Question 1

Provide an appropriate response.
-Histograms and Pareto charts are both bar charts. What is the significant difference between the two?

Accepted Answer

Histograms and Pareto charts are indeed both types of bar charts, but they serve different purposes and present data in distinct ways.

A histogram is a graphical representation of the distribution of numerical data. It is an estimate of the probability distribution of a continuous variable and is used to depict the underlying frequency distribution (shape) of a set of continuous data. This allows the viewer to see patterns that are difficult to detect in a simple table of numbers. The data is grouped into ranges (bins), and the frequency of data points within each range is represented by the height of the corresponding bar. The bars in a histogram are adjacent to each other, with no gaps, because the data is continuous.

Key characteristics of histograms:
- They display the distribution of a continuous variable over a set of intervals (bins).
- The bars are usually of equal width, representing the interval of the data.
- The height of each bar reflects the frequency or count of data points within each bin.
- Histograms do not have to be in any particular order as they represent intervals of the x-axis variable.

On the other hand, a Pareto chart is a type of chart that contains both bars and a line graph, where individual values are represented in descending order by bars, and the cumulative total is represented by the line. The purpose of the Pareto chart is to highlight the most important among a (typically large) set of factors. In terms of quality control, it is used to depict the most common causes of defects, the highest occurring type of defect, or the most frequent reasons for customer complaints, etc. The principle behind the Pareto chart is the Pareto principle or the 80/20 rule, which states that roughly 80% of the effects come from 20% of the causes.

Key characteristics of Pareto charts:
- They display categorical data (not continuous).
- The bars are arranged in descending order of height from left to right, so the most significant factors are seen first.
- The cumulative total is represented by a line graph, which can help identify the "vital few" from the "trivial many."
- The chart is used for prioritization and identifying the most significant factors in a set of data.

In summary, the significant difference between histograms and Pareto charts lies in the type of data they represent and their purpose: histograms are for continuous data and show the distribution of a variable, while Pareto charts are for categorical data and are used to prioritize factors contributing to a particular effect.

Question 2

Provide an appropriate response.
-One purpose of displaying data graphically is to provide clues about trends. The given values are weights (ounces)of steaks listed on a restaurant menu as "20 ounce porterhouse" steaks. The weights are supposed to be 21 ounces because they supposedly lose an ounce when cooked. Create a frequency distribution with 5 classes. Based on your distribution, comment on the advertised "20 ounce" steaks. $17 20 21 18 20 20 20 18 19 19 20 19 21 20 18 20 20 19 18 19$

Accepted Answer

The frequency distribution with 5 classes for the given weights of steaks is as follows:

17-18: 5
19-20: 9
21-22: 4

Based on this distribution, it is clear that the majority of the steaks actually weigh between 19-20 ounces, with only a few outliers at 17-18 and 21-22 ounces. This suggests that the advertised "20 ounce" steaks may actually be closer to the 21 ounce mark, as they are supposed to lose an ounce when cooked.

Question 3

Provide an appropriate response.
-Suppose you are comparing frequency data for two different groups, 25 managers and 150 blue collar workers. Why would a relative frequency distribution be better than a frequency distribution?

Accepted Answer

When comparing frequency data for two different groups of different sizes, such as 25 managers and 150 blue collar workers, a relative frequency distribution is often more informative than a simple frequency distribution. This is because a relative frequency distribution takes into account the size of the group when presenting the data, which allows for a more accurate comparison between groups of different sizes.

A frequency distribution simply counts the number of occurrences of each value or category within the data set. For example, if you were looking at the frequency of certain decisions made by managers versus blue collar workers, the frequency distribution would show the raw counts. However, because the group of blue collar workers is much larger (150) than the group of managers (25), the raw counts would likely be higher for the blue collar workers simply due to their greater number, not necessarily because their behavior or choices are inherently different.

On the other hand, a relative frequency distribution calculates the proportion or percentage of the total that each category represents. This is done by dividing the frequency of each category by the total number of observations in the group. By using relative frequencies, you can compare the behavior or choices of the two groups on a more equal footing, as each group's data is scaled according to its size.

