poisson distribution\nscore: 1/4 answered: 1/4\nprogress saved done\nquestion 2\ndetails\ndandelions are…

poisson distribution\nscore: 1/4 answered: 1/4\nprogress saved done\nquestion 2\ndetails\ndandelions 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.\nfind the probability of no dandelions in an area of 1 m².\np(x = 0) =\nfind the probability of at least one dandelion in an area of 1 m².\np(at least one) =\nfind the probability of at most two dandelions in an area of 1 m².\np(x ≤ 2) =\nquestion help: message instructor

poisson distribution\nscore: 1/4 answered: 1/4\nprogress saved done\nquestion 2\ndetails\ndandelions 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.\nfind the probability of no dandelions in an area of 1 m².\np(x = 0) =\nfind the probability of at least one dandelion in an area of 1 m².\np(at least one) =\nfind the probability of at most two dandelions in an area of 1 m².\np(x ≤ 2) =\nquestion help: message instructor

Answer

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: [ \begin{align*} P(X = 0)&=\frac{8.2^{0}\times e^{- 8.2}}{0!}\ &=e^{-8.2}\approx0.00039 \end{align*} ]

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))

[ \begin{align*} P(X\leq2)&=P(X = 0)+P(X = 1)+P(X = 2)\ P(X = 1)&=\frac{8.2^{1}\times e^{-8.2}}{1!}=8.2e^{-8.2}\approx8.2\times0.00039 = 0.003198\ 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}=0.01316\ P(X\leq2)&=0.00039+0.003198 + 0.01316\approx0.01675 \end{align*} ]

Answer:

(P(X = 0)\approx0.00039) (P(\text{at least one})\approx0.99961) (P(X\leq2)\approx0.01675)