/*! 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 29 The distributor of a machine for... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 29

The distributor of a machine for cytogenics has developed a new model. The company estimates that when it is introduced into the market, it will be very successful with a probability \(0.6,\) moderately successful with a probability \(0.3,\) and not successful with probability \(0.1 .\) The estimated yearly profit associated with the model being very successful is \(\$ 15\) million and with it being moderately successful is \(\$ 5\) million; not successful would result in a loss of \(\$ 500,000\). Let \(X\) be the yearly profit of the new model. Determine the probability mass function of \(X\).

Short Answer

Expert verified
The PMF of \(X\) is: \(P(X = 15) = 0.6\), \(P(X = 5) = 0.3\), \(P(X = -0.5) = 0.1\).

Step by step solution

01

Define Random Variable and Events

Let the random variable \(X\) represent the yearly profit associated with the new machine model. There are three events: \(X = 15\) million (very successful), \(X = 5\) million (moderately successful), and \(X = -0.5\) million (not successful).
02

List the Probabilities

The probabilities for each event are given by the company: \(P(X = 15) = 0.6\), \(P(X = 5) = 0.3\), and \(P(X = -0.5) = 0.1\). These probabilities reflect the likelihood of the respective profits.
03

Express the Probability Mass Function (PMF)

The probability mass function (PMF) represents the probabilities for each possible value of \( X \). We have: \[ \begin{cases}P(X = 15) = 0.6, \P(X = 5) = 0.3, \P(X = -0.5) = 0.1.\end{cases} \]

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.

Random Variable
In probability and statistics, a random variable is a fundamental concept. It is essentially a variable whose value is subject to variations due to random occurrences. In simpler terms, a random variable is a way to map outcomes of a random process to numerical values.

In the exercise, the random variable is denoted as \(X\) and represents the yearly profit from a new machine model. Here, \(X\) can take on three possible values, each corresponding to a certain level of success. These values are:\
  • \(X = 15\) million, indicating a very successful outcome,
  • \(X = 5\) million, indicating a moderately successful outcome,
  • \(X = -0.5\) million, indicating that the model was not successful.
Random variables help in modeling real-world situations in a structured format, allowing us to analyze the outcomes probabilistically.
Events and Probabilities
Understanding events and probabilities is crucial in probability theory. An event is any set of outcomes of a random process. Probabilities are numerical values assigned to these events, showing their likelihood.

In this context, the events are the different levels of success the machine model may achieve. We have three distinct events with their probabilities provided:
  • \(P(X = 15) = 0.6\) - This implies there is a 60% chance of the model being very successful.
  • \(P(X = 5) = 0.3\) - There is a 30% chance of a moderate success.
  • \(P(X = -0.5) = 0.1\) - This represents a 10% likelihood of failure.
These probabilities are integral as they provide a forecast on the model's potential performance, helping stakeholders make informed decisions.
Yearly Profit Analysis
Carrying out a yearly profit analysis using a probability mass function allows companies to anticipate potential financial outcomes in various scenarios.

The provided data enables marketers and analysts to calculate expected profits using the probability mass function (PMF). PMF is a function that gives the probability that a discrete random variable is exactly equal to some value. In our example:
\[ P(X = 15) = 0.6, \quad P(X = 5) = 0.3, \quad P(X = -0.5) = 0.1 \]

By analyzing this, companies can evaluate risks and returns and strategize accordingly. It helps in-budget planning, risk management, and forecasting, allowing companies to navigate uncertainties more effectively.

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 company performs inspection on shipments from suppliers to detect nonconforming products. The company's policy is to use a sample size that is always \(10 \%\) of the lot size. Comment on the effectiveness of this policy as a general rule for all sizes of lots.

Let \(X\) denote the number of bits received in error in a digital communication channel, and assume that \(X\) is a binomial random variable with \(p=0.001\). If 1000 bits are transmitted, determine the following: (a) \(P(X=1)\) (b) \(P(X \geq 1)\) (c) \(P(X \leq 2)\) (d) mean and variance of \(X\)

A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day for sharpness. If any dull blade is found, the assembly is replaced with a newly sharpened set of blades. (a) If 10 of the blades in an assembly are dull, what is the probability that the assembly is replaced the first day it is evaluated? (b) If 10 of the blades in an assembly are dull, what is the probability that the assembly is not replaced until the third day of evaluation? [Hint: Assume that the daily decisions are independent, and use the geometric distribution. (c) Suppose that on the first day of evaluation, 2 of the blades are dull; on the second day of evaluation, 6 are dull; and on the third day of evaluation, 10 are dull. What is the probability that the assembly is not replaced until the third day of evaluation? [Hint: Assume that the daily decisions are independent. However, the probability of replacement changes every day.]

Four identical computer servers operate a Web site. Only one is used to operate the site; the others are spares that can be activated in case the active server fails. The probability that a request to the Web site generates a failure in the active server is \(0.0001 .\) Assume that each request is an independent trial. What is the mean time until all four computers fail?

In a manufacturing process that laminates several ceramic layers, \(1 \%\) of the assemblies are defective. Assume that the assemblies is independent. (a) What is the mean number of assemblies that need to be checked to obtain five defective assemblies? (b) What is the standard deviation of the number of assemblies that need to be checked to obtain five defective assemblies? (c) Determine the minimum number of assemblies that need to be checked so that the probability that at least one defective assembly is obtained exceeds \(0.95 .\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

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.