To graph: A stem-and-leaf plot for the data of the age distribution.

Question

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Answer 1

Question

Chapter 2, Problem 8CR

(a)

To determine

To graph: A stem-and-leaf plot for the data of the age distribution.

(a)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct stem-and-leaf plot by using the Minitab, the steps are as follows:

Step 1: Enter the data in C1.

Step 2: Go to Graph > Stem-and-Leaf plot and select ‘C1’ in Graph variable.

Step 3: Click on OK.

The Stem-and-Leaf plot for these data is obtained as:

Stem-and-leaf of the age distribution N = 50
	1	6 8
	2	0 1 1 2 2 2 3 4 4 5 6 6 6 7 7 7 9
	3	0 0 1 1 2 3 4 4 5 5 6 7 8 9
	4	0 0 1 3 5 6 7 7 9 9
	5	1 3 5 6 8
	6	3 4

(b)

To determine

To find: The frequency table for the data..

(b)

Expert Solution

Answer to Problem 8CR

Solution: The complete frequency table is as:

Class Limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	19
30-36	29.5-36.5	33	11	0.22	30
37-43	36.5-43.5	40	7	0.14	37
44-50	43.5-50.5	47	6	0.12	43
51-57	50.5-57.5	54	4	0.08	47
58-64	57.5-64.5	61	3	0.06	50

Explanation of Solution

Calculation: To find the class width for the whole data of 50 values, it is observed that largest value of the data set is 64 and the smallest value is 16 in the data. Using 7 classes, the class width calculated in the following way:

Class width=Largest data value - smallest data valueDesired number of classes

Class width=64−167=6.9≈7

The value is round up to the nearest whole number. Hence, the class width of the data set is 7. The class width for the data is 7 and the lowest data value (16) will be the lower class limit of the first class. Because the class width is 7, it must add 7 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 7 desired classes. Hence, the class limits are 16-22, 23-29, 30-36, 37-43, 44-50, 51-57, and 58-64. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 15.5-22.5, 22.5-29.5, 29.5-36.5, 36.5-43.5, 43.5-50.5, 50.5-57.5, and 57.5-64.5.

Next to find the midpoint of the class is calculated by using formula,

Midpoint=Lower class limit + Upper class limit2

Midpoint of first class is calculated as:

Midpoint=15.5+ 22.52=19

The frequencies for respective classes are 8, 11, 11, 7, 6, 4 and 3.

Relative frequency is calculated by using the formula,

Relative frequency=fn=frequencyTotal of all frequencies

The frequency for 1^st class is 8 and total frequencies are 50 so the relative frequency is 850=0.16(Same calculation for other class).

The calculated frequency table is as follows:

Class limits	Class boundaries	midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Interpretation: Hence, the complete frequency table is as:

Class limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

(c)

To determine

To graph: A histogram for the data of the age distribution.

(c)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct the histogram by using the MINITAB, the steps are as follows:

Step 1: Enter the class boundaries in C1 and frequency in C2.

Step 2: Go to Graph > Histogram > Simple.

Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.

Step 4: Click on OK.

The obtained histogram is

UNDERSTANDING BASIC STAT LL BUND >A< F, Chapter 2, Problem 8CR

(d)

To determine

The shape of the histogram of age distribution..

(d)

Expert Solution

Answer to Problem 8CR

Solution: The shape of histogram of age distribution is skewed to the right.

Explanation of Solution

A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions. That’s because there is a long tail in the positive direction on the number line.

From above histogram, there are two class boundaries (22.5-29.5 and 29.5-36) have higher frequencies 11 on left side and most of the data values fall on the left side of the graph. The data to lean towards right side of the graph and also there is tail on right side.

Hence, the shape of histogram of age distribution is skewed to the right.

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Answer 2

Question

Chapter 2, Problem 8CR

(a)

To determine

To graph: A stem-and-leaf plot for the data of the age distribution.

(a)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct stem-and-leaf plot by using the Minitab, the steps are as follows:

Step 1: Enter the data in C1.

Step 2: Go to Graph > Stem-and-Leaf plot and select ‘C1’ in Graph variable.

Step 3: Click on OK.

The Stem-and-Leaf plot for these data is obtained as:

Stem-and-leaf of the age distribution N = 50
	1	6 8
	2	0 1 1 2 2 2 3 4 4 5 6 6 6 7 7 7 9
	3	0 0 1 1 2 3 4 4 5 5 6 7 8 9
	4	0 0 1 3 5 6 7 7 9 9
	5	1 3 5 6 8
	6	3 4

(b)

To determine

To find: The frequency table for the data..

