exercise:8018332246: Difference between revisions
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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> It is often assumed that the auto traffic that arrives at the intersection during a unit time period has a Poisson distribution with expected value <math>m</math>. Assume that the number of cars <math>X</math> that arrive at an intersection from...") |
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It is often assumed that the auto traffic that arrives at the intersection during a unit time period has a Poisson distribution with expected | |||
intersection during a unit time period has a Poisson distribution with expected | |||
value <math>m</math>. Assume that the number of cars <math>X</math> that arrive at an intersection from the | value <math>m</math>. Assume that the number of cars <math>X</math> that arrive at an intersection from the | ||
north in unit time has a Poisson distribution with parameter <math>\lambda = m</math> and the | north in unit time has a Poisson distribution with parameter <math>\lambda = m</math> and the | ||
Latest revision as of 00:13, 14 June 2024
It is often assumed that the auto traffic that arrives at the intersection during a unit time period has a Poisson distribution with expected value [math]m[/math]. Assume that the number of cars [math]X[/math] that arrive at an intersection from the north in unit time has a Poisson distribution with parameter [math]\lambda = m[/math] and the number [math]Y[/math] that arrive from the west in unit time has a Poisson distribution with parameter [math]\lambda = \bar m[/math]. If [math]X[/math] and [math]Y[/math] are independent, show that the total number [math]X + Y[/math] that arrive at the intersection in unit time has a Poisson distribution with parameter [math]\lambda = m + \bar m[/math].