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A store owner has overstocked a certain item and decides to use the following promotion to decrease the supply. The item has a marked price of \(100 .\) For each customer purchasing the item during a particular day, the owner will reduce the price by a factor of one-half. Thus, the first customer will pay \(50\) for the item, the second will pay \(25,\) and so on. Suppose that the number of customers who purchase the item during the day has a Poisson distribution with mean 2 . Find the expected cost of the item at the end of the day. [Hint: The cost at the end of the day is \(100(1 / 2)^{Y}\), where \(Y\) is the number of customers who have purchased the item.]

Short Answer

Expert verified
The expected cost of the item at the end of the day is approximately $36.79.

Step by step solution

01

Calculate Probability Distribution

The number of customers purchasing the item follows a Poisson distribution with parameter \( \lambda = 2 \). The probability of observing \( y \) customers is given by the Poisson probability mass function (PMF): \[ P(Y = y) = \frac{e^{-\lambda} \lambda^y}{y!} \]. For \( \lambda = 2 \), this becomes \[ P(Y = y) = \frac{e^{-2} \cdot 2^y}{y!} \].
02

Define the Cost Function

The cost of the item at the end of the day is given by the function \( C(Y) = 100 \left( \frac{1}{2} \right)^Y \), where \( Y \) is the number of customers.
03

Calculate Expected Cost

The expected value of the cost function, \( E[C(Y)] \), is calculated by summing over all possible values of \( Y \): \[ E[C(Y)] = \sum_{y=0}^{\infty} 100 \left( \frac{1}{2} \right)^y \cdot \frac{e^{-2} \cdot 2^y}{y!} \]. This simplifies further to: \[ E[C(Y)] = 100 \cdot e^{-2} \sum_{y=0}^{\infty} \frac{(1)^y}{y!} \].
04

Simplify Expected Value Sum

Recognize that the series \( \sum_{y=0}^{\infty} \frac{1^y}{y!} \) is the Maclaurin series expansion for \( e^1 \), which equals \( e \). Therefore, \[ E[C(Y)] = 100 \cdot e^{-2} \cdot e = 100 \cdot e^{-1} \].
05

Calculate Final Expected Cost

Calculate the numerical value of \( E[C(Y)] \): \[ E[C(Y)] = \frac{100}{e} \approx 36.79 \].

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Key Concepts

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

Understanding Expected Value in Poisson Distribution
The "expected value" is a fundamental concept in probability and statistics, describing the average outcome of a random event. In the context of a Poisson distribution, the expected value is particularly important to determine outcomes over a long period. The Poisson distribution is ideal for modeling the number of times an event happens in a set time. Here, with an expected number of customers (events) mean, \( \lambda = 2 \), the expected value can be applied to predict the average final cost of an item after a series of transactions.

For our exercise, we calculate the expected value of the cost of an item, which is subject to a discount calculated proportionally to the number of customers purchasing it. This formula is expressed as \( E[C(Y)] = \sum_{y=0}^{\infty} \)\( 100 \left( \frac{1}{2} \right)^y \cdot \frac{e^{-2} \cdot 2^y}{y!} \). The expected value here acts as a "weighted" average, where each potential outcome \( y \) is weighted by its probability of occurrence. This average is vital because it gives a single, understandable figure representing potential results.
  • In this problem, the expected value helps predict the most likely final price per item.
  • This insight can assist businesses in making strategy decisions regarding pricing and sales forecasts.
The Role of Probability Mass Function (PMF)
The probability mass function (PMF) in the Poisson distribution provides the probability of exactly \( y \) events happening within a fixed period given a known average number of events. For this exercise, it indicates the likelihood that a certain number of customers will purchase items on a specific day. By understanding how likely each possible outcome is, we can effectively gauge the likely changes in the item price.

The given PMF for the Poisson distribution is \( P(Y = y) = \frac{e^{-\lambda} \lambda^y}{y!} \). Here, the mean \( \lambda = 2 \), making it \( P(Y = y) = \frac{e^{-2} \cdot 2^y}{y!} \). This formula is central to calculating probabilities because it provides a discrete probability model for planning.
  • It allows the store owner to understand the frequency of customers more accurately.
  • Provides foundational data for other calculations, like cost expectations.
Defining the Cost Function for the Item
The cost function in this context is a formula that determines the final price of the item based on the number of customers who purchase it. It's a crucial representation of how customer numbers impact sales figures sequentially. Our function is defined as \( C(Y) = 100 \left( \frac{1}{2} \right)^Y \), where \( Y \) denotes the number of customers.

Understanding this function helps in visualizing how each customer's purchase affects pricing. Each new customer triggers a price cut by half, showing a pattern and decreasing the price in a predictable way. This formula effectively ties into broader business strategies facilitating:
  • Forecasting revenue impacts based on varying day sales.
  • Demonstrating the effect of a sales strategy quantitatively.
By incorporating this function, we can navigate complex business scenarios with more clarity and make more informed decisions based on logical calculations.

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Most popular questions from this chapter

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