If a drop of water is placed on a slide and examined under a microscope, the num

Question

If a drop of water is placed on a slide and examined under a microscope, the number x of a specific type of bacteria present has been found to have a Poisson probability distribution. Suppose the maximum permissible count per water specimen for this type of bacteria is six. If the mean count for your water supply is three and you test a single specimen, is it likely that the count will exceed the maximum permissible count? Explain. (Round your bacteria value to the nearest whole number, and round your probability to three decimal places.) No, at least three-quarters of the observations should fall between 0 bacteria and  __?__ bacteria present, inclusive, and the exact Poisson probability is P(x > 6) = __?__. Please show work. If a drop of water is placed on a slide and examined under a microscope, the number x of a specific type of bacteria present has been found to have a Poisson probability distribution. Suppose the maximum permissible count per water specimen for this type of bacteria is six. If the mean count for your water supply is three and you test a single specimen, is it likely that the count will exceed the maximum permissible count? Explain. (Round your bacteria value to the nearest whole number, and round your probability to three decimal places.) No, at least three-quarters of the observations should fall between 0 bacteria and  __?__ bacteria present, inclusive, and the exact Poisson probability is P(x > 6) = __?__. Please show work. bacteria and  __?__ bacteria present, inclusive, and the exact Poisson probability is P(x > 6) = __?__. Please show work.

Explanation / Answer

POSSION DISTRIBUTION
pmf of P.D is = f ( k ) = e- x / x!
where
= parameter of the distribution.   
x = is the number of independent trials
I.
mean =
= 3
GREATER THAN
P( X < = 6) = P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= e^-3 * 3 ^ 6 / 6! + e^-3 * 0 ^ 5 / 5! + e^-3 * ^ 4 / 4! + e^-3 * ^ 3 / 3! + e^-3 * ^ 2 / 2! + e^-3 * ^ 1 / 1! + e^-3 * ^ 0 / 0!
= 0.96649

P( X > 6) = 1 -P ( X <= 6) = 1 - 0.9665 = 0.03351

which is less than 0.05 and we conclude,

no evidence that the count will exceed the maximum permissible count

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