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Chapter 8: Problem 15

Which of the following functions makes the most sense as a model for the probability density representing the time (in minutes, starting from \(t=0\) ) that the next customer walks into a store? (a) \(p(t)=\left\\{\begin{array}{ll}\cos t & 0 \leq t \leq 2 \pi \\ e^{t-2 \pi} & t \geq 2 \pi\end{array}\right.\) (b) \(p(t)=3 e^{-3 t}\) for \(t \geq 0\) (c) \(p(t)=e^{-3 t}\) for \(t \geq 0\) (d) \(p(t)=1 / 4\) for \(0 \leq t \leq 4\)

Short Answer

Expert verified
Option (b) is the most suitable model for the probability density function.

Step by step solution

01

Understanding the Problem

We are tasked with identifying which function serves as a probability density function (pdf) for modeling the time a customer takes to enter a store, starting from time \( t = 0 \). A pdf must satisfy two conditions: it must be non-negative, and its integral over its domain must equal 1.
02

Evaluate Option (a): Trigonometric and Exponential Combination

Option (a) consists of a piecewise function with \( \cos t \) over \( [0, 2\pi] \) and an exponential function for \( t \geq 2\pi \). The function \( \cos t \) is not always non-negative over \( [0, 2\pi] \); thus, it cannot be a valid probability density function.
03

Evaluate Option (b): Exponential Function

Option (b) presents the function \( p(t) = 3e^{-3t} \) for \( t \geq 0 \). This function is non-negative and exponential functions are typically used for modeling waiting times. We check the integral from 0 to infinity: \[\int_{0}^{\infty} 3e^{-3t} \, dt = \left[-e^{-3t}\right]_{0}^{\infty} = 1\] which confirms it is a valid pdf.
04

Evaluate Option (c): Exponential Function Without Scaling

Option (c) gives \( p(t) = e^{-3t} \) for \( t \geq 0 \). The function is non-negative, but when checking the integral: \[\int_{0}^{\infty} e^{-3t} \, dt = \left[-\frac{1}{3}e^{-3t}\right]_{0}^{\infty} = \frac{1}{3}\] which is not equal to 1; hence, it cannot be a valid pdf.
05

Evaluate Option (d): Uniform Distribution

Option (d) describes a uniform distribution \( p(t) = \frac{1}{4} \) for \( 0 \leq t \leq 4 \). This is non-negative, and its integral is: \[\int_{0}^{4} \frac{1}{4} \, dt = \left[\frac{1}{4}t\right]_{0}^{4} = 1\] verifying it as a valid pdf over this limited domain.
06

Selecting the Most Appropriate Model

Considering that the exponential distribution is commonly used to model the time between events in a Poisson process, option (b) \( p(t) = 3e^{-3t} \) is the most suitable model for the probability density of the time when the next customer walks into the store, starting from \( t = 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Distribution
The exponential distribution is often employed to model the time between events in a continuous-time setting. One of its key applications is in scenarios like predicting the time until the next customer arrives at a store. It is memoryless, meaning the probability of an event occurring in the next moment is independent of past events.
  • The probability density function (pdf) of an exponential distribution is typically given as \( p(t) = \lambda e^{-\lambda t} \), where \( \lambda \) is the rate parameter.
  • An important property is that the integral from 0 to infinity of its pdf equals 1, ensuring it's a valid pdf.
It's essential that the rate parameter \( \lambda \) is positive. The larger \( \lambda \), the faster the expected event happens. In the given exercise, option (b) \( 3e^{-3t} \) was identified as exponential, with a rate \( \lambda = 3 \), correctly serving as a valid pdf.
Poisson Process
A Poisson process is a stochastic process used to predict the occurrence of events over time. It finds applications in areas like queueing theory, telecommunications, and traffic flow analysis. When events occur independently and randomly over a continuum, this process is a good modeling tool.
  • It comprises an exponential distribution that models the time interval between consecutive events.
  • The gap or interval between events is memoryless, complementing the properties of the exponential distribution.
Each event follows a Poisson distribution, and these processes can be either homogeneous (constant rate \( \lambda \)) or non-homogeneous (varying rate). By utilizing an exponential distribution like \( 3e^{-3t} \), we are leveraging a Poisson process nature to determine the arrival time of customers.
Uniform Distribution
The uniform distribution, in its simplest form, suggests that every interval of a certain length has the same probability of containing an event. It is best used in situations where every outcome within a given range is equally likely. This distribution has a constant pdf value across the domain.
  • The probability density function of the uniform distribution over an interval \([a, b]\) is \( p(t) = \frac{1}{b-a} \) for \( a \leq t \leq b \).
  • The property of having a constant pdf ensures that the integral over its domain equals 1.
In the exercise provided, option (d) \( p(t) = \frac{1}{4} \) for \( 0 \leq t \leq 4 \) meets these criteria, confirming it as a valid uniform pdf. However, uniform distribution may not always be ideal for scenarios where the memoryless property of event timing is critical, such as in modeling customer arrivals.

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Most popular questions from this chapter

Suppose \(F(x)\) is the cumulative distribution function for heights (in meters) of trees in a forest. (a) Explain in terms of trees the meaning of the statement \(F(7)=0.6\) (b) Which is greater, \(F(6)\) or \(F(7)\) ? Justify your answer in terms of trees.

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Write an integral that represents the arc length of the portion of the graph of \(f(x)=-x(x-4)\) that lies above the \(x\) -axis. Do not evaluate the integral.

If we think of an electron as a particle, the function $$P(r)=1-\left(2 r^{2}+2 r+1\right) e^{-2 r}$$ is the cumulative distribution function of the distance, \(r\) of the electron in a hydrogen atom from the center of the atom. The distance is measured in Bohr radii. (1 Bohr radius \(=5.29 \times 10^{-11} \mathrm{m} .\) Niels Bohr \((1885-1962)\) was a Danish physicist.) For example, \(P(1)=1-5 e^{-2} \approx 0.32\) means that the electron is within 1 Bohr radius from the center of the atom \(32 \%\) of the time. (a) Find a formula for the density function of this distribution. Sketch the density function and the cumulative distribution function. (b) Find the median distance and the mean distance. Near what value of \(r\) is an electron most likely to be found? (c) The Bohr radius is sometimes called the "radius of the hydrogen atom." Why?

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