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Chapter 3: Problem 19

Due to antiquated equipment and frequent windstorms, the town of Oak City often suffers power outages. The following data give the numbers of power outages for each of the past 12 months. \(\begin{array}{llllllllllll}4 & 5 & 7 & 3 & 2 & 0 & 2 & 3 & 2 & 1 & 2\end{array}\) 4 Compute the mean, median, and mode for these data.

Short Answer

Expert verified
After performing the necessary calculations, the mean is \(3.25\), the median is \(2.5\) and the mode is \(2\). These values reflect the average, middle point and most frequently occurred number of power outages in the given 12 months respectively.

Step by step solution

01

Compute the Mean

Add all the numbers of power outages together to obtain the total number of outages over the 12-month period. This total is then divided by 12 (which is the total number of months) to compute the mean. This method allows to calculate the average or mean number of power outages per month.
02

Compute the Median

Arrange the numbers in ascending order. For this dataset, since there are 12 data points, the median would be the average of the 6th and 7th numbers in the list. Median gives the midpoint of the dataset.
03

Compute the Mode

Identify which value in the dataset occurs most frequently. The mode represents the number of outages that occurred most frequently in a single month in the past year. If no number repeats, then there is no mode.

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Key Concepts

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

Mean
The mean gives us an idea of the "average" situation by calculating the sum of all observations and dividing it by the number of observations. In our exercise on power outages, we start by adding up the outages for each of the 12 months. To find the mean:
  • Add up all the values: \(4 + 5 + 7 + 3 + 2 + 0 + 2 + 3 + 2 + 1 + 2 + 4 = 35\)
  • Divide the sum by the number of months: \(35 \div 12 \approx 2.92\)
This result means, on average, the town of Oak City experienced about 2.92 power outages per month over the past year. Understanding the mean helps provide a general overview of the data and smoothes out fluctuations from month to month. It gives a general snapshot of what one can expect under typical conditions.
Median
The median finds the middle value when data is organized in order. It tells us how data is spread out and provides insight into the structure of the data. Unlike the mean, the median is less affected by very high or very low values. For our dataset:
  • First, arrange the data in ascending order: \(0, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 7\)
  • Since there are 12 numbers, calculate the median by taking the average of the 6th and 7th values. Here, these are both \(2\) and \(3\).
  • The median is \((2+3) \div 2 = 2.5\).
This value indicates the middle of our dataset, meaning half of the months had fewer than 2.5 outages and half had more. The median provides a balanced way to see the central trend without being skewed by outliers.
Mode
Mode identifies the most frequently occurring number in a dataset. It tells us about the most common value and can be especially useful for understanding trends. In the power outages dataset:
  • Examine which number appears most often: \(2\) is the value that appears the most.
  • The mode is therefore \(2\).
  • If no number repeats, there would be no mode, but that's not the case here.
The mode helps us see what "typical" looks like in a specific term of frequency. Knowing the mode, we can say that "2" outages was the most common monthly occurrence in Oak City, reflecting a common pattern during the year.

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Most popular questions from this chapter

The following table gives the frequency distribution of the number of errors committed by a college baseball team in all of the 45 games that it played during the \(2011-12\) season. \begin{tabular}{cc} \hline Number of Errors & Number of Games \\ \hline 0 & 11 \\ 1 & 14 \\ 2 & 9 \\ 3 & 7 \\ 4 & 3 \\ 5 & 1 \\ \hline \end{tabular} Find the mean, variance, and standard deviation. (Hint: The classes in this example are single valued. These values of classes will be used as values of \(m\) in the formulas for the mean, variance, and standard deviation.)

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According to the National Center for Education Statistics (www.nces.ed.gov), the amounts of all loans, including Federal Parent PLUS loans, granted to students during the \(2007-2008\) academic year had 芒 distribution with a mean of \(\$ 8109.65\). Suppose that the standard deviation of this distribution is \(\$ 2412 .\) a. Using Chebyshev's theorem, find at least what percentage of students had \(2007-2008\) such loans between i. \(\$ 2079.65\) and \(\$ 14,139.65\) ii. \(\$ 3285.65\) and \(\$ 12,933.65\) *b. Using Chebyshev's theorem, find the interval that contains the amounts of \(2007-2008\) such loans for at least \(89 \%\) of all students.

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The following data give the 2009 estimates of crude oil reserves (in billions of barrels) of Saudi Arabia, Iran, Iraq, Kuwait, Venezuela, the United Arab Emirates, Russia, Libya, Nigeria, Canada, the United States, China, Brazil, and Mexico (source: www.eia.gov). \(\begin{array}{rrrrrrr}266.7 & 136.2 & 115.0 & 107.0 & 99.4 & 97.8 & 60.0 \\ 43.7 & 36.2 & 27.7 & 21.3 & 16.0 & 12.6 & 10.5\end{array}\) Prepare a box-and-whisker plot. Is the distribution of these data symmetric or skewed? Are there any outliers? If so, classify them as mild or extreme.
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