/*! 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 106 A heavy-equipment salesman can c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 106

A heavy-equipment salesman can contact either one or two customers per day with probabilities \(1 / 3\) and \(2 / 3,\) respectively. Each contact will result in either no sale or a \(\$ 50,000\) sale with probabilities \(9 / 10\) and \(1 / 10,\) respectively. What is the expected value of his daily sales?

Short Answer

Expert verified
Answer: The expected value of daily sales is approximately $10,667.

Step by step solution

01

Identify possible outcomes

First, we will identify the possible outcomes and their corresponding probabilities. There are three possible outcomes: no sale (\(0), one sale (\)50,000), and two sales ($100,000).
02

Calculate probabilities of each outcome

Next, we need to calculate the probability of each outcome. We will do this using the probabilities provided for contacting customers and making sales. Probability of no sale: $$\frac{1}{3}\times \frac{9}{10} + \frac{2}{3}\times \frac{9}{10}\times \frac{9}{10} = \frac{9}{30} + \frac{162}{270} = \frac{9}{10}$$ Probability of one sale: $$\frac{1}{3}\times \frac{1}{10} + 2\times \frac{2}{3}\times \frac{1}{10}\times \frac{9}{10} = \frac{1}{30} + 2\times \frac{18}{270} = \frac{6}{30} = \frac{1}{5}$$ Probability of two sales: $$\frac{2}{3}\times \frac{1}{10}\times \frac{1}{10} = \frac{2}{300} = \frac{1}{150}$$
03

Calculate the expected value

Now that we have the probabilities for each outcome, we can calculate the expected value of daily sales by multiplying each outcome by its respective probability and summing the results: Expected value = $$0\times \frac{9}{10} + 50{,}000\times \frac{1}{5} + 100{,}000\times \frac{1}{150}$$ Expected value = $$0 + 10{,}000 + \frac{100{,}000}{150}$$ Expected value = $$10{,}000 + \frac{2{,}000}{3} = 10{,}000 + 666.67 \approx 10{,}667$$ Therefore, the expected value of the heavy-equipment salesman's daily sales is approximately $10,667.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different potential outcomes. In this exercise, the probability distribution describes the likelihood of the salesman making different numbers of sales each day.

There are certain key points about probability distributions:
  • They help in understanding how probabilities are spread across different outcomes.
  • Every outcome must have a probability between 0 and 1.
  • The sum of all probabilities for all possible outcomes must equal 1.
The probability distribution here is based on contacts and sales:
  • Contact 1 or 2 customers per day.
  • Make a sale or not with each customer contacted.
For this salesman, the probabilities of contacting customers and making a sale create the framework for calculating the expected sales outcomes.
Sales Outcome
Sales outcomes in this context refer to the possible results of the salesman's efforts in a single day: making no sales, one sale, or two sales. Each outcome is associated with a monetary value, either $0, $50,000, or $100,000.

Calculated probabilities for these outcomes are determined by considering combinations of contacts and sales:
  • No sale: Happens when all contacts result in no sales. This has a high probability because each sale has a small 1/10 chance.
  • One sale: Occur when exactly one contact results in a sale, while others do not. This combines probabilities of one contact being successful and others failing.
  • Two sales: The least likely, as this requires both contacts to be successful, a rarity given the individual low likelihood of a sale.
Understanding sales outcomes helps in predicting performance and planning strategies to improve success rates.
Heavy-Equipment Sales
Heavy-equipment sales are significant because each successful sale nets a large amount of revenue. As shown in this problem, selling a piece of heavy equipment yields $50,000 per unit sold.

This class of sales has unique characteristics:
  • High stakes: Each sale involves large sums of money.
  • Low frequency: Given the high individual cost, the probability of any given contact resulting in a sale is lower.
  • Great impact on earnings: Despite low frequency, successful sales significantly impact daily earnings due to their high value.
Calculating expected value can help sales leaders make informed decisions about where to allocate resources and how to focus sales efforts. For this salesman, determining the expected value of $10,667 provides insight into average daily earnings given typical situation probabilities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A French restaurant in Riverside, California, offers a special summer menu in which, for a fixed dinner cost, you can choose from one of two salads, one of two entrees, and one of two desserts. How many different dinners are available?

Your family vacation involves a cross-country air flight, a rental car, and a hotel stay in Boston. If you can choose from four major air carriers, five car rental agencies, and three major hotel chains, how many options are available for your vacation accommodations?

Choosing a Mate Men and women often disagree on how they think about selecting a mate. Suppose that a poll of 1000 individuals in their twenties gave the following responses to the question of whether it is more important for their future mate to be able to communicate their feelings ( \(F\) ) than it is for that person to make a good living \((G)\). $$\begin{array}{lccc} & \text { Feelings }(F) & \text { Good Living }(G) & \text { Totals } \\\\\hline \text { Men }(M) & .35 & .20 & .55 \\\\\text { Women }(W) & .36 & .09 & .45 \\\\\hline \text { Totals } & 71 & 29 & 1.00\end{array}$$ If an individual is selected at random from this group of 1000 individuals, calculate the following probabilities: a. \(P(F)\) b. \(P(G)\) c. \(P(F \mid M)\) d. \(P(F \mid W)\) e. \(P(M \mid F)\) f. \(P(W \mid G)\)

The maximum patent life for a new drug is 17 years. Subtracting the length of time required by the FDA for testing and approval of the drug provides the actual patent life of the drug \(-\) that is, the length of time that a company has to recover research and development costs and make a profit. Suppose the distribution of the lengths of patent life for new drugs is as shown here: $$\begin{array}{l|cccccc}\text { Years, } x & 3 & 4 & 5 & 6 & 7 & 8 \\\\\hline p(x) & .03 & .05 & .07 & .10 & .14 & .20 \\\\\text { Years, } x & 9 & 10 & 11 & 12 & 13 & \\\\\hline p(x) & .18 & .12 & .07 & .03 & .01 &\end{array}$$ a. Find the expected number of years of patent life for a new drug. b. Find the standard deviation of \(x\). c. Find the probability that \(x\) falls into the interval \(\mu \pm 2 \sigma\).

An investor has the option of investing in three of five recommended stocks. Unknown to her, only two will show a substantial profit within the next 5 years. If she selects the three stocks at random (giving every combination of three stocks an equal chance of selection), what is the probability that she selects the two profitable stocks? What is the probability that she selects only one of the two profitable stocks?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.