Question

A pathologist has been studying the frequency of bacterial colonies within the field of a microscope using samples of throat cultures from healthy adults. Long-term history indicates that there is an average of 2.90 bacteria colonies per field. Let r be a random variable that represents the number of bacteria colonies per field. Let O represent the number of observed bacteria colonies per field for throat cultures from healthy adults. A random sample of 100 healthy adults gave the following information:
r 0 1 2 3 4 5 or more
O 11 14 30 17 22 6

The pathologist wants to use a Poisson distribution to represent the probability of r, the number of bacteria colonies per field. The Poisson distribution is given below.
P(r) = e^−λλr / r!
Here λ = 2.90 is the average number of bacteria colonies per field.

Required:
Compute P(r) for r = 0, 1, 2, 3, 4, and 5 or more.

Answer 1

Answer:

P(0) = 0.055

P(1) = 0.16

P(2) = 0.231

P(3) = 0.224

P(4) = 0.162

P(5 or more) = 0.168

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Here λ = 2.90 is the average number of bacteria colonies per field.

This means that $\mu = 2.90$

Compute P(r) for r = 0, 1, 2, 3, 4, and 5 or more.

$P(0) = \frac{e^{-2.9}*(2.9)^{0}}{(0)!} = 0.055$

$P(1) = \frac{e^{-2.9}*(2.9)^{1}}{(1)!} = 0.16$

$P(2) = \frac{e^{-2.9}*(2.9)^{2}}{(2)!} = 0.231$

$P(3) = \frac{e^{-2.9}*(2.9)^{3}}{(3)!} = 0.224$

$P(4) = \frac{e^{-2.9}*(2.9)^{4}}{(4)!} = 0.162$

5 or more:

This is

$P(X \geq 5) - 1 - P(X < 5)$

In which:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.055 + 0.16 + 0.231 + 0.224 + 0.162 = 0.832$

$P(X \geq 5) - 1 - P(X < 5) = 1 - 0.832 = 0.168$

So

P(5 or more) = 0.168