(b)

Expert Solution

Answer to Problem 8CR

Solution: The complete frequency table is as:

Class Limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	19
30-36	29.5-36.5	33	11	0.22	30
37-43	36.5-43.5	40	7	0.14	37
44-50	43.5-50.5	47	6	0.12	43
51-57	50.5-57.5	54	4	0.08	47
58-64	57.5-64.5	61	3	0.06	50

Explanation of Solution

Calculation: To find the class width for the whole data of 50 values, it is observed that largest value of the data set is 64 and the smallest value is 16 in the data. Using 7 classes, the class width calculated in the following way:

Class width=Largest data value - smallest data valueDesired number of classes

Class width=64−167=6.9≈7

The value is round up to the nearest whole number. Hence, the class width of the data set is 7. The class width for the data is 7 and the lowest data value (16) will be the lower class limit of the first class. Because the class width is 7, it must add 7 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 7 desired classes. Hence, the class limits are 16-22, 23-29, 30-36, 37-43, 44-50, 51-57, and 58-64. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 15.5-22.5, 22.5-29.5, 29.5-36.5, 36.5-43.5, 43.5-50.5, 50.5-57.5, and 57.5-64.5.

Next to find the midpoint of the class is calculated by using formula,

Midpoint=Lower class limit + Upper class limit2

Midpoint of first class is calculated as:

Midpoint=15.5+ 22.52=19

The frequencies for respective classes are 8, 11, 11, 7, 6, 4 and 3.

Relative frequency is calculated by using the formula,

Relative frequency=fn=frequencyTotal of all frequencies

The frequency for 1^st class is 8 and total frequencies are 50 so the relative frequency is 850=0.16(Same calculation for other class).

The calculated frequency table is as follows:

Class limits	Class boundaries	midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Interpretation: Hence, the complete frequency table is as:

Class limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

(c)

To determine

To graph: A histogram for the data of the age distribution.

(c)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct the histogram by using the MINITAB, the steps are as follows:

Step 1: Enter the class boundaries in C1 and frequency in C2.

Step 2: Go to Graph > Histogram > Simple.

Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.

Step 4: Click on OK.

The obtained histogram is

UNDERSTANDING BASIC STAT LL BUND >A< F, Chapter 2, Problem 8CR

(d)

To determine

The shape of the histogram of age distribution..

(d)

Expert Solution

Answer to Problem 8CR

Solution: The shape of histogram of age distribution is skewed to the right.

Explanation of Solution

A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions. That’s because there is a long tail in the positive direction on the number line.

From above histogram, there are two class boundaries (22.5-29.5 and 29.5-36) have higher frequencies 11 on left side and most of the data values fall on the left side of the graph. The data to lean towards right side of the graph and also there is tail on right side.

Hence, the shape of histogram of age distribution is skewed to the right.

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Answer 3

Question

Chapter 2, Problem 8CR

(a)

To determine

To graph: A stem-and-leaf plot for the data of the age distribution.

(a)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct stem-and-leaf plot by using the Minitab, the steps are as follows:

Step 1: Enter the data in C1.

Step 2: Go to Graph > Stem-and-Leaf plot and select ‘C1’ in Graph variable.

Step 3: Click on OK.

The Stem-and-Leaf plot for these data is obtained as:

Stem-and-leaf of the age distribution N = 50
	1	6 8
	2	0 1 1 2 2 2 3 4 4 5 6 6 6 7 7 7 9
	3	0 0 1 1 2 3 4 4 5 5 6 7 8 9
	4	0 0 1 3 5 6 7 7 9 9
	5	1 3 5 6 8
	6	3 4

(b)

To determine

To find: The frequency table for the data..

(b)

Expert Solution

Answer to Problem 8CR

Solution: The complete frequency table is as:

Class Limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	19
30-36	29.5-36.5	33	11	0.22	30
37-43	36.5-43.5	40	7	0.14	37
44-50	43.5-50.5	47	6	0.12	43
51-57	50.5-57.5	54	4	0.08	47
58-64	57.5-64.5	61	3	0.06	50

Explanation of Solution

Calculation: To find the class width for the whole data of 50 values, it is observed that largest value of the data set is 64 and the smallest value is 16 in the data. Using 7 classes, the class width calculated in the following way:

Class width=Largest data value - smallest data valueDesired number of classes

Class width=64−167=6.9≈7

The value is round up to the nearest whole number. Hence, the class width of the data set is 7. The class width for the data is 7 and the lowest data value (16) will be the lower class limit of the first class. Because the class width is 7, it must add 7 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 7 desired classes. Hence, the class limits are 16-22, 23-29, 30-36, 37-43, 44-50, 51-57, and 58-64. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 15.5-22.5, 22.5-29.5, 29.5-36.5, 36.5-43.5, 43.5-50.5, 50.5-57.5, and 57.5-64.5.

