The distribution of the number of printing mistakes per page of a book is Poisson with mean 3. Given that , the probability that there is some mistake in a page is
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The distribution of the number of printing mistakes per page of a book is Poisson with mean 3. Given that , the probability that there is some mistake in a page is
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The distribution of the number of printing mistakes per page of a book is Poisson with mean 3. Given that e⁻³ = 0.049787, the probability that there is some mistake in a page is (A) 0.049787 (B) 0.950212 (C) 0.905020 (D) None of the above
Also written as: The distribution of the number of printing mistakes per page of a book is Poisson with mean 3. Given that e^(-3) = 0.049787, the probability that there is some mistake in a page is
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This question is on The Poisson distribution, part of Random variables and standard distributions in Statistics.
Counts of rare events, where mean equals variance.
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Question from the JNU SIS 2012 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.