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In a sample survey, 1800 senior citizens were asked whether or not they have ever been victimized by a dishonest telemarketer. The following table gives the responses by age group. $$\begin{array}{l|llcc} & & & \begin{array}{c} \text { Have Been } \\ \text { Victimized } \end{array} & \begin{array}{c} \text { Have Never } \\ \text { Been Victimized } \end{array} \\ \hline & 60-69 & \text { (A) } & 106 & 698 \\ \text { Age } & 70-79 & \text { (B) } & 145 & 447 \\ & 80 \text { or over (C) } & 61 & 343 \\ \hline \end{array}$$ Suppose one person is randomly selected from these senior citizens. Find the following probabilities a. \(P(\) have been victimized or \(\mathrm{B}\) ) b. \(P(\) have never been victimized or \(\mathrm{C}\) )

Short Answer

Expert verified
The probabilities are as follows:\na. \(P(\) have been victimized or \(\mathrm{B}\) ) = \(759 / 1800\).\nb. \(P(\) have never been victimized or \(\mathrm{C}\) ) = \(1549 / 1800\).

Step by step solution

01

Organize and Understand the Data

First, organize the data into a more understandable format. The total number of surveyed senior citizens is 1800. For age groups A (60-69), B (70-79), and C (80 or over), the counts of those who have and have not been victimized by telemarketers are given respectively.
02

Compute Total Probabilities

Calculate the total numbers for each category. Total number of victimized people across all age groups is \(106+145+61 = 312\). Total number of people in group B is \(145 + 447 = 592\). The total number of people who have never been victimized is \(698 + 447 + 343 = 1488\). Total number of people in group C is \(61 + 343 = 404\).
03

Calculation for Part a

To find \(P(\) have been victimized or \(\mathrm{B}\) ), sum the total number of people who have been victimized and the total number of people in group B, then subtract the number of people in group B who have been victimize (to avoid double counting). So, \(P(\) have been victimized or \(\mathrm{B}\) ) = \((312 + 592 - 145) / 1800 = 759 / 1800\).
04

Calculation for Part b

To find \(P(\) have never been victimized or \(\mathrm{C}\) ), sum the total number of people who have never been victimized and the total number of people in group C, then subtract the number of people in group C who have never been victimize (to avoid double counting). So, \(P(\) have never been victimized or \(\mathrm{C}\) ) = \((1488 + 404 - 343) / 1800 = 1549 / 1800\).

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Key Concepts

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

Conditional Probability
Conditional probability is a fundamental concept in statistics and probability theory, focusing on finding the probability of an event given that another event has already occurred. In simpler words, it's about narrowing down probabilities based on known information.
For example, in our exercise, you have to determine probabilities involving being victimized or belonging to a certain age group. The probability of being victimized or being in age group B is an example where one must consider overlapping events, also known as non-mutually exclusive events.
  • To calculate this type of probability, sum the probabilities of individual events.
  • Then, subtract the probability of both events occurring at the same time, to avoid counting it twice.
This method shines when dealing with complex, dependent events, making it a trusted tool in scenarios like sample surveys and contingency tables.
Sample Survey
A sample survey is a research method used to gather information from a part of a population, aiming to make inferences about the entire population. In this case, our sample consists of senior citizens. By sampling only 1800 people from a potentially larger population, this survey captures a snapshot of how many have been victimized by telemarketers across different age groups. The significance lies in the survey design:
  • It ensures a representative sample, allowing results to reflect the broader population.
  • It provides data useful for estimating probabilities and detecting trends.
Sample surveys are valuable for collecting qualitative and quantitative information. They are efficient and insightful for studies requiring significant data collection over broad demographics.
Age Groups
Age groups are distinct categories within a population, based on age ranges. They help stratify data for detailed analysis.
In our exercise, age groups are divided into three: 60-69, 70-79, and 80 or over. These divisions are critical:
  • They provide insight into different behavioral or statistical trends among various age demographics.
  • Help identify variations in how often individuals within these groups fall victim to telemarketers.
Analyzing data by distinct age groups allows researchers to observe age-specific patterns that might remain hidden in aggregated data, enriching the study outcome by providing more granulated context.
Contingency Table
A contingency table is a statistical tool used to showcase the frequency distribution of variables. It organizes the given data, making analysis more straightforward. In this exercise, the contingency table shows responses regarding victimization across different age groups.
  • Rows represent different age groups.
  • Columns represent whether individuals have been victimized or not.
This tool not only simplifies data visualization but is instrumental in calculating joint probabilities like the ones we dealt with in the exercise. It aids in understanding relationships between categorical variables, making it a staple in data analysis and probability estimation.

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Most popular questions from this chapter

Briefly explain the difference between the marginal and conditional probabilities of events. Give one example of each.

Powerball is a game of chance that has generated intense interest because of its large jackpots. To play this game, a player selects five different numbers from 1 through 59 , and then picks a Powerball number from 1 through 39 . The lottery organization randomly draws 5 different white balls from 59 balls numbered 1 through 59 , and then randomly picks a Powerball number from 1 through \(39 .\) Note that it is possible for the Powerball number to be the same as one of the first five numbers. a. If the player's first five numbers match the numbers on the five white balls drawn by the lottery organization and the player's Powerball number matches the Powerball number drawn by the lottery organization, the player wins the jackpot. Find the probability that a player who buys one ticket will win the jackpot. (Note that the order in which the five white balls are drawn is unimportant.) b. If the player's first five numbers match the numbers on the five white balls drawn by the lottery organization, the player wins about \(\$ 200,000\). Find the probability that a player who buys one ticket will win this prize.

Let \(A\) be the event that a number less than 3 is obtained if we roll a die once. What is the probability of \(A ?\) What is the complementary event of \(A\), and what is its probability?

A hat contains 40 marbles. Of them, 18 are red and 22 are green. If one marble is randomly selected out of this hat, what is the probability that this marble is \(\begin{array}{ll}\text { a. red? } & \text { b. green? }\end{array}\)

Suppose a randomly selected passenger is about to go through the metal detector at JFK Airport in New York City. Consider the following two outcomes: The passenger sets off the metal detector, and the passenger does not set off the metal detector. Are these two outcomes equally likely? Explain why or why not. If you are to find the probability of these two outcomes, would you use the classical approach or the relative frequency approach? Explain why

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