/*! 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 139 In the daily production of a cer... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In the daily production of a certain kind of rope, the number of defects per foot \(Y\) is assumed to have a Poisson distribution with mean \(\lambda=2\). The profit per foot when the rope is sold is given by \(X,\) where \(X=50-2 Y-Y^{2} .\) Find the expected profit per foot.

Short Answer

Expert verified
The expected profit per foot is 40.

Step by step solution

01

Define the Profit Equation

The profit per foot is determined by the equation \( X = 50 - 2Y - Y^2 \), where \( Y \) is the number of defects per foot following a Poisson distribution with a mean of \( \lambda = 2 \). Our job is to find the expected value of \( X \), denoted as \( E(X) \).
02

Express the Expectation as an Integral

The expected value of \( X \) can be calculated as \( E(X) = E(50 - 2Y - Y^2) = 50 - 2E(Y) - E(Y^2) \). Since \( Y \) follows a Poisson distribution, we use its properties to determine \( E(Y) \) and \( E(Y^2) \).
03

Calculate \( E(Y) \) for Poisson Distribution

For a Poisson distribution with mean \( \lambda \), the expected value \( E(Y) \) is equal to \( \lambda \). Hence, \( E(Y) = 2 \).
04

Determine \( E(Y^2) \) Using a Known Identity

For a Poisson distribution, \( E(Y^2) = \lambda + \lambda^2 \). Substituting \( \lambda = 2 \) gives us \( E(Y^2) = 2 + 4 = 6 \).
05

Calculate Expected Profit Per Foot

Substitute \( E(Y) \) and \( E(Y^2) \) into the profit expectation formula: \[ E(X) = 50 - 2(2) - 6 = 50 - 4 - 6 = 40. \] Thus, the expected profit per foot is \( 40 \).

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 91Ó°ÊÓ!

Key Concepts

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

Expected Value
When analyzing different types of statistical data, the expected value is an essential concept to understand. It is, in simple terms, the average value you would anticipate if an experiment is repeated numerous times. For a random variable, this involves looking at all possible outcomes, each with their probability of occurrence.

To calculate the expected value for a particular situation like this rope production scenario, we look at the formula for the profit per foot, which is: \[ X = 50 - 2Y - Y^2 \] Here, \( Y \) is the number of defects, calculated as a Poisson random variable with mean \( \lambda = 2 \).

The expected value of the profit, \( E(X) \), then becomes a straightforward calculation since the properties of the Poisson distribution allow us to easily find \( E(Y) \) and \( E(Y^2) \). * \( E(Y) = \lambda \) for Poisson, thus \( E(Y) = 2 \).* \( E(Y^2) = \lambda + \lambda^2 = 6 \).Substituting these into the profit equation: \[ E(X) = 50 - 2 \times E(Y) - E(Y^2) = 40 \]. Hence, 40 is the expected profit per foot.
Profit Calculation
Profit calculation in statistical terms can sometimes be complex, but it becomes easier once we break down the components. Here, the profit per foot, \( X \), is dependent on the number of defects \( Y \) in the rope. This relationship is guided by the formula: \[ X = 50 - 2Y - Y^2 \] This equation shows how the profit changes based on the defect count \( Y \). Understanding that \( Y \) follows a Poisson distribution with a mean of \( \lambda = 2 \) allows us to determine the expected values, often the main components in profit calculations.

In the equation, * **50** represents the base profit.* **-2Y** decreases the profit linearly based on defects.* **-Y²** suggests that as defects increase, the loss increases exponentially.

Profit calculation in such scenarios is simply about inserting expected defect outcomes into the profit equation, which yields the expected profits after considering potential defects.
Statistical Distribution Properties
Statistical distributions like the Poisson distribution help us understand how variables behave in repeated random experiments. The Poisson distribution is particularly useful when dealing with counts of events that occur independently within a fixed interval.

For this exercise, with defects in rope production, the number of defects per foot \( Y \) follows the Poisson distribution with mean \( \lambda = 2 \). This defines how often you expect to find defects in the rope. Key properties of the Poisson distribution that we leverage include:
  • The mean, or average defect count, is equal to \( \lambda \). Here, it's 2.
  • The variance is also \( \lambda \) when using Poisson.
  • Properties allow us to find \( E(Y) \) and \( E(Y^2) \) directly, which simplifies calculations.
Understanding these properties helps us simplify complex equations, like profit calculations, by reducing unknowns. It's these properties that protect the integrity of the statistical models and permit accurate predictions based on expected values.

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

In southern California, a growing number of individuals pursuing teaching credentials are choosing paid internships over traditional student teaching programs. A group of eight candidates for three local teaching positions consisted of five who had enrolled in paid internships and three who enrolled in traditional student teaching programs. All eight candidates appear to be equally qualified, so three are randomly selected to fill the open positions. Let \(Y\) be the number of internship trained candidates who are hired. a. Does \(Y\) have a binomial or hypergeometric distribution? Why? b. Find the probability that two or more internship trained candidates are hired. c. What are the mean and standard deviation of \(Y\) ?

A type of bacteria cell divides at a constant rate \(\lambda\) over time. (That is, the probability that a cell divides in a small interval of time \(t\) is approximately \(\lambda t\).) Given that a population starts out at time zero with \(k\) cells of this bacteria and that cell divisions are independent of one another, the size of the population at time \(t, Y(t),\) has the probability distribution $$P[Y(t)=n]=\left(\begin{array}{l}n-1 \\\k-1\end{array}\right) e^{-\lambda k t}\left(1-e^{-\lambda t}\right)^{n-k}, \quad n=k, k+1, \ldots$$ b. If, for a type of bacteria cell, \(\lambda=.1\) per second and the population starts out with two cells at time zero, find the expected value and variance of the population after five seconds.

Toss a balanced die and let \(Y\) be the number of dots observed on the upper face. Find the mean and variance of \(Y\). Construct a probability histogram, and locate the interval \(\mu \pm 2 \sigma\). Verify that Tchebysheff's theorem holds.

a. We observe a sequence of independent identical trials with two possible outcomes on each trial, \(S\) and \(F,\) and with \(P(S)=p .\) The number of the trial on which we observe the fifth success, \(Y\), has a negative binomial distribution with parameters \(r=5\) and \(p\). Suppose that we observe the fifth success on the eleventh trial. Find the value of \(p\) that maximizes \(P(Y=11\) ). b. Generalize the result from part (a) to find the value of \(p\) that maximizes \(P\left(Y=y_{0}\right)\) when \(Y\) has a negative binomial distribution with parameters \(r\) (known) and \(p\)

Cards are dealt at random and without replacement from a standard 52 card deck. What is the probability that the second king is dealt on the fifth card?

See all solutions

Recommended explanations on Math Textbooks

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.