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Chapter 3: Problem 35

Americans spend an average of 3 hours per day online. If the standard deviation is 32 minutes, find the range in which at least \(88.89 \%\) of the data will lie. Use Chebyshev's theorem.

Short Answer

Expert verified
The data lies between 84 and 276 minutes.

Step by step solution

01

Understanding the Question

We need to find the range where at least 88.89% of Americans' online time falls. We know the average is 3 hours (180 minutes), and the standard deviation is 32 minutes. We'll use Chebyshev's theorem for this.
02

Chebyshev's Theorem Formula

Chebyshev's theorem states that at least \(1 - \frac{1}{k^2}\) of the data falls within \(k\) standard deviations from the mean. For our problem, this percentage is given as 88.89%.
03

Set Up Chebyshev's Equation

Convert 88.89% to decimals to match the formula: \(1 - \frac{1}{k^2} = 0.8889\). Solve for \(k^2\) to find the number of standard deviations.
04

Solving for k

Calculate \(k\) by rearranging: \(\frac{1}{k^2} = 1 - 0.8889 = 0.1111\). Thus, \(k^2 = \frac{1}{0.1111}\) and \(k = \sqrt{\frac{1}{0.1111}}\).
05

Calculate k Value

Compute \(k = \sqrt{9} = 3\). This means at least 88.89% of data lies within 3 standard deviations of the mean.
06

Determine the Range

Each standard deviation is 32 minutes, so for 3 standard deviations: \(32 \times 3 = 96\) minutes. Since the average (mean) is 180 minutes, compute the range as \(180 - 96\) to \(180 + 96\) resulting in 84 to 276 minutes.

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Key Concepts

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

Standard Deviation
To understand standard deviation, let's think about how spread out your set of data points are around their mean (average). It's a number that tells you, "On average, how far is each number in your list from the mean?" So, the smaller the standard deviation, the closer the numbers are to the mean. Conversely, a larger standard deviation means the numbers are more spread out.

Mathematically, it's calculated by taking the square root of the variance. The formula for standard deviation (2_{ ext{sd}}2) is:
  • Find the mean of the data set.
  • Subtract the mean from each data point and square the result.
  • Find the average of these squared differences. That's the variance.
  • Take the square root of the variance to get the standard deviation.
In our original problem, the standard deviation is 32 minutes, meaning the average time spent online deviates by this amount from the overall average of 180 minutes.
Mean
The mean, often just called the average, is a measure of central tendency. It helps you find the average value of a set of numbers. You add all the numbers up and divide by how many numbers there are.

For example, to find the mean time Americans spend online per day, you would add up the time for each individual and divide by the total number of individuals.

The formula for the mean is: \[ \text{Mean} = \frac{\text{Sum of all data points}}{\text{Number of data points}} \]In our exercise, the mean time Americans spend online is 3 hours (or 180 minutes). This is the 'center' around which we try to see how spread out other times are using standard deviation.
Data Range
The data range gives us a quick view of how spread out or varied the data points in our set are. It鈥檚 simply the difference between the smallest and largest data points in your set.

The formula is: \[ \text{Data Range} = \text{Maximum data point} - \text{Minimum data point} \]

In context, when Chebyshev's theorem tells us that most data lies within a specific range around the mean, it's using the standard deviations to help us find that range. In this exercise, that range is calculated by going 96 minutes below and above the mean, giving us a total range from 84 minutes to 276 minutes.
Probability
Probability represents the likelihood or chance of an event occurring. When using Chebyshev's theorem, probability helps us understand how much data we expect to find within a certain number of standard deviations from the mean.

Chebyshev鈥檚 theorem tells us, with certainty, the fraction of data that's expected to be within any specified number of standard deviations from the mean. Specifically, for any data set:
  • At least \[1 - \frac{1}{k^2}\] of the data points fall within \[k\] standard deviations from the mean.
  • This is valid for any \[k > 1\], regardless of the data's shape.
In our problem, we used it to find that at least 88.89% of the data falls within 3 standard deviations of the mean. Thus, helping us identify how the majority of data points are grouped around the mean value.

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Most popular questions from this chapter

All values of a data set must be within \(s \sqrt{n-1}\) of the mean. If a person collected 25 data values that had a mean of 50 and a standard deviation of 3 and you saw that one data value was 67 , what would you conclude?

Using Chebyshev's theorem, complete the table to find the minimum percentage of data values that fall within \(k\) standard deviations of the mean. $$ \begin{array}{l|lllll} k & 1.5 & 2 & 2.5 & 3 & 3.5 \\ \hline \text { Percent } & & & & & \end{array} $$

The data show the maximum wind speeds for a sample of 40 states. Find the mean and modal class for the data. $$\begin{array}{lc}\text { Class boundaries } & \text { Frequency } \\\\\hline 47.5-54.5 & 3 \\\54.5-61.5 & 2 \\\61.5-68.5 & 9 \\\68.5-75.5 & 13 \\\75.5-82.5 & 8 \\\82.5-89.5 & 3 \\\89.5-96.5 & 2\end{array}$$

Harmonic Mean The harmonic mean (HM) is defined as the number of values divided by the sum of the reciprocals of each value. The formula is $$\mathrm{HM}=\frac{n}{\Sigma(1 / X)}$$ For example, the harmonic mean of \(1,4,5,\) and 2 is $$\mathrm{HM}=\frac{4}{1 / 1+1 / 4+1 / 5+1 / 2} \approx 2.051$$ This mean is useful for finding the average speed. Suppose a person drove 100 miles at 40 miles per hour and returned driving 50 miles per hour. The average miles per hour is not 45 miles per hour, which is found by adding 40 and 50 and dividing by 2 . The average is found as shown. Since Time \(=\) distance \(\div\) rate then Time \(1=\frac{100}{40}=2.5\) hours to make the trip Time \(2=\frac{100}{50}=2\) hours to return Hence, the total time is 4.5 hours, and the total miles driven are \(200 .\) Now, the average speed is $$\text { Rate }=\frac{\text { distance }}{\text { time }}=\frac{200}{4.5} \approx 44.444 \text { miles per hour }$$ This value can also be found by using the harmonic mean formula $$\mathrm{HM}=\frac{2}{1 / 40+1 / 50} \approx 44.444$$ Using the harmonic mean, find each of these. a. A salesperson drives 300 miles round trip at 30 miles per hour going to Chicago and 45 miles per hour returning home. Find the average miles per hour. b. A bus driver drives the 50 miles to West Chester at 40 miles per hour and returns driving 25 miles per hour. Find the average miles per hour. c. A carpenter buys \(\$ 500\) worth of nails at \(\$ 50\) per pound and \(\$ 500\) worth of nails at \(\$ 10\) per pound. Find the average cost of 1 pound of nails.

The average U.S. yearly per capita consumption of citrus fruit is 26.8 pounds. Suppose that the distribution of fruit amounts consumed is bell-shaped with a standard deviation equal to 4.2 pounds. What percentage of Americans would you expect to consume more than 31 pounds of citrus fruit per year?
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