Statistics and Probability

Statistics and Probability - mean 4 So the chance that they...

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Part (a): No. of bacteria in bottle per liter follows Poisson distribution, with mean 2. A 20cc bottle has more than 2 bacteria means it has more than 2 bacteria per liter. This probability is equal to 1 - e - 2 (1 + 2 + 2) = 1 - 5 e 2 Part (b): number of bacteria in 3 bottles per liter follow poisson distribution with mean 6. 3 bottles contain more than 3 bacteria means the bottles have more than 3 bacteria per liter among them. So the required probability is 1 - e - 6 (1 + 6 + 18) = 1 - 25 e 6 Part (c): Take p = 5 e 2 = probability that a bottle contains less than 4 bac- teria. For different bottles the process follows negative binomial distribution. To obtain ’success’ i.e. less than 4 bacteria third time in 20th draw, one has to get it twice in the first 19 draws, which can be done in (19 - choose - 2) ways. Furthermore there are 3 ’succeses’ and 17 ’failure’-s in total, so the final probability is 19! 2! . 17! p 3 . (2 - p ) 17 Part (d): Number of bacteria in 2 bottles follow Poisson distribution with
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Unformatted text preview: mean 4. So, the chance that they contain 10 bacteria in them equals e-4 4 10 / 10!. Now, we can choose the bottle having ¡ 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 e-2 2 k k ! .e-2 2 10-k (10-k )! e-4 4 10 10! = 1 2 10 2 X k =0 10! k !(10-k )! Part (e): Say the bottle has n bacteria per liter. Probability that n > 5 bacteria per liter is there equals e-2 2 n n ! Among them, exactly 2 bacteria are in the bottle, that probability equals ( n-choose-2). So summing over all n > 5 the total probability equals ∞ X n =6 ( n-choose-2) e-2 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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