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Sagot :
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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
Here λ = 2.90 is the average number of bacteria colonies per field.
This means that [tex]\mu = 2.90[/tex]
Compute P(r) for r = 0, 1, 2, 3, 4, and 5 or more.
[tex]P(0) = \frac{e^{-2.9}*(2.9)^{0}}{(0)!} = 0.055[/tex]
[tex]P(1) = \frac{e^{-2.9}*(2.9)^{1}}{(1)!} = 0.16[/tex]
[tex]P(2) = \frac{e^{-2.9}*(2.9)^{2}}{(2)!} = 0.231[/tex]
[tex]P(3) = \frac{e^{-2.9}*(2.9)^{3}}{(3)!} = 0.224[/tex]
[tex]P(4) = \frac{e^{-2.9}*(2.9)^{4}}{(4)!} = 0.162[/tex]
5 or more:
This is
[tex]P(X \geq 5) - 1 - P(X < 5)[/tex]
In which:
[tex]P(X < 5) Â = 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[/tex]
[tex]P(X \geq 5) - 1 - P(X < 5) = 1 - 0.832 = 0.168[/tex]
So
P(5 or more) = 0.168
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