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The following table gives the 2002 age distribution of the U.S. population: $$\begin{array}{lcccccc}\hline \text { Group } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \begin{array}{l}\text { Age } \\\\\text { (in years) } \end{array} & \text { Under } 5 & 5-19 & 20-24 & 25-44 & 45-64 & 65 \text { and over } \\\\\hline \begin{array}{l} \text { Number (in } \\\\\text { thousands) }\end{array} & 19,527 & 59,716 & 18,611 & 83,009 & 66,088 & 33,590 \\ \hline\end{array}$$ Let the random variable \(X\) denote a randomly chosen age group within the population. Find the probability distribution associated with these data.

Short Answer

Expert verified
The probability distribution of the random variable X for each age group is: P(X=1) ≈ 0.0696 P(X=2) ≈ 0.2131 P(X=3) ≈ 0.0664 P(X=4) ≈ 0.2960 P(X=5) ≈ 0.2355 P(X=6) ≈ 0.1197

Step by step solution

01

Calculate the total population

To find the probability distribution, we need to find the total population. We can do this by summing the number of people in each age group. Total population = 19,527 + 59,716 + 18,611 + 83,009 + 66,088 + 33,590
02

Calculate the probability for each age group

Using the total population calculated in step 1, we will divide the number of people in each age group by the total population to find the probability of each age group. Let's denote the probabilities for age groups 1 to 6 as P(X=1), P(X=2), P(X=3), P(X=4), P(X=5), and P(X=6) respectively. P(X=1) = \(\frac{19,527}{Total\ population}\) P(X=2) = \(\frac{59,716}{Total\ population}\) P(X=3) = \(\frac{18,611}{Total\ population}\) P(X=4) = \(\frac{83,009}{Total\ population}\) P(X=5) = \(\frac{66,088}{Total\ population}\) P(X=6) = \(\frac{33,590}{Total\ population}\)
03

Calculate the total population and the probabilities

Let's now calculate the total population and the probability for each age group. Total population = 19,527 + 59,716 + 18,611 + 83,009 + 66,088 + 33,590 = 280,541 P(X=1) = \(\frac{19,527}{280,541}\) ≈ 0.0696 P(X=2) = \(\frac{59,716}{280,541}\) ≈ 0.2131 P(X=3) = \(\frac{18,611}{280,541}\) ≈ 0.0664 P(X=4) = \(\frac{83,009}{280,541}\) ≈ 0.2960 P(X=5) = \(\frac{66,088}{280,541}\) ≈ 0.2355 P(X=6) = \(\frac{33,590}{280,541}\) ≈ 0.1197
04

Write out the probability distribution associated with these data

Now that we have calculated the probabilities for each age group, we can write out the probability distribution for the random variable X. The probability distribution of X is: P(X=1) ≈ 0.0696 P(X=2) ≈ 0.2131 P(X=3) ≈ 0.0664 P(X=4) ≈ 0.2960 P(X=5) ≈ 0.2355 P(X=6) ≈ 0.1197

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Key Concepts

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

Applied Mathematics
When we talk about applied mathematics, we are referring to the use of mathematical methods by different fields such as science, engineering, business, computer science, and industry. In essence, applied mathematics is not just about abstract theories; it is about practical problem-solving.

A relevant applied mathematics concept in this exercise is probability. Probability is employed to predict how likely events are to occur. This is crucial in a multitude of areas, from risk assessment in finance to making predictions in weather forecasts or determining outcomes in genetics.

This textbook exercise leverages probability to analyze a real-world scenario, which is the age distribution of the U.S. population. Such analysis can aid in planning for the educational needs of younger age groups, health care facilities for older populations, and economic forecasting.
Age Distribution Analysis
Analyzing age distribution involves understanding the various age categories within a population and their respective proportions. This is a key demographic analysis that can provide insight into the social structure and future trends of a society.

