Question

poisson distribution
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question 2
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dandelions are studied for their effects on crop production and lawn growth. in one region, the mean number of dandelions per square meter was found to be 8.2.
find the probability of no dandelions in an area of 1 m².
p(x = 0) =
find the probability of at least one dandelion in an area of 1 m².
p(at least one) =
find the probability of at most two dandelions in an area of 1 m².
p(x ≤ 2) =
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Explanation:

Step1: Recall the Poisson probability formula

The Poisson probability formula is \(P(X = k)=\frac{\lambda^{k}e^{-\lambda}}{k!}\), where \(\lambda\) is the mean number of events, \(k\) is the actual number of events, and \(e\approx2.71828\). Here, \(\lambda = 8.2\).

Step2: Calculate \(P(X = 0)\)

Substitute \(k = 0\) and \(\lambda=8.2\) into the formula:

$$
LATEXBLOCK0
$$

Step3: Calculate \(P(\text{at least one})\)

Use the complement rule \(P(X\geq1)=1 - P(X = 0)\). Since \(P(X = 0)\approx0.00039\), then \(P(X\geq1)=1 - 0.00039 = 0.99961\)

Step4: Calculate \(P(X\leq2)\)

\(P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)\)
For \(k = 1\): \(P(X = 1)=\frac{8.2^{1}\times e^{-8.2}}{1!}=8.2e^{-8.2}\approx8.2\times0.00039 = 0.003198\)
For \(k = 2\): \(P(X = 2)=\frac{8.2^{2}\times e^{-8.2}}{2!}=\frac{67.24\times e^{-8.2}}{2}\approx\frac{67.24\times0.00039}{2}=0.0131118\)
\(P(X\leq2)=0.00039+0.003198 + 0.0131118=0.0166998\approx0.0167\)

Answer:

\(P(X = 0)\approx0.00039\)
\(P(\text{at least one})\approx0.99961\)
\(P(X\leq2)\approx0.0167\)