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Chapter 3: Problem 18

These data represent the net worth (in millions of dollars) of 45 national corporations. Find the mean and modal class for the data. Class limits \(\quad\) Frequency \(\begin{array}{lc}{10-20} & {2} \\ {21-31} & {8} \\ {32-42} & {15} \\\ {43-53} & {7} \\ {54-64} & {10} \\ {65-75} & {3}\end{array}\)

Short Answer

Expert verified
The mean is approximately 42.87, and the modal class is 32-42.

Step by step solution

01

Calculate the Midpoint of Each Class

To find the mean, determine the midpoint for each class, also called the class mark. The midpoint is calculated as the average of the lower and upper class limits. For example, for the class 10-20, the midpoint is \(\frac{10 + 20}{2} = 15\). Calculate midpoints for all classes:\[\begin{align*}10-20 & : 15 \21-31 & : 26 \32-42 & : 37 \43-53 & : 48 \54-64 & : 59 \65-75 & : 70 \end{align*}\]
02

Multiply Midpoints by Frequencies

For each class, multiply the class midpoint by the class frequency. This gives us the total for each class:\[\begin{align*}15 \times 2 &= 30 \26 \times 8 &= 208 \37 \times 15 &= 555 \48 \times 7 &= 336 \59 \times 10 &= 590 \70 \times 3 &= 210 \end{align*}\]
03

Calculate the Mean

The mean can be found by dividing the sum of all the products from Step 2 by the total frequency (45). First, find the sum of all the products and then divide:\[\text{Mean} = \frac{30 + 208 + 555 + 336 + 590 + 210}{45} = \frac{1929}{45} \approx 42.87\]
04

Identify the Modal Class

The modal class is the class with the highest frequency. From the given data, the class 32-42 has the highest frequency of 15. Therefore, the modal class is \(32-42\).

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Key Concepts

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

Frequency Distribution
A frequency distribution is a summary of how often each value or range of values occurs in a data set. In this exercise, we organized the net worth values of 45 corporations into intervals or classes, each with its own frequency. This method is incredibly useful when dealing with large data sets, as it allows us to see the overall pattern of the data. By grouping the data into classes such as "10-20", we simplify the analysis and visualization process. Each class has a frequency, representing how many data points fall within that range. This clarifies which ranges are densest, helping identify trends or outliers in the data set.
Modal Class
The modal class refers to the class interval within a frequency distribution that contains the highest frequency. Essentially, it's the range where most data points fall. Finding the modal class is about locating which class appears most frequently in the data. In this particular example, the class "32-42" is the modal class since it has the highest frequency, which is 15. Knowing the modal class provides insight into the most common range of the distribution, highlighting where concentrated data points lie.
Class Midpoint
The class midpoint, or class mark, is the average of the upper and lower limits of a class interval. It's a representative value for that class. We calculate it by adding the smallest and largest values of the class and dividing by two. For example, for the class "10-20," the midpoint is calculated as \((10 + 20) / 2 = 15\). Each class in the frequency distribution has its own midpoint, and these midpoints help in calculating the mean of the entire data set. Midpoints serve as estimates of the central value of data in each class, simplifying the computation of various statistics.
Step by Step Solution
Working through statistical problems often involves multiple steps, which should be clear and logical. In this exercise, the step-by-step solution involved calculating the class midpoints, multiplying them by their respective frequencies, summing these products, and finally, using this sum to calculate the mean. Here's the process in brief:
  • Compute midpoints for each class.
  • Multiply each midpoint by its class frequency.
  • Add all these products together.
  • Divide the total by the overall frequency to find the mean.
This systematic approach ensures accuracy and thorough understanding of how individual data classes contribute to collective measures like the mean.

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Most popular questions from this chapter

The average college student produces 640 pounds of solid waste each year. If the standard deviation is approximately 85 pounds, within what weight limits will at least 88.89% of all students鈥� garbage lie?

The average of the number of trials it took a sample of mice to learn to traverse a maze was 12. The standard deviation was 3. Using Chebyshev鈥檚 theorem, find the minimum percentage of data values that will fall in the range of 4鈥�20 trials.

In a study of reaction times to a specific stimulus, a psychologist recorded these data (in seconds). Find the variance and standard deviation for the data. \(\begin{array}{lc}{\text { Class limits }} & {\text { Frequency }} \\ \hline 2.1-2.7 & {12} \\ {2.8-3.4} & {13} \\ {3.5-4.1} & {7} \\ {4.2-4.8} & {5} \\\ {4.9-5.5} & {2} \\ {5.6-6.2} & {1}\end{array}\)

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 $$\text { Time }=\text { distance } \div \text { rate }$$ then $$\begin{array}{l}{\text { Time } 1=\frac{100}{40}=2.5 \text { hours to make the trip }} \\ {\text { Time } 2=\frac{100}{50}=2 \text { hours to return }}\end{array}$$ 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.

Another measure of the average is called the midquartile; it is the numerical value halfway between \(Q_{1}\) and \(Q_{3},\) and the formula is $$\text { Midquartile }=\frac{Q_{1}+Q_{3}}{2}$$ Using this formula and other formulas, find \(Q_{1},\) \(Q_{2}, Q_{3},\) the midquartile, and the interquartile range for each data set. a. \(5,12,16,25,32,38\) b. \(53,62,78,94,96,99,103\)
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