Unformatted text preview: mean 4. So, the chance that they contain 10 bacteria in them equals e4 4 10 / 10!. Now, we can choose the bottle having Â¡ 3 bacteria in 2 ways, and it can have 0,1 or 2 bacteria. Summing over the 3 cases, we get that the required probability equals 2 X k =0 e2 2 k k ! .e2 2 10k (10k )! e4 4 10 10! = 1 2 10 2 X k =0 10! k !(10k )! Part (e): Say the bottle has n bacteria per liter. Probability that n > 5 bacteria per liter is there equals e2 2 n n ! Among them, exactly 2 bacteria are in the bottle, that probability equals ( nchoose2). So summing over all n > 5 the total probability equals âˆž X n =6 ( nchoose2) e2 2 n n ! Part (f): For poisson distribution mean equals variance. Number of bacteria per liter in 10 bottles follows poisson distribution with mean 20. So this will be the required variance. 1...
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 Spring '08
 CELMIN
 Statistics, Normal Distribution, Poisson Distribution, Probability, Probability theory, Binomial distribution, Discrete probability distribution, Negative binomial distribution

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