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Chapter 3: Q87E (page 136)

The number of requests for assistance received by a towing service is a Poisson process with rate \({\rm{\alpha = 4}}\) per hour. a. Compute the probability that exactly ten requests are received during a particular \({\rm{2}}\)-hour period. b. If the operators of the towing service take a \({\rm{30}}\)-min break for lunch, what is the probability that they do not miss any calls for assistance? c. How many calls would you expect during their break?

Short Answer

Expert verified

(a) The probability is obtained as:\({\rm{0}}{\rm{.099}}\).

(b) The probability is obtained as: \({\rm{0}}{\rm{.135}}\).

(c) The calls expected during the break is: \({\rm{2}}\).

Step by step solution

01

Define Discrete random variables

A discrete random variable is one that can only take on a finite number of different values.

02

Step 2:Evaluating the probability

(a) The parameter that is valid for a two-hour period is:

\(\begin{array}{c}{\rm{\mu = 4 \times 2}}\\{\rm{ = 8}}\end{array}\)

Let X represent the number of requests received in the last two hours.

\(\begin{array}{c}{\rm{P(X = 10) = }}\frac{{{\rm{e}}{}^{{\rm{ - 8}}}{\rm{ \times }}{{\rm{8}}^{{\rm{10}}}}}}{{{\rm{10!}}}}\\{\rm{ = 0}}{\rm{.099}}\end{array}\)

Therefore, the value is: \({\rm{0}}{\rm{.099}}\).

03

Step 3:Evaluating the probability

(b) Fora period of \({\rm{30}}\) minutes:

\(\begin{array}{c}{\rm{\mu = }}\frac{{\rm{4}}}{{\rm{2}}}\\{\rm{ = 2}}\end{array}\)

Let Y equal the number of requests received in the last\({\rm{30}}\)minutes.

\(\begin{array}{c}{\rm{P(Y = 0) = }}\frac{{{\rm{e}}{}^{{\rm{ - 2}}}{\rm{ \times }}{{\rm{2}}^{\rm{0}}}}}{{{\rm{0!}}}}\\{\rm{ = 0}}{\rm{.135}}\end{array}\)

Therefore, the value is: \({\rm{0}}{\rm{.135}}\).

04

Step 4:Evaluating the calls during the break

(c) The expected value is \({\rm{\mu }}\).

It is for\({\rm{30}}\)minutes.

The period then will be\({\rm{2}}\).

Therefore, the calls expected will be:\({\rm{2}}\).

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