Question

In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample. To determine the number of CFUs, the sample is prepared, spread uniformly on an agar plate, and then incubated at some suitable temperature. Suppose that the number of CFUs that appear after incubation follows a Poisson distribution with μ = 15.

(a) If the area of the agar plate is 75 square centimeters (cm²), what is the probability of observing fewer than 4 CFUs in a 25 cm² area of the plate? (Round your answer to four decimal places.)

(b) If you were to count the total number of CFUs in 5 plates, what is the probability you would observe more than 85 CFUs? Use the Poisson distribution to obtain this probability. (Round your answer to four decimal places.)


(c) Repeat the probability calculation in part (b) but now use the normal approximation. (Round your answer to four decimal places.)


Find the difference between this value and your answer in part (b). (Round your answer to four decimal places.)

Answer 1

a)expected number of CFUs in a 25 cm² area =25*15/75 =5

probability of observing fewer than 4 CFUs in a 25 cm² area of the plate :

P(X<4)=

∑x=0x   {e-λ*λx/x!}=

0.2650

b)

expected number of f CFUs in 5 plates =5*15 =75

probability you would observe more than 85 CFUs :

P(X>=86)=1-P(X<=85)=

1-∑x=0x e-λ*λx/x!=

0.1142

c)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	75
std deviation   =σ=	8.660

therefore from normal approximation of poisson distribution and continuity correction:

probability =P(X>85.5)=P(Z>(85.5-75)/8.66)=P(Z>1.21)=1-P(Z<1.21)=1-0.8873=0.1127

(please try 0.1131 if this comes wrong)

difference between this value and your answer =0.1142-0.1127 =0.0015 (please try 0.0011 if this comes wrong)