/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A small market orders copies of ... [FREE SOLUTION] | 91Ó°ÊÓ

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A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf \begin{tabular}{c|cccccc} \(x\) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(\frac{1}{15}\) & \(\frac{2}{15}\) & \(\frac{3}{15}\) & \(\frac{4}{15}\) & \(\frac{3}{15}\) & \(\frac{2}{15}\) \end{tabular} Suppose the store owner actually pays \(\$ 2.00\) for each copy of the magazine and the price to customers is \(\$ 4.00\). If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand \(X\), and then compute the expected revenue.]

Short Answer

Expert verified
Order 4 copies; it increases expected revenue to approximately 5.33 dollars.

Step by step solution

01

Define Net Revenue Function

When ordering 3 copies, the net revenue function can be described as \( R_3(X) \).- If \( X \leq 3 \), then all demand can be met, and revenue equals \( 4X - 2 \times 3 = 4X - 6 \) dollars.- If \( X > 3 \), then revenue equals \( 4 \times 3 - 2 \times 3 = 12 - 6 = 6 \) dollars, since only 3 copies are sold.When ordering 4 copies, the net revenue function can be described as \( R_4(X) \).- If \( X \leq 4 \), then revenue equals \( 4X - 2 \times 4 = 4X - 8 \) dollars.- If \( X > 4 \), then revenue equals \( 4 \times 4 - 2 \times 4 = 16 - 8 = 8 \) dollars.
02

Compute Expected Revenue for 3 Copies

First, calculate \( E[R_3(X)] \):- \( R_3(1) = 4 \times 1 - 6 = -2 \), with probability \( \frac{1}{15} \).- \( R_3(2) = 4 \times 2 - 6 = 2 \), with probability \( \frac{2}{15} \).- \( R_3(3) = 4 \times 3 - 6 = 6 \), with probability \( \frac{3}{15} \).- \( R_3(4) = 6 \), with probability \( \frac{4}{15} \).- \( R_3(5) = 6 \), with probability \( \frac{3}{15} \).- \( R_3(6) = 6 \), with probability \( \frac{2}{15} \). Calculate:\[ E[R_3(X)] = \sum (R_3(x) \times p(x)) = -2 \times \frac{1}{15} + 2 \times \frac{2}{15} + 6 \left(\frac{3}{15} + \frac{4}{15} + \frac{3}{15} + \frac{2}{15}\right) = \frac{60}{15} = 4.0 \]
03

Compute Expected Revenue for 4 Copies

Now calculate \( E[R_4(X)] \):- \( R_4(1) = 4 \times 1 - 8 = -4 \), with probability \( \frac{1}{15} \).- \( R_4(2) = 4 \times 2 - 8 = 0 \), with probability \( \frac{2}{15} \).- \( R_4(3) = 4 \times 3 - 8 = 4 \), with probability \( \frac{3}{15} \).- \( R_4(4) = 4 \times 4 - 8 = 8 \), with probability \( \frac{4}{15} \).- \( R_4(5) = 8 \), with probability \( \frac{3}{15} \).- \( R_4(6) = 8 \), with probability \( \frac{2}{15} \). Calculate:\[ E[R_4(X)] = \sum (R_4(x) \times p(x)) = -4 \times \frac{1}{15} + 0 \times \frac{2}{15} + 4 \times \frac{3}{15} + 8 \left(\frac{4}{15} + \frac{3}{15} + \frac{2}{15}\right) = \frac{80}{15} \approx 5.33 \]
04

Compare Expected Revenues

Compare the expected revenues found:- Expected revenue from ordering 3 copies: 4.0 dollars.- Expected revenue from ordering 4 copies: approximately 5.33 dollars.Since \( 5.33 > 4.0 \), the owner earns more if 4 copies are ordered.

