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Chapter 3: Problem 20

This frequency distribution represents the commission earned (in dollars) by 100 salespeople employed at several branches of a large chain store. Find the mean and modal class for the data. $$\begin{array}{lr}\text { Class limits } & \text { Frequency } \\\\\hline 150-158 & 5 \\\159-167 & 16 \\\168-176 & 20 \\\177-185 & 21 \\\186-194 & 20 \\\195-203 & 15 \\\204-212 & 3\end{array}$$

Short Answer

Expert verified
The mean commission is $180.28, and the modal class is 177-185.

Step by step solution

01

Identify the Midpoint for Each Class

To calculate the mean, you need the midpoint of each class. The midpoint is calculated by averaging the upper and lower limits of each class. For example, the first class 150-158 has a midpoint of \( (150 + 158) / 2 = 154 \). Repeat this for each class.Here are all the midpoints:- 150-158: 154- 159-167: 163- 168-176: 172- 177-185: 181- 186-194: 190- 195-203: 199- 204-212: 208
02

Calculate the Frequency-Midpoint Product

For each class, multiply the midpoint by the frequency. This product helps in calculating the mean.Here are the products:- 150-158: \( 154 \times 5 = 770 \)- 159-167: \( 163 \times 16 = 2608 \)- 168-176: \( 172 \times 20 = 3440 \)- 177-185: \( 181 \times 21 = 3801 \)- 186-194: \( 190 \times 20 = 3800 \)- 195-203: \( 199 \times 15 = 2985 \)- 204-212: \( 208 \times 3 = 624 \)
03

Sum All Frequency-Midpoint Products

Add up all the frequency-midpoint products to find the total sum needed for the mean. Total: \( 770 + 2608 + 3440 + 3801 + 3800 + 2985 + 624 = 18028 \)
04

Calculate the Mean

The mean is found by dividing the sum of the frequency-midpoint products by the total number of observations (100, in this case).Mean = \( \frac{18028}{100} = 180.28 \)
05

Determine the Modal Class

The modal class is the class with the highest frequency. Look for the highest number in the frequency column. From the table, the class 177-185 has the highest frequency of 21, making it the modal class.

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Key Concepts

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

Mean Calculation
The mean is an average that helps us understand the central value in a set of data. In frequency distribution, calculating the mean involves a few steps:
  • **Midpoint Calculation:** First, find the midpoint for each class by averaging the upper and lower class limits. For example, for the class 150-158, the midpoint is \( \frac{150 + 158}{2} = 154 \).
  • **Frequency-Midpoint Product:** Next, multiply each class's midpoint by its frequency. This gives you a product that reflects the contribution of each class to the overall mean.
  • **Summation:** Then, add up all these frequency-midpoint products to get a total sum.
  • **Mean Formula:** Finally, divide this sum by the total number of observations. In this dataset, the formula for the mean would be \( \text{Mean} = \frac{18028}{100} = 180.28 \).
All these steps ensure that the mean reflects the whole dataset, using the midpoint as a representative value for each class.
Modal Class
The modal class is the class interval with the highest frequency. It tells us which class has the most data points. Finding it is crucial in understanding the characteristics of the data distribution.
To identify the modal class, you simply look at the frequency column and find the highest value. This represents the class with the most data entries. In our example, the class 177-185 stands out with a frequency of 21. Therefore, this is our modal class, highlighting that most of the sales commissions fall within this range. The modal class gives insights into where data is concentrated in a distribution.
Midpoint Formula
The midpoint is crucial in summarizing class data, especially when dealing with grouped frequency distributions. It provides a representative value for each class and is used to simplify calculations like the mean.
The midpoint for a class is the average of its lower and upper limits. For the class 150-158, the midpoint calculation is \( \frac{150 + 158}{2} = 154 \). This value functions as a singular point that represents the entire class. Calculating the midpoint allows for streamlined data processing and bridges the gap between intervals and more accessible forms like single numbers.
  • Helps in computing the mean more effectively.
  • Simplifies the representation of a class interval.
Using midpoints is a way to simplify the complexity of grouped data in frequency distribution, making subsequent calculations clearer and more manageable.

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Most popular questions from this chapter

Pearson Coefficient of Skewness A measure to determine the skewness of a distribution is called the Pearson coefficient \((P C)\) of skewness. The formula is $$\mathrm{PC}=\frac{3(\bar{X}-\mathrm{MD})}{s}$$ The values of the coefficient usually range from -3 to +3 . When the distribution is symmetric, the coefficient is zero; when the distribution is positively skewed, it is positive; and when the distribution is negatively skewed, it is negative. Using the formula, find the coefficient of skewness for each distribution, and describe the shape of the distribution. a. Mean \(=10,\) median \(=8,\) standard deviation \(=3\) b. Mean \(=42,\) median \(=45,\) standard deviation \(=4\). c. Mean \(=18.6,\) median \(=18.6,\) standard deviation \(=1.5 .\) d. Mean \(=98,\) median \(=97.6,\) standard deviation \(=4\).

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.

In a distribution of 160 values with a mean of \(72,\) at least 120 fall within the interval \(67-77\). Approximately what percentage of values should fall in the interval \(62-82 ?\) Use Chebyshev's theorem.

The frequency distribution shows a sample of the waterfall heights, in feet, of 28 waterfalls. Find the variance and standard deviation for the data. $$ \begin{array}{rr} \text { Class boundaries } & \text { Frequency } \\ \hline 52.5-185.5 & 8 \\ 185.5-318.5 & 11 \\ 318.5-451.5 & 2 \\ 451.5-584.5 & 1 \\ 584.5-717.5 & 4 \\ 717.5-850.5 & 2 \end{array} $$

Another instructor gives four 1 -hour exams and one final exam, which counts as two 1 -hour exams. Find a student's grade if she received \(62,83,97,\) and 90 on the 1 -hour exams and 82 on the final exam.
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