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Grasshoppers are distributed at random in a large field according to a Poisson process with parameter a \({\rm{\alpha = 2}}\) per square yard. How large should the radius \({\rm{R}}\) of a circular sampling region be taken so that the probability of finding at least one in the region equals \({\rm{.99}}\)?

Short Answer

Expert verified

The radius \({\rm{R}}\) of a circular sampling region should be taken \({\rm{R = 0}}{\rm{.85612}}\) so that the probability of finding at least one in the region equals \({\rm{.99}}\).

Step by step solution

01

Concept Introduction

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

Proposition: Number of events during a time interval of length\({\rm{t}}\)can be modelled using Poisson Random Variable with parameter\({\rm{\mu = \alpha t}}\). This indicates that –

\({{\rm{P}}_{\rm{k}}}{\rm{(t) = }}{{\rm{e}}^{{\rm{ - \alpha t}}}}\frac{{{{{\rm{(\alpha t)}}}^{\rm{k}}}}}{{{\rm{x!}}}}\)

It is also known as Poisson process (not formally defined).

02

Radius of the region

It is given that –

\({\rm{t = }}\pi {{\rm{R}}^{\rm{2}}}\)(circular sampling region).

This means that –

\({\rm{\mu = \alpha }} \cdot {\rm{t = 2}} \cdot \pi {{\rm{R}}^{\rm{2}}}\)

Therefore, it is obtained –

\(\begin{array}{c}{\rm{P(X}} \ge {\rm{1) = 1 - P(X = 0)}}\\{\rm{ = 1 - }}{{\rm{P}}_{\rm{0}}}\left( {\pi {{\rm{R}}^{\rm{2}}}} \right)\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - 2}} \cdot \pi {{\rm{R}}^{\rm{2}}}}} \cdot \frac{{{{\left( {{\rm{2}} \cdot \pi {{\rm{R}}^{\rm{2}}}} \right)}^{\rm{0}}}}}{{{\rm{0!}}}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - 2}} \cdot \pi {{\rm{R}}^{\rm{2}}}}}\end{array}\)

From the relation –

\({\rm{P(X}} \ge {\rm{1) = 0}}{\rm{.99}}\)

It is obtained that –

\(\begin{array}{c}{\rm{1 - }}{{\rm{e}}^{{\rm{ - 2}} \cdot \pi {{\rm{R}}^{\rm{2}}}}}{\rm{ = 0}}{\rm{.99}}\\{{\rm{e}}^{{\rm{ - 2}} \cdot \pi {{\rm{R}}^{\rm{2}}}}}{\rm{ = 0}}{\rm{.01}}\\{\rm{ - 2}} \cdot \pi {{\rm{R}}^{\rm{2}}}{\rm{ = ln0}}{\rm{.01}}\\{{\rm{R}}^{\rm{2}}}{\rm{ = - }}\frac{{{\rm{ln0}}{\rm{.01}}}}{{{\rm{2\pi }}}}\\{\rm{R = 0}}{\rm{.85612}}\end{array}\)

Therefore, the value is obtained as \({\rm{R = 0}}{\rm{.85612}}\).

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Most popular questions from this chapter

Forty percent of seeds from maize (modern-day corn) ears carry single spikelets, and the other \({\rm{60\% }}\) carry paired spikelets. A seed with single spikelets will produce an ear with single spikelets \({\rm{29\% }}\) of the time, whereas a seed with paired spikelets will produce an ear with single spikelets \({\rm{26\% }}\) of the time. Consider randomly selecting ten seeds.

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