91Ó°ÊÓ

A political action group is planning to interview home owners to assess the impact caused by a recent slump in housing prices. According to a Wall Street Journal/Harris Interactive Personal Finance poll, \(26 \%\) of individuals aged \(18-34,50 \%\) of individuals aged \(35-44,\) and \(88 \%\) of individuals aged 55 and over are home owners (All Business website, January 23,2008 ). a. How many people from the \(18-34\) age group must be sampled to find an expected number of at least 20 home owners? b. How many people from the \(35-44\) age group must be sampled to find an expected number of at least 20 home owners? c. How many people from the 55 and over age group must be sampled to find an expected number of at least 20 home owners? d. If the number of \(18-34\) year olds sampled is equal to the value identified in part (a), what is the standard deviation of the number who will be home owners? e. If the number of \(35-44\) year olds sampled is equal to the value identified in part what is the standard deviation of the number who will be home owners?

Step by step solution

01

Understand the Problem

First, recognize the main task is to determine how many individuals from each age group need to be sampled to have an expected number of at least 20 home owners, using the given probabilities for each group.

02

Setting Up Equations for Expected Home Owners

For each age group, use the formula for expected value: \( E = n \cdot p \), where \( n \) is the number of individuals sampled and \( p \) is the probability of being a homeowner. Set this equal to 20 to find \( n \).

03

Solve for 18-34 Age Group

For the 18-34 age group, \( p = 0.26 \). Thus, solve \( n \cdot 0.26 = 20 \). Therefore, \( n = \frac{20}{0.26} \approx 76.92 \). Round up to ensure at least 20 home owners, resulting in \( n = 77 \).

04

Solve for 35-44 Age Group

For the 35-44 age group, \( p = 0.50 \). Solve \( n \cdot 0.50 = 20 \). Therefore, \( n = \frac{20}{0.50} = 40 \).

05

Solve for 55 and Over Age Group

For the 55 and over age group, \( p = 0.88 \). Solve \( n \cdot 0.88 = 20 \). Therefore, \( n = \frac{20}{0.88} \approx 22.73 \). Round up to obtain \( n = 23 \).

06

Compute Standard Deviation for 18-34 Age Group

Use the formula for standard deviation of a binomial distribution: \( \sigma = \sqrt{n \cdot p \cdot (1-p)} \). For \( n = 77 \) and \( p = 0.26 \), calculate \( \sigma = \sqrt{77 \cdot 0.26 \cdot 0.74} \approx 3.68 \).

07

Compute Standard Deviation for 35-44 Age Group

Using \( n = 40 \) and \( p = 0.50 \), calculate \( \sigma = \sqrt{40 \cdot 0.50 \cdot 0.50} = 3.16 \).

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Key Concepts

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

Binomial Distribution

The Binomial Distribution is a key concept in statistics, particularly useful when you deal with situations that have two possible outcomes. Think of it like a coin flip, where each flip results in either heads or tails. For the Binomial Distribution:

• Each trial is independent.
• There are only two possible outcomes, often referred to as success or failure.
• The probability of success remains constant across trials.

The task in the problem involves determining how many people we need to poll in different age groups to achieve a certain number of successes (homeowners, in this case). For this, we use the probability given for each age group (like 0.26 for the 18-34 age group) and calculate the required sample size using the formula for Expected Value. Understanding this distribution is crucial since many real-world scenarios fit into this category. It allows you to predict the probability of a given number of successes over a set number of trials, given a constant probability of success.

Expected Value

The Expected Value (EV) is a concept used widely in probability to anticipate outcomes in uncertain scenarios. It's essentially the weighted average of all possible outcomes where each outcome is multiplied by its probability. In mathematical terms for a binomial situation: The formula is given by:\[ E(X) = n imes p \] where:

• \( n \) is the number of trials.
• \( p \) is the probability of success on each trial.

For instance, to expect at least 20 homeowners in a sample from the 18-34 age group with a probability of 0.26 of being a home owner, you set up the equation like this:\[ 20 = n imes 0.26 \]Solving gives you \( n = \frac{20}{0.26} \approx 77 \). By understanding expected value, you can make informed predictions about your sample outcomes and plan accordingly.

Standard Deviation

Standard Deviation is a measure that quantifies the amount of variation or dispersion in a set of data. In the context of a Binomial Distribution, it tells you how spread out the results are likely to be from the mean (which in many cases is the Expected Value). The formula for the standard deviation in a binomial distribution is:\[ \sigma = \sqrt{n imes p imes (1-p)} \]where:

• \( n \) is the total number of trials.
• \( p \) is the probability of success in a single trial.
• \( (1-p) \) is the probability of failure.

So, using the 18-34 age group from the problem as an example:If we sample 77 individuals and expect a success probability of 0.26, the calculation would be:\[ \sigma = \sqrt{77 \times 0.26 \times 0.74} \approx 3.68 \]This indicates the average amount by which the number of homeowners in the sample might differ from the expected value of 20. Understanding standard deviation helps in gauging the reliability and variability of your expected result, which is critical in making social predictions or policy plans.