For example, if 10 out of 25 managers (40%) made a certain decision, and 60 out of 150 blue collar workers (40%) made the same decision, the relative frequency distribution would show that the proportion is the same across both groups, despite the raw frequency being different. This allows for a fair comparison and can provide insights that might be overlooked if only raw frequencies were considered.

In summary, a relative frequency distribution is better than a frequency distribution for comparing groups of different sizes because it normalizes the data, allowing for a direct comparison of the proportions or percentages across the groups, which is a more accurate reflection of the underlying patterns or trends.

Question 4

Solve the problem.
-The following frequency distribution analyzes the scores on a math test. Find the indicated class midpoint or boundaries. $$\begin{array} { c c } 
\hline \text {score}& \begin{array} { c } 
\text { Number of } \
\text { Students }
\end{array} \
\hline 40 - 59 & 2 \
60 - 75 & 4 \
76 - 82 & 6 \
83 - 94 & 15 \
95 - 99 & 5
\end{array}$$ The class boundaries of scores interval 95-99
A)95.5, 99.5
B)95.5, 100.5
C)94.5, 99.5
D)94.5, 100.5

Accepted Answer

The answer of Solve the problem.
-The following frequency distribution analyzes...

Question 5

Solve the problem.
-The following frequency distribution analyzes the scores on a math test. Find the indicated class midpoint or boundaries. $$\begin{array} { c c } 
\hline \text {score}& \begin{array} { c } 
\text { Number of } \
\text { Students }
\end{array} \
\hline 40 - 59 & 2 \
60 - 75 & 4 \
76 - 82 & 6 \
83 - 94 & 15 \
95 - 99 & 5
\end{array}$$ The class midpoint of scores interval 95-99
A)96.5
B)97.0
C)97.5
D)98.0

Accepted Answer

The answer of Solve the problem.
-The following frequency distribution analyzes...

Question 6

Solve the problem.
-The following frequency distribution analyzes the scores on a math test. Find the indicated class midpoint or boundaries. 
$$\begin{array} { c c } 
\hline \text {score}& \begin{array} { c } 
\text { Number of } \
\text { Students }
\end{array} \
\hline 40 - 59 & 2 \
60 - 75 & 4 \
76 - 82 & 6 \
83 - 94 & 15 \
95 - 99 & 5
\end{array}$$ The class boundaries of scores interval 40-59
A)39.5, 59.5
B)40.5, 59.5
C)40.5, 58.5
D)39.5, 58.5

Accepted Answer

Class boundaries are calculated by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of the class interval. For the interval 40-59, the boundaries are 39.5 and 59.5.

Question 7

Solve the problem.
-The following frequency distribution analyzes the scores on a math test. Find the indicated class midpoint or boundaries. $$\begin{array} { c c } 
\hline \text {score}& \begin{array} { c } 
\text { Number of } \
\text { Students }
\end{array} \
\hline 40 - 59 & 2 \
60 - 75 & 4 \
76 - 82 & 6 \
83 - 94 & 15 \
95 - 99 & 5
\end{array}$$ The class midpoint of scores interval 40-59
A)48.5
B)50.5
C)49.0
D)49.5

Accepted Answer

The class midpoint is calculated by averaging the lower and upper boundaries of the class interval: (40 + 59) / 2 = 49.5.

Question 8

Solve the problem.
-Using the employment information in the table on Alpha Corporation, determine the width of each class. $$\begin{array}{l}
\text { Years employed at Alpha Corporation }\
\begin{array} { c | c } 
\begin{array} { c } 
\text { Class Limits } \
\text { (years of service) }
\end{array} & \begin{array} { c } 
\text { Frequency } \
\text { (No. of employees) }
\end{array} \
\hline 1 - 5 & 5 \
6 - 10 & 20 \
11 - 15 & 25 \
16 - 20 & 10 \
21 - 25 & 5 \
26 - 30 & 3
\end{array}
\end{array}$$
A)5
B)10
C)6
D)4

Accepted Answer

The width of each class is determined by subtracting the lower limit of a class from the upper limit and adding one. For the first class, 5 - 1 + 1 = 5. This calculation is consistent across all classes, indicating a class width of 5 years.