Next to find the midpoint of the class is calculated by using formula,

Midpoint=Lower class limit + Upper class limit2

Midpoint of first class is calculated as:

Midpoint=15.5+ 22.52=19

The frequencies for respective classes are 8, 11, 11, 7, 6, 4 and 3.

Relative frequency is calculated by using the formula,

Relative frequency=fn=frequencyTotal of all frequencies

The frequency for 1^st class is 8 and total frequencies are 50 so the relative frequency is 850=0.16(Same calculation for other class).

The calculated frequency table is as follows:

Class limits	Class boundaries	midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Interpretation: Hence, the complete frequency table is as:

Class limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

(c)

To determine

To graph: A histogram for the data of the age distribution.

(c)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct the histogram by using the MINITAB, the steps are as follows:

Step 1: Enter the class boundaries in C1 and frequency in C2.

Step 2: Go to Graph > Histogram > Simple.

Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.

Step 4: Click on OK.

The obtained histogram is

UNDERSTANDING BASIC STAT LL BUND >A< F, Chapter 2, Problem 8CR

(d)

To determine

The shape of the histogram of age distribution..

(d)

Expert Solution

Answer to Problem 8CR

Solution: The shape of histogram of age distribution is skewed to the right.

Explanation of Solution

A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions. That’s because there is a long tail in the positive direction on the number line.

From above histogram, there are two class boundaries (22.5-29.5 and 29.5-36) have higher frequencies 11 on left side and most of the data values fall on the left side of the graph. The data to lean towards right side of the graph and also there is tail on right side.

Hence, the shape of histogram of age distribution is skewed to the right.

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Answer 4

Question

Chapter 2, Problem 8CR

(a)

To determine

To graph: A stem-and-leaf plot for the data of the age distribution.

(a)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct stem-and-leaf plot by using the Minitab, the steps are as follows:

Step 1: Enter the data in C1.

Step 2: Go to Graph > Stem-and-Leaf plot and select ‘C1’ in Graph variable.

Step 3: Click on OK.

The Stem-and-Leaf plot for these data is obtained as:

Stem-and-leaf of the age distribution N = 50
	1	6 8
	2	0 1 1 2 2 2 3 4 4 5 6 6 6 7 7 7 9
	3	0 0 1 1 2 3 4 4 5 5 6 7 8 9
	4	0 0 1 3 5 6 7 7 9 9
	5	1 3 5 6 8
	6	3 4

(b)

To determine

To find: The frequency table for the data..

(b)

Expert Solution

Answer to Problem 8CR

Solution: The complete frequency table is as:

Class Limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	19
30-36	29.5-36.5	33	11	0.22	30
37-43	36.5-43.5	40	7	0.14	37
44-50	43.5-50.5	47	6	0.12	43
51-57	50.5-57.5	54	4	0.08	47
58-64	57.5-64.5	61	3	0.06	50

Explanation of Solution

Calculation: To find the class width for the whole data of 50 values, it is observed that largest value of the data set is 64 and the smallest value is 16 in the data. Using 7 classes, the class width calculated in the following way:

Class width=Largest data value - smallest data valueDesired number of classes

Class width=64−167=6.9≈7

The value is round up to the nearest whole number. Hence, the class width of the data set is 7. The class width for the data is 7 and the lowest data value (16) will be the lower class limit of the first class. Because the class width is 7, it must add 7 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 7 desired classes. Hence, the class limits are 16-22, 23-29, 30-36, 37-43, 44-50, 51-57, and 58-64. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 15.5-22.5, 22.5-29.5, 29.5-36.5, 36.5-43.5, 43.5-50.5, 50.5-57.5, and 57.5-64.5.

Next to find the midpoint of the class is calculated by using formula,

Midpoint=Lower class limit + Upper class limit2

Midpoint of first class is calculated as:

Midpoint=15.5+ 22.52=19

The frequencies for respective classes are 8, 11, 11, 7, 6, 4 and 3.

Relative frequency is calculated by using the formula,

Relative frequency=fn=frequencyTotal of all frequencies

The frequency for 1^st class is 8 and total frequencies are 50 so the relative frequency is 850=0.16(Same calculation for other class).

