/*! 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 An individual who has automobile... [FREE SOLUTION] | 91Ó°ÊÓ

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An individual who has automobile insurance from a company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pme of \(Y\) is \begin{tabular}{l|cccc} \(y\) & 0 & 1 & 2 & 3 \\ \hline\(p(y)\) & \(.60\) & \(.25\) & \(.10\) & \(.05\) \end{tabular} a. Compute \(E(Y)\). b. Suppose an individual with \(Y\) violations incurs a surcharge of \(\$ 100 Y^{2}\). Calculate the expected amount of the surcharge.

Short Answer

Expert verified
a. \( E(Y) = 0.60 \). b. Expected surcharge is \($110\).

Step by step solution

01

Calculate Expected Value (E(Y))

The expected value formula for a discrete random variable is given by \[ E(Y) = \sum_{y} y \cdot p(y) \]. Substituting the given values: \[ E(Y) = 0 \cdot 0.60 + 1 \cdot 0.25 + 2 \cdot 0.10 + 3 \cdot 0.05 \]. Calculate each term: - \( 0 \cdot 0.60 = 0 \),- \( 1 \cdot 0.25 = 0.25 \),- \( 2 \cdot 0.10 = 0.20 \), - \( 3 \cdot 0.05 = 0.15 \). Add these results: \( 0 + 0.25 + 0.20 + 0.15 = 0.60 \). Thus, \( E(Y) = 0.60 \).
02

Calculate Expected Surcharge

The surcharge is given as \(\(100Y^2\). To find the expected surcharge, compute \[ E(S) = \sum_{y} (100y^2) \cdot p(y) \]. Substitute the values:- For \(y = 0\): \( 100 \cdot 0^2 \cdot 0.60 = 0 \),- For \(y = 1\): \( 100 \cdot 1^2 \cdot 0.25 = 25 \),- For \(y = 2\): \( 100 \cdot 2^2 \cdot 0.10 = 40 \),- For \(y = 3\): \( 100 \cdot 3^2 \cdot 0.05 = 45 \).Now, add these values: \( 0 + 25 + 40 + 45 = 110 \). Thus, the expected surcharge is \(\)110\).

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Key Concepts

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

Discrete Random Variable
Discrete random variables are an essential concept in statistics and probability, especially when dealing with countable outcomes. In the given exercise, the discrete random variable, denoted as \(Y\), represents the number of moving violations an individual received over the last three years. This variable can take on specific, separate values such as 0, 1, 2, or 3.
A discrete random variable contrasts with a continuous random variable, which can take on any value in a range. Understanding discrete random variables is crucial because they help us determine probabilities for different outcomes. They are often used in real-life situations, like counting the number of events happening over a fixed period. By analyzing these variables, we can make predictions and informed decisions based on likelihoods.
Probability Mass Function
The probability mass function (PMF) is a powerful tool for dealing with discrete random variables. A PMF assigns probabilities to each possible value that a discrete random variable can take. In our exercise, the PMF for \(Y\) is given by the values \(p(y)\): 0.60 for \(y=0\), 0.25 for \(y=1\), 0.10 for \(y=2\), and 0.05 for \(y=3\).
A PMF is particularly useful because it provides a complete description of the probability distribution of a discrete random variable. The probabilities in a PMF always sum up to 1. This is necessary because one of the values must occur — a fundamental property of probability distributions. Using the PMF, we can calculate important statistical metrics, such as the expected value of the variable.
Expected Surcharge
Calculating an expected surcharge involves applying the concept of expected value to monetary outcomes. In our exercise, the surcharge is determined using the formula \(100Y^2\), meaning the surcharge increases with the square of the number of violations.
To find the expected surcharge, we need to evaluate the expected value of \(100Y^2\). This is done by summing up the products of \(100y^2\) and the respective probabilities from the PMF. The results for \(y=0, 1, 2,\) and \(3\) are combined to give us the expected surcharge of 110 units.
Understanding the expected surcharge is important because it allows insurance companies to predict average costs associated with customer claims. This information can be critical for setting premiums and managing financial risk.
Statistics Problem Solving
Statistics problem solving often involves identifying the right methods and models to apply to a given situation. In this exercise, we are tasked with calculating both an expected value and an expected surcharge based on a discrete random variable and its PMF.
A structured approach is key to tackling these problems efficiently:
  • Identify the type and nature of the random variables involved.
  • Choose the appropriate statistical model or formula, like the PMF or expected value formula.
  • Substitute the given data into these formulas.
  • Calculate each term methodically, checking work at each step.
  • Sum the results to find the final answer.
By practicing these steps and understanding the underlying concepts, students can improve their problem-solving skills in statistics. This structured approach is applicable to a wide range of statistical and probabilistic scenarios.

