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Chapter 5: Problem 60

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 (http://www.zogby.com, 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; b) 12.96; c) 12.96.

Step by step solution

01

Calculate expected number of people

Given that 30% of the sample believes their local radio stations "always" play the music they like, multiply this percentage by the total sample size to find the expected number. Calculate as follows: \[ \text{Expected number} = 0.30 \times 800 = 240 \] Therefore, we expect 240 people to say that their local radio stations always play the kind of music they like.
02

Compute the standard deviation for respondents who agree

To find the standard deviation, first recognize that the number of people who agree follows a binomial distribution, where the number of trials is 800 and the probability of success (agreeing) is 0.30. The standard deviation for a binomial distribution is calculated using the formula:\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \] Substitute the given values into the formula:\[ \sigma = \sqrt{800 \times 0.30 \times (1 - 0.30)} = \sqrt{800 \times 0.30 \times 0.70} \approx 12.96 \] Thus, the standard deviation is approximately 12.96.
03

Determine the standard deviation for respondents who disagree

Since the total sample size and the event probabilities remain the same, the standard deviation for those who do not agree can use the same formula: \[ \sigma_{disagree} = \sqrt{n \cdot p_{disagree} \cdot (1 - p_{disagree})} \] Here, the probability of disagreeing is \((1 - 0.30) = 0.70\).\[ \sigma_{disagree} = \sqrt{800 \times 0.70 \times (1 - 0.70)} = \sqrt{800 \times 0.70 \times 0.30} \approx 12.96 \] Since the variance of agreeing when knowing how many disagree (or vice versa) is based on complementary probability, the same standard deviation calculation results. So the standard deviation remains approximately 12.96 for both.

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Key Concepts

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

Statistics
Statistics is a fascinating field that helps us make sense of data in a meaningful way. In the context of our original exercise, we use statistical concepts to understand the likes and preferences of people based on survey data.
When we say a sample of 800 people, we are referring to a portion of a larger population that we study to draw conclusions about that population. The use of the percentage, such as the 30% mentioned, is a statistical expression indicating the proportion of interest to us. This allows us to easily extrapolate how a specific characteristic or behavior (like radio preferences) is distributed across the broader group.
  • Statistics involves sampling, a process of selecting a subset of data from a population to make inferences about the whole.
  • Percentages in statistics help describe ratios and relate the part of interest to the whole sample or population.
  • Understanding statistical representations like mean, median, and mode is crucial.
Overall, statistics empowers us to interpret trends and patterns from numbers, guiding informed decision-making.
Standard Deviation
Standard deviation is a statistical measurement of variation or dispersion within a dataset. It tells us how much the values in our data sample differ from the mean (average) value. In our exercise, standard deviation helps measure the variability of respondents' opinions about local radio stations.
When working with binomial distributions as we are in this context, the standard deviation provides insight into how much variation or spread we might expect from the number of people agreeing or disagreeing when another poll is conducted.
  • The formula for standard deviation in a binomial distribution is \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} \).
  • Here, \( n \) represents the number of trials (800 people), \( p \) the probability of success (0.30 for agreeing), and \( 1 - p \) the probability of failure (0.70 for disagreeing).
  • The resulting value in our exercise was approximately 12.96, indicating a typical difference of around 13 respondents from the expected number of 240.
Knowing standard deviation helps us anticipate variability and understand the precision of our sample results.
Probability
Probability is a fundamental concept in statistics that measures the likelihood of an event occurring. It ranges from 0 (impossibility) to 1 (certainty), and provides a quantitative description of how likely a particular event is to happen.
In our original exercise, probability helps us understand how likely it is for a random person in the sample to agree that their local radio station plays the music they like.
  • The given probability is 0.30 or 30%, indicating a 30% chance of a random individual from the sample saying 'yes'.
  • In binomial distributions, such probabilities help define the model we use for calculating expectations and variances.
  • By using probability, statisticians can predict outcomes and the confidence they might have in those predictions.
Probability allows us to grasp the concept of chance in everyday life and makes data-driven predictions more reliable and informed.

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Most popular questions from this chapter

Fifty percent of Americans believed the country was in a recession, even though technically the economy had not shown two straight quarters of negative growth (BusinessWeek, July 30,2001 ). For a sample of 20 Americans, make the following calculations. a. Compute the probability that exactly 12 people believed the country was in a recession. b. Compute the probability that no more than five people believed the country was in a recession. c. How many people would you expect to say the country was in a recession? d. Compute the variance and standard deviation of the number of people who believed the country was in a recession.

The following table provides a probability distribution for the random variable \(x\). $$\begin{array}{cc} x & f(x) \\ 3 & .25 \\ 6 & .50 \\ 9 & .25 \end{array}$$ a. \(\quad\) Compute \(E(x),\) the expected value of \(x\). b. Compute \(\sigma^{2},\) the variance of \(x\). c. Compute \(\sigma,\) the standard deviation of \(x\) .

According to a survey conducted by TD Ameritrade, one out of four investors have exchange-traded funds in their portfolios (USA Today, January 11,2007 ). For a sample of 20 investors, answer the following questions: a. Compute the probability that exactly four investors have exchange-traded funds in their portfolio. b. Compute the probability that at least two of the investors have exchange- traded funds in their portfolio. c. If you found that exactly twelve of the investors have exchange-traded funds in their portfolio, would you doubt the accuracy of the survey results? d. Compute the expected number of investors who have exchange-traded funds in their portfolio.

A survey conducted by the Bureau of Transportation Statistics (BTS) showed that the average commuter spends about 26 minutes on a one-way door-to-door trip from home to work. In addition, \(5 \%\) of commuters reported a one-way commute of more than one hour (http://www.bts.gov, January 12,2004 ). a. If 20 commuters are surveyed on a particular day, what is the probability that three will report a one-way commute of more than one hour? b. If 20 commuters are surveyed on a particular day, what is the probability that none will report a one-way commute of more than one hour? c. If a company has 2000 employees, what is the expected number of employees that have a one-way commute of more than one hour? d. If a company has 2000 employees, what is the variance and standard deviation of the number of employees that have a one-way commute of more than one hour?

Consider a binomial experiment with \(n=10\) and \(p=.10\). a. Compute \(f(0)\). b. Compute \(f(2)\). c. Compute \(P(x \leq 2)\). d. Compute \(P(x \geq 1)\). e. Compute \(E(x)\). f. Compute \(\operatorname{Var}(x)\) and \(\sigma\) .
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