Biologists tagged 196 fish in a lake on January 1. On February 1, they returned and collected a random sample of 42 fish, 14 o
1 answer:
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Answer:
(a) 0.2650
(b) 0.0111
(c) 0.0105
(d) 0.0006
Step-by-step explanation:
Given that:
In microbiology, colony-forming units (CFUs) are used to measure the number of microorganisms present in a sample.
Suppose that the number of CFUs that appear after incubation follows a Poisson distribution with mean μ = 15. &;
If the area of the agar plate is 75cm²;
what is the probability of observing fewer than 4 CFUs in a 25 cm² area of the plate.
We can determine the mean number of CFUs that appear on a 25cm² area of the plate as follows;
75cm²/25cm² = 3
Since;
mean μ = 15
mean number of CFUs that appear on a 25cm² = 15/3 = 5 CFUs
Thus ; the probability of observing fewer than 4 CFUs in a 25 cm² area of the plate is estimated as:
= P(X < 4)
Using the EXCEL FUNCTION ( = poisson.dist(3, 5, TRUE) )
we have ;
P(X < 4) = 0.2650
b) If you were to count the total number of CFUs in 5 plates, what is the probability you would observe more than 95 CFUs?
Given that the total number of CFUs = 5 plates; then the mean number of CFUs in 5 plates = 15×5 = 75 CFUs
The probability is therefore = P( X > 95 )
= 1 - P(X ≤ 95)
= 1 - poisson.dist(95,75,TRUE) ( by using the excel function)
= 0.0111
c) Repeat the probability calculation in part (b) but now use the normal approximation.
Let assume that the mean and the variance of the poisson distribution are equal
Then;
We are to repeated the probability calculation in part (b) from above;
So:
P( X > 95 )
use the normal approximation
From standard normal variable table:
P(Z > 2.3094)
Using normal table
P(Z > 2.3094) = 0.0105
(d) Find the difference between this value and your answer in part (b).
So;
the difference between the value in part c and part b is;
= 0.0111 - 0.0105
= 0.0006 to four decimal places