Answer:
(a) The probability of catching 5 individuals in the pond is 0.1429.
(b) The probability of catching at least 2 individuals in the pond is 0.8838.
Step-by-step explanation:
Let X = number of water fleas caught by sweeping the water a single time.
The random variable X follows a Poisson distribution with parameter λ = 3.7.
The probability mass function of the Poisson distribution is:
[tex]P(X=x)=\frac{e^{-3.7}3.7^{x}}{x!}[/tex]
(a)
Compute the value of P (X = 5) as follows:
[tex]P(X=5)=\frac{e^{-3.7}3.7^{5}}{5!}=\frac{17.1443}{120} =0.142869\approx0.1429[/tex]
Thus, the probability of catching 5 individuals in the pond is 0.1429.
(b)
Compute the value of P (X ≥ 2) as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)
[tex]=1-\frac{e^{-3.7}3.7^{0}}{0!}-\frac{e^{-3.7}3.7^{1}}{1!}\\=1-0.0247-0.0915\\=0.8838[/tex]
Thus, the probability of catching at least 2 individuals in the pond is 0.8838.