The calculated frequency table is as follows:

Class limits	Class boundaries	midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Interpretation: Hence, the complete frequency table is as:

Class limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

(c)

To determine

To graph: A histogram for the data of the age distribution.

(c)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct the histogram by using the MINITAB, the steps are as follows:

Step 1: Enter the class boundaries in C1 and frequency in C2.

Step 2: Go to Graph > Histogram > Simple.

Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.

Step 4: Click on OK.

The obtained histogram is

UNDERSTANDING BASIC STAT LL BUND >A< F, Chapter 2, Problem 8CR

(d)

To determine

The shape of the histogram of age distribution..

(d)

Expert Solution

Answer to Problem 8CR

Solution: The shape of histogram of age distribution is skewed to the right.

Explanation of Solution

A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions. That’s because there is a long tail in the positive direction on the number line.

From above histogram, there are two class boundaries (22.5-29.5 and 29.5-36) have higher frequencies 11 on left side and most of the data values fall on the left side of the graph. The data to lean towards right side of the graph and also there is tail on right side.

Hence, the shape of histogram of age distribution is skewed to the right.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 2, Problem 8CR

(a)

To determine

To graph: A stem-and-leaf plot for the data of the age distribution.

(a)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct stem-and-leaf plot by using the Minitab, the steps are as follows:

Step 1: Enter the data in C1.

Step 2: Go to Graph > Stem-and-Leaf plot and select ‘C1’ in Graph variable.

Step 3: Click on OK.

The Stem-and-Leaf plot for these data is obtained as:

Stem-and-leaf of the age distribution N = 50
	1	6 8
	2	0 1 1 2 2 2 3 4 4 5 6 6 6 7 7 7 9
	3	0 0 1 1 2 3 4 4 5 5 6 7 8 9
	4	0 0 1 3 5 6 7 7 9 9
	5	1 3 5 6 8
	6	3 4

(b)

To determine

To find: The frequency table for the data..

(b)

Expert Solution

Answer to Problem 8CR

Solution: The complete frequency table is as:

Class Limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	19
30-36	29.5-36.5	33	11	0.22	30
37-43	36.5-43.5	40	7	0.14	37
44-50	43.5-50.5	47	6	0.12	43
51-57	50.5-57.5	54	4	0.08	47
58-64	57.5-64.5	61	3	0.06	50

Explanation of Solution

Calculation: To find the class width for the whole data of 50 values, it is observed that largest value of the data set is 64 and the smallest value is 16 in the data. Using 7 classes, the class width calculated in the following way:

Class width=Largest data value - smallest data valueDesired number of classes

Class width=64−167=6.9≈7

The value is round up to the nearest whole number. Hence, the class width of the data set is 7. The class width for the data is 7 and the lowest data value (16) will be the lower class limit of the first class. Because the class width is 7, it must add 7 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 7 desired classes. Hence, the class limits are 16-22, 23-29, 30-36, 37-43, 44-50, 51-57, and 58-64. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 15.5-22.5, 22.5-29.5, 29.5-36.5, 36.5-43.5, 43.5-50.5, 50.5-57.5, and 57.5-64.5.

Next to find the midpoint of the class is calculated by using formula,

Midpoint=Lower class limit + Upper class limit2

Midpoint of first class is calculated as:

Midpoint=15.5+ 22.52=19

The frequencies for respective classes are 8, 11, 11, 7, 6, 4 and 3.

Relative frequency is calculated by using the formula,

Relative frequency=fn=frequencyTotal of all frequencies

The frequency for 1^st class is 8 and total frequencies are 50 so the relative frequency is 850=0.16(Same calculation for other class).

The calculated frequency table is as follows:

Class limits	Class boundaries	midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Interpretation: Hence, the complete frequency table is as:

Class limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

(c)

To determine

To graph: A histogram for the data of the age distribution.

(c)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct the histogram by using the MINITAB, the steps are as follows:

Step 1: Enter the class boundaries in C1 and frequency in C2.

Step 2: Go to Graph > Histogram > Simple.

Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.

Step 4: Click on OK.

The obtained histogram is

UNDERSTANDING BASIC STAT LL BUND >A< F, Chapter 2, Problem 8CR

(d)

To determine

The shape of the histogram of age distribution..

(d)

Expert Solution

Answer to Problem 8CR

Solution: The shape of histogram of age distribution is skewed to the right.

Explanation of Solution

A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions. That’s because there is a long tail in the positive direction on the number line.

From above histogram, there are two class boundaries (22.5-29.5 and 29.5-36) have higher frequencies 11 on left side and most of the data values fall on the left side of the graph. The data to lean towards right side of the graph and also there is tail on right side.

