poison distribution is used when avg probability of some event is given while binomial distribution is used when exact probabilty is given...

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Shubhanshu
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in Mathematical Logic
Aug 1, 2017

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$\lambda =\large ( \frac{4}{20} \times 0 ) + ( \frac{3}{20} \times 1) + ( \frac{5}{20} \times 2 )+ ( \frac{2}{20} \times 3 ) + ( \frac{4}{20} \times 4) + (\frac{5}{20} \times 5 )+ ( \frac{1}{20} \times 6 ) = 2.3 $

$P\{X = i\} = \Large \frac{e^{\lambda}\lambda^{i}}{i!}$

$P\{X = i + 1\} = P(i) \times \Large \frac{\lambda}{i+1}$

$P\{X = 0\} = \Large \frac{e^{-2.3}\times 2.3^{0}}{0!} = 0.100$

$P\{X = 1\} = P(0) \times \Large \frac{2.3}{0+1} = 0.230$

$P\{X = 2\} = P(1) \times \Large \frac{2.3}{1+1} = 0.264$

$P\{X > 2\} = 1 - 0.100 - 0.230 - 0.264 = 0.406$

$P\{X = i\} = \Large \frac{e^{\lambda}\lambda^{i}}{i!}$

$P\{X = i + 1\} = P(i) \times \Large \frac{\lambda}{i+1}$

$P\{X = 0\} = \Large \frac{e^{-2.3}\times 2.3^{0}}{0!} = 0.100$

$P\{X = 1\} = P(0) \times \Large \frac{2.3}{0+1} = 0.230$

$P\{X = 2\} = P(1) \times \Large \frac{2.3}{1+1} = 0.264$

$P\{X > 2\} = 1 - 0.100 - 0.230 - 0.264 = 0.406$