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Chapter 3: Problem 126

Refer to Exercise \(3.122 .\) Assume that arrivals occur according to a Poisson process with an average of seven per hour. What is the probability that exactly two customers arrive in the twohour period of time between a. 2: 00 P.M. and 4: 00 P.M. (one continuous two-hour period)? b. 1: 00 P.M. and 2: 00 p.M. or between 3: 00 p.M. and 4: 00 P.M. (two separate one-hour periods that total two hours)?

Short Answer

Expert verified
a. The probability is approximately \( 8.1 \times 10^{-5} \). b. The probability is the sum of all valid combinations calculated in part b.

Step by step solution

01

Identify the Problem Type

This is a problem involving a Poisson process, which is used to model random events that occur independently at a constant average rate. In this case, the events are customer arrivals, and the average rate is 7 per hour.
02

Define Parameters for Part a

For part (a), we consider a continuous two-hour period. The rate of arrival \( \lambda \) per hour is 7, so for two hours, \( \lambda = 7 \times 2 = 14 \). We want to find the probability of exactly two arrivals, \( X = 2 \).
03

Use the Poisson Probability Formula for Part a

The Poisson probability formula is given by \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \). Substituting \( \lambda = 14 \) and \( k = 2 \), the probability is \( P(X=2) = \frac{e^{-14} \cdot 14^2}{2!} \).
04

Calculate Probability for Part a

Substitute the known values into the formula: \( e^{-14} \approx 8.315 \times 10^{-7} \), and \( 14^2 = 196 \). So, \( \frac{8.315 \times 10^{-7} \cdot 196}{2} \approx 8.1 \times 10^{-5} \). Thus the probability is approximately \( 8.1 \times 10^{-5} \).
05

Identify Separate Periods for Part b

For part (b), we consider two separate one-hour periods: 1:00-2:00 PM and 3:00-4:00 PM. Since arrivals in these periods are independent, we use the rate for each one-hour period by itself, \( \lambda = 7 \).
06

Find Probability for Each Period in Part b

Use the Poisson probability formula for each one-hour period: \( P(X=0) = \frac{e^{-7} \cdot 7^0}{0!} = e^{-7} \approx 0.000911 \) and \( P(X=1) = \frac{e^{-7} \cdot 7^1}{1!} \approx 0.006705 \). Use the same formula for no arrivals needed when you're targeting at least two different outcomes.
07

Combine Probabilities for Part b

For exactly two arrivals across two periods, we need either one arrival in each period or two in one and none in the other. Combining possibilities, \( P(1~in~1~hour~and~1~in~next~hour) + P(2~in~one~hour~and~0~in~next~hour) + P(0~in~one~hour~and~2~in~next~hour) \). Calculations for each combination are as follows:- \( 2P(1,1) = (0.006705 \times 0.006705 \times 2) \approx 9.0 \times 10^{-5} \).- \( P(2,0) = 0.0902 \times 0.000911 \) and \( P(0,2) = 0.000911 \times 0.0902 \) which is solved similiraly to the prior calculations when rate of arrival sum up to mean square probabilistic contribution.

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Key Concepts

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

Probability Calculation
Probability calculation is a key component in understanding the Poisson process. In this context, we work with problems where random events, like customer arrivals, occur at a known constant rate. The goal is to determine the likelihood of a specific number of events happening within a given time frame. To do this, we use the Poisson distribution formula:
  • The formula is given by \( P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where \( \lambda \) is the average number of events (arrivals) you expect to occur.
  • The parameter \( k \) represents the exact number of arrivals for which the probability is calculated, such as 2 arrivals in our example.
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
The formula essentially weighs the likelihood of different outcomes based on how often events are known to occur on average. When we compute this for different scenarios, it helps in predicting and modeling the behavior of random processes.
Random Events
Random events are at the heart of many statistical processes, including the Poisson process. These are events that happen independently of each other, with no predictable pattern beyond a maintenance of a fixed rate over time. For example:
  • If customer arrivals at a store follow a Poisson process, each arrival is independent of the previous one.
  • The probability of an arrival happening at any moment remains constant over time.
  • In our problem, arrivals could happen at any time between two fixed periods, but they are statistically independent.
This independence is essential for applying Poisson processes to model real-world problems because it helps simplify complex random occurrences into a single manageable pattern. Thus, even though events are random, their statistical properties like the rate \( \lambda \) make them predictable in a certain sense.
Statistical Modelling
Statistical modeling involves creating mathematical representations of real-world processes. The Poisson process is one such model used to describe and predict how random events occur over time. It is particularly useful for scenarios involving:
  • Predicting the number of occurrences of an event within a specific time period. For instance, how many cars pass a checkpoint.
  • Understanding variance around an average rate, helping businesses and researchers manage expectations about event frequency.
Through models like Poisson, we can compute probabilities not just for the occurrence of events but for any combination of occurrences. In the textbook problem, this allows us to calculate probabilities for different configurations of arrivals over different periods. This power of anticipation makes statistical modeling invaluable across fields from telecommunications to finance.

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Most popular questions from this chapter

Refer to Exercises \(3.142\) and \(3.143 .\) If the number of phone calls to the fire department, \(Y\), in a day has a Poisson distribution with mean \(6,\) show that \(p(5)=p(6)\) so that 5 and 6 are the two most likely values for \(Y\)

Let \(Y\) have a hypergeometric distribution $$p(y)=\frac{\left(\begin{array}{l} r \\ y \end{array}\right)\left(\begin{array}{l} N-r \\ n-y \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)} \quad y=0,1,2, \ldots, n$$ a. Show that $$P(Y=n)=p(n)=\left(\frac{r}{N}\right)\left(\frac{r-1}{N-1}\right)\left(\frac{r-2}{N-2}\right) \cdots\left(\frac{r-n+1}{N-n+1}\right)$$ b. Write \(p(y)\) as \(p(y | r) .\) Show that if \(r_{1}\frac{p\left(y+1 | r_{1}\right)}{p\left(y+1 | r_{2}\right)}$$ c. Apply the binomial expansion to each factor in the following equation: $$(1+a)^{N_{1}}(1+a)^{N_{2}}=(1+a)^{N_{1}+N_{2}}$$ Now compare the coefficients of \(a^{n}\) on both sides to prove that $$\left(\begin{array}{c} N_{1} \\ 0 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n \end{array}\right)+\left(\begin{array}{c} N_{1} \\ 1 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n-1 \end{array}\right)+\cdots+\left(\begin{array}{c} N_{1} \\ n \end{array}\right)\left(\begin{array}{c} N_{2} \\ 0 \end{array}\right)=\left(\begin{array}{c} N_{1}+N_{2} \\ n \end{array}\right)$$ d. Using the result of part ( \(\underline{\mathrm{c}}\) ), conclude that $$\sum_{y=0}^{n} p(y)=1$$

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