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Chapter 5: Problem 78

A store opens at 8 A.M. From 8 until 10 customers arrive at a Poisson rate of four an hour. Between 10 and 12 they arrive at a Poisson rate of eight an hour. From 12 to 2 the arrival rate increases steadily from eight per hour at 12 to ten per hour at 2; and from 2 to 5 the arrival rate drops steadily from ten per hour at 2 to four per hour at \(5 .\) Determine the probability distribution of the number of customers that enter the store on a given day.

Short Answer

Expert verified
The probability distribution of the number of customers that enter the store on a given day is given by a Poisson distribution with 位 = 63 customers. The probability mass function (PMF) for this distribution is P(X=k) = \( \frac{e^{-63}63^{k}}{k!} \), where k is the number of customers.

Step by step solution

01

Identify the time intervals and arrival rates#create_tag_content#Given in the problem, we have four different time intervals with varying Poisson arrival rates: 1. 8 A.M to 10 A.M with 4 customers/hour 2. 10 A.M to 12 P.M with 8 customers/hour 3. 12 P.M to 2 P.M with an arrival rate increasing steadily from 8 customers/hour to 10 customers/hour 4. 2 P.M to 5 P.M with an arrival rate decreasing steadily from 10 customers/hour to 4 customers/hour

Step 2: Calculate the expected number of customers for the first two intervals#create_tag_content#For the first interval, we can multiply the rate by the duration of the interval: Expected number of customers from 8 A.M to 10 A.M = 4 customers/hour * 2 hours = 8 customers For the second interval, we can do the same: Expected number of customers from 10 A.M to 12 P.M = 8 customers/hour * 2 hours = 16 customers
02

Calculate the expected number of customers for the third interval#create_tag_content#For the third interval, we need to find the average rate and then multiply by the interval duration, since the rate varies linearly: Average rate from 12 P.M to 2 P.M = (8 customers/hour + 10 customers/hour) / 2 = 9 customers/hour Expected number of customers from 12 P.M to 2 P.M = 9 customers/hour * 2 hours = 18 customers

Step 4: Calculate the expected number of customers for the fourth interval#create_tag_content#Similarly, for the fourth interval: Average rate from 2 P.M to 5 P.M = (10 customers/hour + 4 customers/hour) / 2 = 7 customers/hour Expected number of customers from 2 P.M to 5 P.M = 7 customers/hour * 3 hours = 21 customers
03

Calculate the total expected number of customers for the entire day#create_tag_content#Adding the expected number of customers from all intervals: Total expected number of customers = 8 customers + 16 customers + 18 customers + 21 customers = 63 customers

Step 6: Determine the Poisson distribution using the total expected number of customers#create_tag_content#Now that we have the total expected number of customers for the entire day, we can define our Poisson distribution: Given 位 (the average rate) = 63 customers, the probability mass function (PMF) P(X=k) for a Poisson distribution is given by: P(X=k) = \( \frac{e^{-\lambda}\lambda^{k}}{k!} \) So the probability distribution of the number of customers that enter the store on a given day is given by the Poisson distribution with 位 = 63 customers.

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Key Concepts

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

Arrival Rate
In a Poisson distribution, the arrival rate is key to understanding how often an event occurs within a specified time frame. For this store's scenario, the arrival rate refers to how many customers enter per hour.

We break down the day into time intervals where the arrival rate varies:
  • 8 A.M to 10 A.M: 4 customers per hour
  • 10 A.M to 12 P.M: 8 customers per hour
  • 12 P.M to 2 P.M: Rate increases from 8 to 10 customers per hour
  • 2 P.M to 5 P.M: Rate decreases from 10 to 4 customers per hour
This fluctuation helps us to calculate the expected number of visits during each period. Understanding these rates is our first step in defining the Poisson distribution.
Probability Mass Function
The probability mass function (PMF) in a Poisson distribution tells us the probability of a given number of events occurring in a fixed interval. For example, it's used to find out the likelihood of having exactly a certain number of customers.

