Question

Please solve this question

Answer 1

Using the Poisson distribution, the probabilities are given as follows:

• $P(X = 0) = e^{-10.6}$

• $P(X \geq 1) = 1 - e^{-10.6}$

• $P(X \leq 2) = 67.78e^{-10.6}$

What is the Poisson distribution?

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

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

The parameters are:

• x is the number of successes

• e = 2.71828 is the Euler number

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

The mean is given as follows:

$\mu = 10.6$

Hence the probability of no dandelions in an area of 1 m² is P(X = 0), hence:

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

$P(X = 0) = \frac{e^{-10.6}(10.6)^{0}}{(0)!} = e^{-10.6}$

The probability of at least one dandelion is:

$P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-10.6}$

The probability of at most two dandelions is:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

Then:

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

$P(X = 0) = \frac{e^{-10.6}(10.6)^{0}}{(0)!} = e^{-10.6}$

$P(X = 1) = \frac{e^{-10.6}(10.6)^{1}}{(1)!} = 10.6e^{-10.6}$

$P(X = 2) = \frac{e^{-10.6}(10.6)^{2}}{(2)!} = 56.18e^{-10.6}$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = e^{-10.6} + 10.6e^{-10.6} + 56.18e^{-10.6} = 67.78e^{-10.6}$

More can be learned about the Poisson distribution at https://brainly.com/question/13971530

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