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Most popular questions from this chapter

A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a chemical, which it sells to customers in 5-lb containers. Let \(X=\) the number of containers ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X\).]

Define a function \(p(x ; \lambda, \mu)\) by $$ \begin{aligned} &p(x ; \lambda, \mu) \\ &=\left\\{\begin{array}{cc} \frac{1}{2} e^{-\lambda} \frac{\lambda^{x}}{x !}+\frac{1}{2} e^{-\mu} \frac{\mu^{x}}{x !} & x=0,1,2, \ldots \\ 0 & \text { otherwise } \end{array}\right. \end{aligned} $$ a. Show that \(p(x, \lambda, \mu)\) satisfies the two conditions necessary for specifying a pmf. [Note: If a firm employs two typists, one of whom makes typographical errors at the rate of ¿ per page and the other at rate \(\mu\) per page and they each do half the firm's typing, then \(p(x ; \lambda, \mu)\) is the pmf of \(X=\) the number of errors on a randomly chosen page.] b. If the first typist (rate \(\lambda\) ) types \(60 \%\) of all pages, what is the penf of \(X\) of part (a)? c. What is \(E(X)\) for \(p(x, \lambda, \mu)\) given by the displayed expression? d. What is \(\sigma^{2}\) for \(p(x ; \lambda, \mu)\) given by that expression?

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of \(5 / \mathrm{h}\). a. What is the probability that exactly four arrivals occur during a particular hour? b. What is the probability that at least four people arrive during a particular hour? c. How many people do you expect to arrive during a 45-min period?

The \(n\) candidates for a job have been ranked 1,2 , \(3, \ldots, n\). Let \(X=\) the rank of a randomly selected candidate, so that \(X\) has pmf $$ p(x)=\left\\{\begin{array}{cc} 1 / n & x=1,2,3, \ldots, n \\ 0 & \text { otherwise } \end{array}\right. $$ (this is called the discrete uniform distribution). Compute \(E(X)\) and \(V(X)\) using the shortcut formula. [Hint: The sum of the first \(n\) positive integers is \(n(n+1) / 2\), whereas the sum of their squares is \(n(n+1)(2 n+1) / 6\).]

Consider a disease whose presence can be identified by carrying out a blood test. Let \(p\) denote the probability thar a randomly selected individual has the disease. Suppose \(n\) individuals are independently selected for testing. One way to proceed is to carry out a separate test on each of the \(n\) blood samples. A potentially more economical approach, group testing, was introduced during World War II to identify syphilitic men among army inductees. First, take a part of each blood sample, combine these specimens, and carry out a single test. If no one has the disease, the result will be negative, and only the one test is required. If at least one individual is diseased, the test on the combined sample will yield a positive result, in which case the \(n\) individual tests are then carried out. If \(p=.1\) and \(n=3\), what is the expected number of tests using this procedure? What is the expected number when \(n=5\) ? [The article "Random Multiple-Access Communication and Group Testing" (IEEE Trans. Commun., 1984: 769-774) applied these ideas to a communication system in which the dichotomy was active/ idle user rather than diseased/nondiseased.]

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