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A large population has a mean of 230 and a standard deviation of 41 . Using Chebyshev's theorem, find at least what percentage of the observations fall in the intervals \(\mu \pm 2 \sigma, \mu \pm 2.5 \sigma\), and \(\mu \pm 3 \sigma\).

Short Answer

Expert verified
At least 75% of the observations will fall within \(\mu \pm 2 \sigma\), 84% within \(\mu \pm 2.5 \sigma\), and 89% within \(\mu \pm 3 \sigma\).

Step by step solution

01

Identify the Mean and Standard Deviation

Since the mean (\(\mu\)) is given as 230 and the standard deviation (\(\sigma\)) as 41, no calculations are necessary here.
02

Applying Chebyshev’s Theorem for \(\mu \pm 2 \sigma\)

Chebyshev's theorem states that at least \(1 - 1/k^2\) percent of data lies within \(k\) standard deviations of the mean. Here, \(k = 2\). Therefore, substitute \(k = 2\) into the theorem to find the percentage: \(1 - 1/k^2 = 1 - 1/2^2 = 1 - 1/4 = 0.75 = 75%\). So, at least 75% of the data lies within \(\mu \pm 2 \sigma [230 \pm (2 * 41)]\).
03

Applying Chebyshev’s Theorem for \(\mu \pm 2.5 \sigma\)

This time, we have \(k = 2.5\). Substituting into the theorem: \(1 - 1/k^2 = 1 - 1/2.5^2 = 1 - 1/6.25 \approx 0.84 = 84%\). So, at least 84% of the observations fall within \(\mu \pm 2.5 \sigma [230 \pm (2.5 * 41)]\).
04

Applying Chebyshev’s Theorem for \(\mu \pm 3 \sigma\)

Finally, we have \(k = 3\). Substituting into the theorem: \(1 - 1/k^2 = 1 - 1/3^2 = 1 - 1/9 \approx 0.89 = 89%\). So, at least 89% of the observations fall within \(\mu \pm 3 \sigma [230 \pm (3 * 41)]\).

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Key Concepts

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

Understanding Mean and Standard Deviation
Mean, often represented as \( \mu \), is a measure of the central tendency of a data set. It is calculated by summing all the data points and dividing by the number of points. Think of it as the "average" value that gives you a general idea of where most data points in a set are clustered. In our example, the mean is 230.

Standard deviation, denoted by \( \sigma \), measures how spread out the numbers in a data set are. A low standard deviation means that most numbers are close to the mean, while a high standard deviation indicates that the numbers are more spread out. In the problem we are discussing, the standard deviation is 41.

Both mean and standard deviation are crucial for understanding data distribution as they help us know where data values lie and how much variation there is within a data set.
Exploring Statistical Intervals Using Chebyshev’s Theorem
Statistical intervals are ranges in which we can expect a certain percentage of data points to lie. Chebyshev’s theorem helps us estimate these intervals when dealing with non-normal distributions.

Chebyshev’s theorem states that at least \( 1 - \frac{1}{k^2} \) of data falls within \( k \) standard deviations from the mean. This theorem is versatile because it applies to any data distribution, unlike the Empirical Rule which is specific to normal distributions.

For instance:
  • In the interval \( \mu \pm 2 \sigma \), the theorem tells us that at least 75% of data points will fall within this range
  • In the interval \( \mu \pm 2.5 \sigma \), at least 84% of data points are expected to fall
  • In the interval \( \mu \pm 3 \sigma \), the coverage rises to at least 89%
In our example, this theorem translates to real numbers around the mean of 230, assisting in understanding where most data points might fall.
Making Sense of Population Data Analysis
Population data analysis involves examining a full set of data points to understand patterns or trends. By analyzing a whole population, rather than a sample, we avoid uncertainties that arise in sampling.

In the given problem, you have data from a large population with defined parameters: a mean of 230 and a standard deviation of 41. Population analysis helps determine where most of these data points lie using statistical methods like Chebyshev’s theorem.

Using these tools, we can:
  • Estimate percentages of observations within specific intervals (e.g., \( \mu \pm 2 \sigma \))
  • Evaluate the spread and concentration of data points around the mean
  • Inform decisions based on statistical deductions from population insights
This comprehensive understanding affords more reliable insights, crucial for practical applications and statistical decision-making.

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Most popular questions from this chapter

The mean 2011 income for five families was \(\$ 99.520 .\) What was the total 2011 income of these five families?

Refer to Exercise 3.24, which listed the number of women from each of 12 countries who were on the Rolex Women's World Golf Rankings Top 50 list as of July \(18.2011\). Those data are reproduced here: $$ \begin{array}{llllllllllll} 3 & 1 & 1 & 1 & 10 & 1 & 1 & 1 & 18 & 2 & 3 & 8 \end{array} $$ Calculate the range, variance, and standard deviation.

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Suppose that there are 150 freshmen engineering majors at a college and each of them will take the same five courses next semester. Four of these courses will be taught in small sections of 25 students each, whereas the fifth course will be taught in one section containing all 150 freshmen. To accommodate all 150 students, there must be six sections of each of the four courses taught in 25 -student sections. Thus, there are 24 classes of 25 students each and one class of 150 students. a. Find the mean size of these 25 classes. b. Find the mean class size from a student's point of view, noting that each student has five classes containing \(25,25,25,25\), and 150 students, respectively. Are the means in parts a and b equal? If not, why not?

According to the Kaiser Family Foundation, U.S. workers who had employer- provided health insur. ance paid an average premium of \(\$ 4129\) for family coverage during 2011 (USA TODAY, October 10,2011 ). Suppose that the premiums for such family coverage paid this year by all such workers have a bell-shaped distribution with a mean of \(\$ 4129\) and a standard deviation of \(\$ 600\). Using the empirical rule, find the ap: proximate percentage of such workers who pay premiums for such family coverage between a. \(\$ 2329\) and \(\$ 5929\) b. \(\$ 3529\) and \(\$ 4729\) c. \(\$ 2929\) and \(\$ 5329\)

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