Question 9

Provide an appropriate response.
-Suppose that a data set has a minimum value of 28 and a max of 73 and that you want 5 classes. Explain how to find the class width for this frequency table. What happens if you mistakenly use a class width of 9 instead of 10?

Accepted Answer

The answer of Provide an appropriate response.
-Suppose that a data...

Question 10

Provide an appropriate response.
-Create an example displaying data in a pie chart. Display the same data in a Pareto chart. Which graph is more effective? List at least two reasons in support of your choice.

Accepted Answer

The answer of Provide an appropriate response.
-Create an example displaying...

Question 11

Provide an appropriate response.
-Describe at least two advantages to using stemplots rather than frequency distributions.

Accepted Answer

The answer of Provide an appropriate response.
-Describe at least two...

Question 12

Solve the problem.
-Using the information in the table on home sale prices in the city of Summerhill for the month of June, determine the width of each class. $$\begin{array} { c | c } 
 \text {Class Limits}& \text {Frequency}\
 \text {(Sale price in thousands)}& \text {(No. of homes sold)}\
\hline 80.0 - 110.9 & 2 \
111.0 - 141.9 & 5 \
142.0 - 172.9 & 7 \
173.0 - 203.9 & 10 \
204.0 - 234.9 & 3 \
235.0 - 265.9 & 1
\end{array}$$
A)31
B)28
C)61
D)30

Accepted Answer

The width of each class is calculated by subtracting the lower limit of a class from the lower limit of the next class: 111.0 - 80.0 = 31.

Question 13

Solve the problem.
-Using the employment information in the table on Alpha Corporation, find the class midpoint for class 1-5. $$\begin{array}{l}
\text { Years employed at Alpha Corporation }\
\begin{array} { c | c } 
\begin{array} { c } 
\text { Class Limits } \
\text { (years of service) }
\end{array} & \begin{array} { c } 
\text { Frequency } \
\text { (No. of employees) }
\end{array} \
\hline 1 - 5 & 5 \
6 - 10 & 20 \
11 - 15 & 25 \
16 - 20 & 10 \
21 - 25 & 5 \
26 - 30 & 3
\end{array}
\end{array}$$
A)3.0
B)3.5
C)5.0
D)2.5

Accepted Answer

The answer of Solve the problem.
-Using the employment information in...

Question 14

Construct the relative frequency distribution that corresponds to the given frequency distribution.
-$\begin{array}{r|c}
\text { Scores } & \text { Frequency } \
\hline 91-100 & 5 \
81-90 & 5 \
71-80 & 7 \
61-70 & 8 \
<61 & 4
\end{array}$
A)
$\begin{array}{r|r}
\text { Scores } & \begin{array}{r}
\text { Relative } \
\text { Frequency }
\end{array} \
\hline 91-100 & 12.5 \% \
81-90 & 20.1 \% \
71-80 & 37.3 \% \
61-70 & 15.2 \% \
<61 & 14.9 \%
\end{array}$
B)
$\begin{array}{r|r}
\text { Scores } & \begin{array}{r}
\text { Relative } \
\text { Frequency }
\end{array} \
\hline 91-100 & 15.5 \% \
81-90 & 22.1 \% \
71-80 & 31.3 \% \
61-70 & 16.2 \% \
<61 & 14.9 \%
\end{array}$
C)
$\begin{array}{r|r}
\text { Scores } & \begin{array}{r}
\text { Relative } \
\text { Frequency }
\end{array} \
\hline 91-100 & 17.24 \% \
81-90 & 17.24 \% \
71-80 & 24.14 \% \
61-70 & 27.59 \% \
<61 & 13.79 \%
\end{array}$
D)
$\begin{array}{r|r}
\text { Scores } & \begin{array}{r}
\text { Relative } \
\text { Frequency }
\end{array} \
\hline 91-100 & 0.28 \% \
81-90 & 0.07 \% \
71-80 & 0.41 \% \
61-70 & 0.17 \% \
<61 & 0.07 \%
\end{array}$

Accepted Answer

To find the relative frequency, divide each frequency by the total number of data points (5+5+7+8+4=29) and multiply by 100 to get percentages. For example, for 91-100: (5/29)*100 ≈ 17.24%. This calculation matches the percentages in option C.

