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Chapter 3: Q110E (page 138)

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 鈥�

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

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

02

Radius of the region

It is given that 鈥�

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

This means that 鈥�

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

Therefore, it is obtained 鈥�

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

From the relation 鈥�

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

It is obtained that 鈥�

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

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

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Most popular questions from this chapter

a. For fixed n, are there values of p(\({\rm{0}} \le {\rm{p}} \le {\rm{1}}\)) for which V(X) \({\rm{ = 0}}\)? Explain why this is so? b. For what value of p is V(X) maximized? (Hint: Either graph V(X) as a function of p or else take a derivative.)

a. Show that b(x; n,\({\rm{1 - }}\)p) = b(n - x; n, p). b. Show that B(x; n,\({\rm{1 - }}\)p) =\({\rm{1 - }}\)B(n - x -\({\rm{1}}\); n, p). (Hint: At most x S鈥檚 is equivalent to at least (n - x) F鈥檚.) c. What do parts (a) and (b) imply about the necessity of including values of p greater than\({\rm{.5}}\)in Appendix Table A\({\rm{.1}}\)?

Consider a communication source that transmits packets containing digitized speech. After each transmission, the receiver sends a message indicating whether the transmission was successful or unsuccessful. If a transmission is unsuccessful, the packet is re-sent. Suppose a voice packet can be transmitted a maximum of \({\rm{10}}\) times. Assuming that the results of successive transmissions are independent of one another and that the probability of any particular transmission being successful is \({\rm{p}}\), determine the probability mass function of the rv \({\rm{X = }}\)the number of times a packet is transmitted. Then obtain an expression for the expected number of times a packet is transmitted.

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls involve fax messages, and consider a sample of 25 incoming calls. What is the probability that

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Individuals\({\rm{A}}\)and\({\rm{B}}\)begin to play a sequence of chess games. Let\({\rm{S = }}\){\({\rm{A}}\)wins a game}, and suppose that outcomes of successive games are independent with\({\rm{P(S) = p}}\)and\({\rm{P(F) = 1 - p}}\)(they never draw). They will play until one of them wins ten games. Let\({\rm{X = }}\)the number of games played (with possible values\({\rm{10, 11, }}...{\rm{, 19}}\)).

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