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A poll conducted by Zogby International showed that of those Americans who said music plays a "very important" role in their lives, \(30 \%\) said their local radio stations "always" play the kind of music they like (Zogby website, January 12,2004 ). Suppose a sample of 800 people who say music plays an important role in their lives is taken. a. How many would you expect to say that their local radio stations always play the kind of music they like? b. What is the standard deviation of the number of respondents who think their local radio stations always play the kind of music they like? c. What is the standard deviation of the number of respondents who do not think their local radio stations always play the kind of music they like?

Short Answer

Expert verified
a: 240 people, b: ≈ 12.96, c: ≈ 12.96.

Step by step solution

01

Identify the Problem

We need to calculate expectations and standard deviations for respondents who feel that their local radio stations "always" play the music they like. Given data: 30% feel this way and we have a sample size of 800.
02

Calculate Expected Value

The expected value, or the mean, is calculated using the formula \( E(X) = n \times p \). Here, \( n = 800 \) and \( p = 0.30 \). So, \( E(X) = 800 \times 0.30 = 240 \). We expect 240 people to say their local radio stations always play the music they like.
03

Calculate Standard Deviation of Positive Respondents

The standard deviation \( \sigma \) for a binomial distribution is calculated using \( \sigma = \sqrt{n \times p \times (1-p)} \). Substituting, \( \sigma = \sqrt{800 \times 0.30 \times 0.70} = \sqrt{168} \approx 12.96 \).
04

Calculate Standard Deviation of Negative Respondents

For respondents who do not agree, use the complement probability \( 1-p = 0.70 \). The standard deviation formula changes to \( \sigma = \sqrt{n \times (1-p) \times p} \), which is the same calculation as above: \( \sigma = \sqrt{800 \times 0.70 \times 0.30} = \sqrt{168} \approx 12.96 \).

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Key Concepts

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

Expected Value
The expected value is a fundamental concept in statistics that helps you understand the average outcome of a random variable over a large number of trials. It provides a measure of the center of a probability distribution.
In the context of a binomial distribution, the expected value can be thought of as the "long-run" average if you were to repeatedly conduct identical experiments. For our sample, where 30% of people like the music played by local radio stations and we sample 800 people, it is calculated using the formula:
\[ E(X) = n imes p \]
  • Where \( n = 800 \) is the sample size.
  • And \( p = 0.30 \) is the probability that an individual in the sample likes the music.
Plugging these values into the formula, we find:
\[ E(X) = 800 imes 0.30 = 240 \]This means that, on average, 240 out of 800 people would expect their local radio stations to "always" play the type of music they enjoy. Expected value gives you a valuable prediction of outcomes in numerous real-life scenarios.
Binomial Distribution
The binomial distribution is a commonly used probability distribution in statistics. It describes the number of successes in a fixed number of independent yes/no experiments.
Each experiment is called a trial, and the result is either a success or a failure. In our example, each person surveyed represents an independent trial. The trial is a success if they state their local radio station always plays the kind of music they like.
Key features of a binomial distribution include:
  • The number of trials \( n \), which in our case is 800 people.
  • The probability of success \( p \), which is 0.30 for liking the radio station's music.
  • The probability of failure \( q = 1 - p \), here 0.70.
This distribution assumes consistency across trials, meaning each person has the same chance of saying they like the music, independent of what others say. The binomial distribution is particularly helpful when working with problems of social behavior and marketing surveys where responses can be categorized into two distinct groups.
Standard Deviation
Standard deviation is a crucial statistical measure that tells you how much variation or dispersion there is from the average (mean).
In a binomial distribution, the standard deviation helps quantify the expected variability in the number of successes.
The formula used to calculate the standard deviation \( \sigma \) in a binomial context is:
\[ \sigma = \sqrt{n \times p \times (1 - p)} \]Where:
  • \( n = 800 \), the number of trials (people surveyed).
  • \( p = 0.30 \), the probability of "success" (radio stations always playing desired music).
  • \((1-p) = 0.70 \), the probability of "failure" (not always playing desired music).
Plugging these into the formula gives:
\[ \sigma = \sqrt{800 \times 0.30 \times 0.70} = \sqrt{168} \approx 12.96 \]This indicates that the number of people who believe their local radio always plays their preferred music can typically vary by about 13 people around the expected value of 240. Understanding standard deviation is key to grasping how much individual data points may differ from the expected outcome in practical, real-world scenarios.

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Most popular questions from this chapter

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?

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