The PMF for a Poisson distribution is given by:\[P(X=k) = \frac{e^{-\lambda}\lambda^{k}}{k!}\]where:
  • \(X\) is the random variable representing the number of events (customers)
  • \(k\) is the actual number of occurrences
  • \(\lambda\) is the average rate of occurrences
With the PMF, we can compute the probability of different customer counts for any given day.
Expected Number of Customers
The expected number of customers allows us to predict how busy the store might be. It's calculated by multiplying the arrival rate by the length of each time interval.

For example:
  • 8 A.M to 10 A.M: 4 customers/hour \(\times\) 2 hours = 8 customers
  • 10 A.M to 12 P.M: 8 customers/hour \(\times\) 2 hours = 16 customers
  • 12 P.M to 2 P.M (average rate 9): 9 customers/hour \(\times\) 2 hours = 18 customers
  • 2 P.M to 5 P.M (average rate 7): 7 customers/hour \(\times\) 3 hours = 21 customers
Adding these gives a total of 63 expected customers for the day.
Time Intervals
Time intervals are periods where the arrival rate is analyzed separately. By splitting the day into intervals, we manage varying arrival rates more accurately.

The intervals for this store are as follows:
  • 8 A.M to 10 A.M
  • 10 A.M to 12 P.M
  • 12 P.M to 2 P.M
  • 2 P.M to 5 P.M
Each interval may have a constant or changing rate, requiring different methods of calculation. For linear changes in rates, averaging the start and end values offers a precise estimate.
Average Rate
The average rate smooths out variations within time intervals where arrival rates change linearly. This rate is essential for calculating expectations in non-constant situations.

For example, between 12 P.M and 2 P.M, the rate varies from 8 to 10 customers per hour. The average rate is:\[(8 + 10)/2 = 9\] customers per hour.
Similarly, for 2 P.M to 5 P.M, the rate changes from 10 to 4 customers per hour, giving an average of:\[(10 + 4)/2 = 7\] customers per hour.
These averages allow easier handling when dealing with calculations involving non-static rates.

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Most popular questions from this chapter

Suppose that people arrive at a bus stop in accordance with a Poisson process with rate \(\lambda\). The bus departs at time \(t\). Let \(X\) denote the total amount of waiting time of all those who get on the bus at time \(t\). We want to determine \(\operatorname{Var}(X)\). Let \(N(t)\) denote the number of arrivals by time \(t .\) (a) What is \(E[X \mid N(t)] ?\) (b) Argue that \(\operatorname{Var}[X \mid N(t)]=N(t) t^{2} / 12\) (c) What is \(\operatorname{Var}(X) ?\)

A two-dimensional Poisson process is a process of randomly occurring events in the plane such that (i) for any region of area \(A\) the number of events in that region has a Poisson distribution with mean \(\lambda A\) and (ii) the number of events in nonoverlapping regions are independent. For such a process, consider an arbitrary point in the plane and let \(X\) denote its distance from its nearest event (where distance is measured in the usual Euclidean manner). Show that (a) \(P\\{X>t\\}=e^{-\lambda \pi t^{2}}\) (b) \(E[X]=\frac{1}{2 \sqrt{\lambda}}\)

Machine 1 is currently working. Machine 2 will be put in use at a time \(t\) from now. If the lifetime of machine \(i\) is exponential with rate \(\lambda_{i}, i=1,2\), what is the probability that machine 1 is the first machine to fail?

Cars pass 'a certain street location according to a Poisson process with ratee A woman who wants to cross the street at that location waits until she can see that no cars will come by in the next \(T\) time units. (a) Find the probability that her waiting time is \(0 .\) (b) Find her expected waiting time. Hint: Condition on the time of the first car.

The number of hours between successive train arrivals at the station is uniformly distributed on \((0,1)\). Passengers arrive according to a Poisson process with rate 7 per hour. Suppose a train has just left the station. Let \(X\) denote the number of people who get on the next train. Find (a) \(E[X]\) (b) \(\operatorname{Var}(X)\)
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