For instance, a high proportion of young people might suggest a need for more educational resources, while a large elderly population may indicate an increased demand for healthcare services. Governments and businesses often use age distribution analysis for planning and decision-making purposes.

By turning age distribution data into probability distributions as seen in the given exercise, we transform raw data into actionable information. Ever-changing demographics shape policy decisions and market strategies, thereby demonstrating the importance of an accurate and comprehensive age distribution analysis.
Random Variable
In statistics, a random variable is a numerical description of the outcomes of a random phenomenon. It can be thought of as a variable whose values depend on the outcomes of a random process.

A random variable is usually denoted by a capital letter, such as 'X' in this exercise, and it can take on multiple possible values, each with its own probability. The collection of all possible values and their associated probabilities is known as the probability distribution of the random variable.

The exercise provided includes a discrete random variable, where specific age groups are associated with various probabilities. Understanding the concept of a random variable is fundamental when dealing with probabilistic models or any form of data that involves randomness. In the context of age distribution, each group represents a potential outcome for the variable 'age group of a randomly chosen individual from the U.S. population'.

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Most popular questions from this chapter

Sally Leonard, a real estate broker. is relocating in a large metropolitan area where she has received job offers from realty company A and realty company B. The number of houses she expects to sell in a year at each firm and the associated probabilities are shown in the following tables. The average price of a house in the locale of company \(\mathrm{A}\) is $$\$ 308,000$$, whereas the average price of a house in the locale of company \(\mathrm{B}\) is $$\$ 474,000$$. If Sally will receive a \(3 \%\) commission on sales at both companies, which job offer should she accept to maximize her expected yearly commission?

Give the range of values that the random variable \(X\) may assume and classify the random variable as finite discrete, infinite discrete, or continuous. \(X=\) The number of defective watches in a sample of eight watches

Sugar packaged by a certain machine has a mean weight of \(5 \mathrm{lb}\) and a standard deviation of \(0.02 \mathrm{lb}\). For what values of \(c\) can the manufacturer of the machinery claim that the sugar packaged by this machine has a weight between \(5-c\) and \(5+c \mathrm{lb}\) with probability at least \(96 \%\) ?

The birthrates in the United States for the years \(1991-2000\) are given in the following table. (The birthrate is the number of live births/1000 population.) $$\begin{array}{lllll}\hline \text { Year } & 1991 & 1992 & 1993 & 1994 \\\\\hline \text { Birthrate } & 16.3 & 15.9 & 15.5 & 15.2 \\\\\hline\end{array}$$ $$\begin{array}{llll}\hline \text { Year } & 1995 & 1996 & 1997 \\ \hline \text { Birthrate } & 14.8 & 14.7 & 14.5 \\ \hline\end{array}$$ $$\begin{array}{llll}\hline \text { Year } & 1998 & 1999 & 2000 \\\\\hline \text { Birthrate } & 14.6 & 14.5 & 14.7 \\ \hline\end{array}$$ a. Describe a random variable \(X\) that is associated with these data. b. Find the probability distribution for the random variable \(X\). c. Compute the mean, variance, and standard deviation of \(X\).

A survey was conducted by the market research department of the National Real Estate Company among 500 prospective buyers in a large metropolitan area to determine the maximum price a prospective buyer would be willing to pay for a house. From the data collected, the distribution that follows was obtained. Compute the mean, variance, and standard deviation of the maximum price \(x\) (in thousands of dollars) that these buyers were willing to pay for a house. $$ \begin{array}{ll} \hline \text { Maximum Price } & \\ \text { Considered, } x & P(X=x) \\ \hline 280 & \frac{10}{500} \\ \hline 290 & \frac{20}{500} \\ \hline 300 & \frac{75}{500} \\ \hline 310 & \frac{85}{506} \\ \hline 320 & \frac{70}{500} \\ \hline 350 & \frac{90}{500} \\ \hline 380 & \frac{90}{500} \\ \hline 400 & \frac{55}{500} \\ \hline 450 & \frac{5}{500} \\ \hline\end{array}$$

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