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...
View
Full
Document
 Spring '08
 CELMIN
 Statistics, Poisson Distribution, Probability

Click to edit the document details