the expected number of calls after 1 second, namely 0.01. Hence the probability that we have at most one call during a 5 minute break equals approximately <math display = "block">1-\exp(-0.01 * 60*5) = 0.95</math>") |
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'''Solution: E''' | |||
The number of calls received after <math>t</math> seconds equals <math>N_t</math> has a Poisson distribution with mean <math>\lambda t </math> where <math>\lambda </math> is the expected number of calls after 1 second, namely 0.01. Hence the probability that we have at most one call during a 5 minute break equals approximately <math display = "block">1-\exp(-0.01 * 60*5) = 0.95</math> | The number of calls received after <math>t</math> seconds equals <math>N_t</math> has a Poisson distribution with mean <math>\lambda t </math> where <math>\lambda </math> is the expected number of calls after 1 second, namely 0.01. Hence the probability that we have at most one call during a 5 minute break equals approximately <math display = "block">1-\exp(-0.01 * 60*5) = 0.95</math> | ||
Latest revision as of 15:42, 26 June 2024
Solution: E
The number of calls received after [math]t[/math] seconds equals [math]N_t[/math] has a Poisson distribution with mean [math]\lambda t [/math] where [math]\lambda [/math] is the expected number of calls after 1 second, namely 0.01. Hence the probability that we have at most one call during a 5 minute break equals approximately
[[math]]1-\exp(-0.01 * 60*5) = 0.95[[/math]]