Question 15

Provide an appropriate response.
-Draw one histogram to illustrate bell-shaped data, another for uniform data, and a third for skewed data. Which of these shapes matches the histogram for the first 100 digits of π? Does changing the number of classes alter your answer? (Below is the frequency table for the first 100 digits of π.) $\begin{array}{r|r|r|r|r|r|r|r|r|r}
0 & 1 & 2 & 3 & 4 & 5 & 6 &7& 8 & 9 \
\hline 8 & 8 & 12 & 11 & 10 & 8 & 9 & 8 & 12 & 14
\end{array}$

Accepted Answer

The answer of Provide an appropriate response.
-Draw one histogram to...

Question 16

Construct a boxplot for the given data. Include values of the 5-number summary in all boxplots.
-$\begin{array}{r|r}
\text { Incomes } & \text { Frequency } \
\hline 200-300 & 75 \
301-400 & 51 \
401-500 & 95 \
501-600 & 92 \
>600 & 14
\end{array}$
A)
$\begin{array}{r|r}
\text { Incomes } & \begin{array}{r}
\text { Relative } \
\text { Frequency }
\end{array} \
\hline 200-300 & 22.94 \% \
301-400 & 15.60 \% \
401-500 & 29.05 \% \
501-600 & 28.13 \% \
>600 & 4.28 \%
\end{array}$
B)
$\begin{array}{r|r}
\text { Incomes } & \begin{array}{r}
\text { Relative } \
\text { Frequency }
\end{array} \
\hline 200-300 & 28.61 \% \
301-400 & 28.40 \% \
401-500 & 4.31 \% \
501-600 & 22.66 \% \
>600 & 28.70 \%
\end{array}$
C)
$\begin{array}{r|r}
\text { Incomes } & \begin{array}{r}
\text { Relative } \
\text { Frequency }
\end{array} \
\hline 200-300 & 12.5 \% \
301-400 & 20.1 \% \
401-500 & 37.3 \% \
501-600 & 15.2 \% \
>600 & 14.9 \%
\end{array}$
D)
$\begin{array}{r|c}
\text { Incomes } & \begin{array}{r}
\text { Relative } \
\text { Frequency }
\end{array} \
\hline 201-300 & 15.5 \% \
301-400 & 22.1 \% \
401-500 & 31.3 \% \
501-600 & 16.2 \% \
>600 & 14.9 \%
\end{array}$

Accepted Answer

To construct a boxplot, the relative frequencies are not directly relevant, but the correct calculation of relative frequencies can indicate an understanding of the data distribution. The correct relative frequencies are calculated by dividing the frequency of each income range by the total number of data points (sum of all frequencies) and then multiplying by 100 to get percentages. The total number of data points is $75 + 51 + 95 + 92 + 14 = 327$. For the 200-300 income range, for example, the relative frequency is $\frac{75}{327} \times 100 \approx 22.94\%$, which matches the values in option A, indicating the calculations there are correct.

Question 17

Solve the problem.
-Using the information in the table on home sale prices in the city of Summerhill for the month of June, find the class midpoint for class 235.0-265.9. $$\begin{array} { c | c } 
 \text {Class Limits}& \text {Frequency}\
 \text {(Sale price in thousands)}& \text {(No. of homes sold)}\
\hline 80.0 - 110.9 & 2 \
111.0 - 141.9 & 5 \
142.0 - 172.9 & 7 \
173.0 - 203.9 & 10 \
204.0 - 234.9 & 3 \
235.0 - 265.9 & 1
\end{array}$$
A)250.55
B)250.50
C)250.40
D)250.45

Accepted Answer

The answer of Solve the problem.
-Using the information in the...