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Key Concepts

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

Probability Mass Function
A Probability Mass Function (PMF) is used to describe the distribution of a discrete random variable over its possible values. For our exercise, the demand for magazines each week is a discrete random variable, represented by \(X\). The PMF given helps us understand the likelihood (or probability) of each possible demand value occurring.

In this case, \(X\) can take values from 1 to 6, each with its own probability indicated in the table:
  • Demand of 1 magazine: Probability = \(\frac{1}{15}\)
  • Demand of 2 magazines: Probability = \(\frac{2}{15}\)
  • Demand of 3 magazines: Probability = \(\frac{3}{15}\)
  • Demand of 4 magazines: Probability = \(\frac{4}{15}\)
  • Demand of 5 magazines: Probability = \(\frac{3}{15}\)
  • Demand of 6 magazines: Probability = \(\frac{2}{15}\)
The sum of these probabilities equals 1, confirming that they are a complete distribution of potential outcomes. PMFs are essential in probability because they provide the precise probabilities needed to calculate expected values in decision making.
Statistical Decision Making
Statistical Decision Making involves analyzing probabilities and expected outcomes to make the best choice under uncertainty. In this exercise, the store owner uses statistical methods to decide how many magazines to order each week.

The choice to order either 3 or 4 magazines is crucial because it affects the net revenue, depending on what the actual demand \(X\) turns out to be. By considering each possible demand along with its probability, the owner can compute expected revenues for ordering different quantities:
  • If all demanded magazines are sold, net revenue increases.
  • For unsold magazines, there’s no gain, which underscores the need for careful decision making.
Ultimately, this decision-making process involves analyzing the expected revenue for each scenario and choosing the option that maximizes profitability based on probable outcomes. This smart decision-making strategy ensures the store owner makes choices aligned with potential profits.
Net Revenue Calculation
Net Revenue Calculation is a critical component in determining profitability from selling magazines. Net revenue is defined as the difference between total sales revenue and the cost incurred for purchasing the magazines.

For this exercise:
  • Each magazine costs the owner \\(2 and sells for \\)4.
  • When 3 magazines are ordered: Different revenue calculations for each demand scenario range from subtracting costs of unsold magazines if demand is less or matching total sales if equal demand exists.
  • When 4 magazines are ordered: Similar calculations apply, varying based on demand but ensuring maximum sales when demand matches or exceeds order quantity.
To find the Expected Revenue, we look at the possible revenues under different demand scenarios and weigh them according to their probabilities, which are derived from the PMF. The overall expectation helps choose the strategy (ordering 4 copies) that results in the highest average revenue over time. This calculation is crucial for making informed decisions that enhance business profitability.

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Most popular questions from this chapter

An article in the Los Angeles Times (Dec. 3 , 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.

Two fair six-sided dice are tossed independently. Let \(M=\) the maximum of the two tosses [thus \(M(1,5)=5, M(3,3)=3\), etc.]. a. What is the pmf of \(M\) ? [Hint: First determine \(p(1)\), then \(p(2)\), and so on.] b. Determine the cdf of \(M\) and graph it.

Let \(X\) have a Poisson distribution with parameter 2. Show that \(E(X)=\lambda\) directly from the definition of expected value. [Hint: The first term in the sum equals 0 , and then \(x\) can be canceled. Now factor out \(\lambda\) and show that what is left sums to 1.]

Suppose that only \(25 \%\) of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible. What is the probability that, of 20 randomly chosen drivers coming to an intersection under these conditions, a. At most 6 will come to a complete stop? b. Exactly 6 will come to a complete stop? c. At least 6 will come to a complete stop? d. How many of the next 20 drivers do you expect to come to a complete stop?

a. Use derivatives of the moment generating function to obtain the mean and variance for the Poisson distribution. b. As discussed in Section 3.4, obtain the Poisson mean and variance from \(R_{X}(t)=\ln\) \(\left[M_{X}(t)\right]\). In terms of effort, how does this method compare with the one in part (a)?

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