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Chapter 3: Problem 30

An individual who has automobile insurance from a certain company is randomly selected. Let \(Y\) be the number of moving violations for which the individual was cited during the last 3 years. The pmf 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
E(Y) = 0.60; Expected surcharge = $110.

Step by step solution

01

Understanding the problem

We are tasked with finding the expected value of the number of moving violations (E(Y)) and the expected surcharge based on these violations.
02

Identifying the given pmf

The given pmf is a probability distribution function for the variable \(Y\), representing the number of violations. The probabilities are: \( p(0) = 0.60 \), \( p(1) = 0.25 \), \( p(2) = 0.10 \), and \( p(3) = 0.05 \).
03

Calculating Expected Value, E(Y)

The expected value \(E(Y)\) is calculated using the formula \( E(Y) = \sum y \, p(y) \). Substituting the values, we have: \[ E(Y) = (0 \times 0.60) + (1 \times 0.25) + (2 \times 0.10) + (3 \times 0.05) \]. This simplifies to \( E(Y) = 0 + 0.25 + 0.20 + 0.15 = 0.60 \).
04

Defining the surcharge formula

According to the problem, the surcharge \( S \) is calculated as \( S = 100Y^2 \). We need to find the expected surcharge, \( E(S) \).
05

Calculating Expected Surcharge, E(S)

Compute \( E(S) = \sum 100 \, y^2 \, p(y) \). The expression becomes \[ E(S) = 100 \times (0^2 \times 0.60) + (1^2 \times 0.25) + (2^2 \times 0.10) + (3^2 \times 0.05) \]. This simplifies to \[ E(S) = 100 \times (0 + 0.25 + 0.40 + 0.45) = 100 \times 1.10 = 110 \].

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Key Concepts

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

Probability Distribution Function
A Probability Distribution Function describes how probabilities are distributed over the values of a random variable. In simpler terms, it tells us the chances of each possible outcome for a particular scenario.
When we look at the exercise, the random variable is the number of moving violations, denoted by \( Y \). Each \( y \) value (0 to 3) has a corresponding probability.
For example:
  • \( p(0) = 0.60 \) implies a 60% chance of having zero violations.
  • \( p(1) = 0.25 \) indicates a 25% chance of having one violation.
  • \( p(2) = 0.10 \) suggests a 10% chance of having two violations.
  • \( p(3) = 0.05 \) highlights a 5% chance of having three violations.
Each probability value is between 0 and 1, and their sum is equal to 1. This ensures all possible outcomes are covered. Understanding this distribution is crucial for calculating expected values and other statistical measures.
Probability Mass Function
A Probability Mass Function (PMF) is a type of probability distribution specifically used for discrete random variables. It indicates the probability of each specific outcome occurring.
In the given exercise, \( Y \) is a discrete random variable representing the number of moving violations. We already know the PMF:
  • \( p(0) = 0.60 \)
  • \( p(1) = 0.25 \)
  • \( p(2) = 0.10 \)
  • \( p(3) = 0.05 \)
These probabilities are essential to calculate the Expected Value, which is a measure of the "average" outcome you expect if the situation were repeated many times.
Here, the Expected Value, \( E(Y) \), was calculated using the formula \( E(Y) = \sum y \cdot p(y) \) and resulted in 0.60, which gives us the average number of violations.
Insurance Surcharge Calculation
Calculating an insurance surcharge involves determining additional costs based on certain conditions or behaviors, such as the number of moving violations in this exercise. Here, the insurance surcharge formula provided is \( S = 100Y^2 \).
This means the surcharge amount depends on the square of the number of violations multiplied by 100. Let's look at how this works with our PMF:
  • When \( Y = 0 \), the surcharge is \( 100 \times 0^2 = 0 \).
  • When \( Y = 1 \), the surcharge is \( 100 \times 1^2 = 100 \).
  • When \( Y = 2 \), the surcharge is \( 100 \times 2^2 = 400 \).
  • When \( Y = 3 \), the surcharge is \( 100 \times 3^2 = 900 \).
To find the Expected Surcharge, \( E(S) \), we use the PMF to calculate \( \sum 100 \cdot y^2 \cdot p(y) \). This sums up as \( 100 \times 1.10 = 110 \).
This value tells us that, on average, one would expect to pay an additional $110 in insurance surcharge due to moving violations.

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Most popular questions from this chapter

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 $$ p(x)= \begin{cases}1 / n & x=1,2,3, \ldots, n \\ 0 & \text { otherwise }\end{cases} $$ (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 \(n(n+1)(2 n+1) / 6 .]\)

The article 'Should You Report That FenderBender?" (Consumer Reports, Sept. 2013: 15) reported that 7 in 10 auto accidents involve a single vehicle (the article recommended always reporting to the insurance company an accident involving multiple vehicles). Suppose 15 accidents are randomly selected. Use Appendix Table A.l to answer each of the following questions. a. What is the probability that at most 4 involve a single vehicle? b. What is the probability that exactly 4 involve a single vehicle? c. What is the probability that exactly 6 involve multiple vehicles? d. What is the probability that between 2 and 4 , inclusive, involve a single vehicle? e. What is the probability that at least 2 involve a single vehicle? f. What is the probability that exactly 4 involve a single vehicle and the other 11 involve multiple vehicles?

If the sample space \(S\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(\rho\) will have an infinite set of possible values? If yes, say why. If no, give an example.

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books in a completely random fashion to each of the four students \((1,2,3\), and 4) who claim to have left books. One possible outcome is that 1 receives 2 's book, 2 receives 4 's book, 3 receives his or her own book, and 4 receives l's book. This outcome can be abbreviated as \((2,4,3,1)\). a. List the other 23 possible outcomes. b. Let \(X\) denote the number of students who receive their own book. Determine the pmf of \(X\).

An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30 , decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. a. What is the probability that exactly 10 of these are from the second section? b. What is the probability that at least 10 of these are from the second section? c. What is the probability that at least 10 of these are from the same section? d. What are the mean value and standard deviation of the number among these 15 that are from the second section? e. What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?
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