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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

According to the article "'Characterizing the Severity and Risk of Drought in the Poudre River, Colorado" (J. of Water Res. Planning and Mgmnt.s 2005: 383-393), the drought length \(Y\) is the number of consecutive time intervals in which the water supply remains below a critical value \(y_{0}\) (a deficit), preceded by and followed by periods in which the supply exceeds this critical value (a surplus). The cited paper proposes a geometric distribution with \(p=.409\) for this random variable. a. What is the probability that a drought lasts exactly 3 intervals? At most 3 intervals? b. What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

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.

An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.

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 \(n(n+1)(2 n+1) / 6 .]\)

A certain brand of upright freezer is available in three different rated capacities: \(16 \mathrm{ft}^{3}, 18 \mathrm{ft}^{3}\), and \(20 \mathrm{ft}^{3}\). Let \(X=\) the rated capacity of a freezer of this brand sold at a certain store. Suppose that \(X\) has pmf \begin{tabular}{l|ccc} \(x\) & 16 & 18 & 20 \\ \hline\(p(x)\) & \(.2\) & \(.5\) & \(.3\) \end{tabular} a. Compute \(E(X), E\left(X^{2}\right)\), and \(V(X)\). b. If the price of a freezer having capacity \(X\) is \(70 X-650\), what is the expected price paid by the next customer to buy a freezer? c. What is the variance of the price paid by the next customer? d. Suppose that although the rated capacity of a freezer is \(X\), the actual capacity is \(h(X)=X-.008 X^{2}\). What is the expected actual capacity of the freezer purchased by the next customer?
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