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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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A manufacturer of integrated circuit chips wishes to control the quality of its product by rejecting any batch in which the proportion of defective chips is too high. To this end, out of each batch (10,000 chips), 25 will be selected and tested. If at least 5 of these 25 are defective, the entire batch will be rejected.

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