Question 18

Solve the problem.
-Using the employment information in the table on Alpha Corporation, find the class boundaries for class 26-30. $$\begin{array}{l}
\text { Years employed at Alpha Corporation }\
\begin{array} { c | c } 
\begin{array} { c } 
\text { Class Limits } \
\text { (years of service) }
\end{array} & \begin{array} { c } 
\text { Frequency } \
\text { (No. of employees) }
\end{array} \
\hline 1 - 5 & 5 \
6 - 10 & 20 \
11 - 15 & 25 \
16 - 20 & 10 \
21 - 25 & 5 \
26 - 30 & 3
\end{array}
\end{array}$$
A)26.5, 30.5
B)25.5, 30.5
C)25.5, 20.5
D)26.5, 29.5

Accepted Answer

The class boundaries for the class 26-30 are found by adding and subtracting 0.5 to the class limits to avoid overlap with adjacent classes. Thus, the boundaries are 25.5 and 30.5.

Question 19

Construct a boxplot for the given data. Include values of the 5-number summary in all boxplots.
-Using the information in the table on home sale prices in the city of Summerhill for the month of June, find the class boundaries for class 80.0-110.9. $$\begin{array} { c | c } 
 \text {Class Limits}& \text {Frequency}\
 \text {(Sale price in thousands)}& \text {(No. of homes sold)}\
\hline 80.0 - 110.9 & 2 \
111.0 - 141.9 & 5 \
142.0 - 172.9 & 7 \
173.0 - 203.9 & 10 \
204.0 - 234.9 & 3 \
235.0 - 265.9 & 1
\end{array}$$
A)79.90, 110.95
B)80.00, 110.95
C)79.95, 110.95
D)79.90, 111.0

Accepted Answer

Class boundaries are calculated by finding the midpoint between the upper limit of one class and the lower limit of the next class. For the class 80.0-110.9, the lower boundary is halfway between 80.0 and the next lower value (which isn't given, but assuming a consistent class width, it would be 79.95), and the upper boundary is halfway between 110.9 and 111.0, which is 110.95.

Question 20

Provide an appropriate response.
-Explain the difference between a frequency distribution and a relative frequency distribution. Comment on the differences on the vertical axis scale. Given the same data set and the same classes, will the shapes of the frequency distribution and the relative frequency distribution be the same? You may draw a diagram to support your answer.

Accepted Answer

The answer of Provide an appropriate response.
-Explain the difference between...

Quiz 2: Summarizing and Graphing Data

Provide an appropriate response. -Histograms and Pareto charts are both bar charts. What is the significant difference between the two?

Provide an appropriate response. -Suppose you are comparing frequency data for two different groups, 25 managers and 150 blue collar workers. Why would a relative frequency distribution be better than a frequency distribution?

Provide an appropriate response. -Suppose that a data set has a minimum value of 28 and a max of 73 and that you want 5 classes. Explain how to find the class width for this frequency table. What happens if you mistakenly use a class width of 9 instead of 10?

Provide an appropriate response. -Create an example displaying data in a pie chart. Display the same data in a Pareto chart. Which graph is more effective? List at least two reasons in support of your choice.

Provide an appropriate response. -Describe at least two advantages to using stemplots rather than frequency distributions.

Construct the relative frequency distribution that corresponds to the given frequency distribution. - $\begin{array}{r|c}\text { Scores } & \text { Frequency } \\\hline 91-100 & 5 \\81-90 & 5 \\71-80 & 7 \\61-70 & 8 \\<61 & 4\end{array}$

Construct a boxplot for the given data. Include values of the 5-number summary in all boxplots. - $\begin{array}{r|r}\text { Incomes } & \text { Frequency } \\\hline 200-300 & 75 \\301-400 & 51 \\401-500 & 95 \\501-600 & 92 \\>600 & 14\end{array}$