Hence, the shape of histogram of age distribution is skewed to the right.

Answer 6

Chapter 2, Problem 8CR

(a)

To determine

To graph: A stem-and-leaf plot for the data of the age distribution.

(a)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct stem-and-leaf plot by using the Minitab, the steps are as follows:

Step 1: Enter the data in C1.

Step 2: Go to Graph > Stem-and-Leaf plot and select ‘C1’ in Graph variable.

Step 3: Click on OK.

The Stem-and-Leaf plot for these data is obtained as:

Stem-and-leaf of the age distribution N = 50
	1	6 8
	2	0 1 1 2 2 2 3 4 4 5 6 6 6 7 7 7 9
	3	0 0 1 1 2 3 4 4 5 5 6 7 8 9
	4	0 0 1 3 5 6 7 7 9 9
	5	1 3 5 6 8
	6	3 4

Answer 7

Chapter 2, Problem 8CR

(a)

To determine

To graph: A stem-and-leaf plot for the data of the age distribution.

Answer 8

(a)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct stem-and-leaf plot by using the Minitab, the steps are as follows:

Step 1: Enter the data in C1.

Step 2: Go to Graph > Stem-and-Leaf plot and select ‘C1’ in Graph variable.

Step 3: Click on OK.

The Stem-and-Leaf plot for these data is obtained as:

Stem-and-leaf of the age distribution N = 50
	1	6 8
	2	0 1 1 2 2 2 3 4 4 5 6 6 6 7 7 7 9
	3	0 0 1 1 2 3 4 4 5 5 6 7 8 9
	4	0 0 1 3 5 6 7 7 9 9
	5	1 3 5 6 8
	6	3 4

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

To find: The frequency table for the data..

(b)

Expert Solution

Answer to Problem 8CR

Solution: The complete frequency table is as:

Class Limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	19
30-36	29.5-36.5	33	11	0.22	30
37-43	36.5-43.5	40	7	0.14	37
44-50	43.5-50.5	47	6	0.12	43
51-57	50.5-57.5	54	4	0.08	47
58-64	57.5-64.5	61	3	0.06	50

Explanation of Solution

Calculation: To find the class width for the whole data of 50 values, it is observed that largest value of the data set is 64 and the smallest value is 16 in the data. Using 7 classes, the class width calculated in the following way:

Class width=Largest data value - smallest data valueDesired number of classes

Class width=64−167=6.9≈7

The value is round up to the nearest whole number. Hence, the class width of the data set is 7. The class width for the data is 7 and the lowest data value (16) will be the lower class limit of the first class. Because the class width is 7, it must add 7 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 7 desired classes. Hence, the class limits are 16-22, 23-29, 30-36, 37-43, 44-50, 51-57, and 58-64. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 15.5-22.5, 22.5-29.5, 29.5-36.5, 36.5-43.5, 43.5-50.5, 50.5-57.5, and 57.5-64.5.

Next to find the midpoint of the class is calculated by using formula,

Midpoint=Lower class limit + Upper class limit2

Midpoint of first class is calculated as:

Midpoint=15.5+ 22.52=19

The frequencies for respective classes are 8, 11, 11, 7, 6, 4 and 3.

Relative frequency is calculated by using the formula,

Relative frequency=fn=frequencyTotal of all frequencies

The frequency for 1^st class is 8 and total frequencies are 50 so the relative frequency is 850=0.16(Same calculation for other class).

The calculated frequency table is as follows:

Class limits	Class boundaries	midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Interpretation: Hence, the complete frequency table is as:

Class limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Answer 13

(b)

To determine

To find: The frequency table for the data..

Answer 14

(b)

Expert Solution

Answer to Problem 8CR

Solution: The complete frequency table is as:

Class Limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	19
30-36	29.5-36.5	33	11	0.22	30
37-43	36.5-43.5	40	7	0.14	37
44-50	43.5-50.5	47	6	0.12	43
51-57	50.5-57.5	54	4	0.08	47
58-64	57.5-64.5	61	3	0.06	50

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

Calculation: To find the class width for the whole data of 50 values, it is observed that largest value of the data set is 64 and the smallest value is 16 in the data. Using 7 classes, the class width calculated in the following way:

Class width=Largest data value - smallest data valueDesired number of classes

Class width=64−167=6.9≈7

The value is round up to the nearest whole number. Hence, the class width of the data set is 7. The class width for the data is 7 and the lowest data value (16) will be the lower class limit of the first class. Because the class width is 7, it must add 7 to the lowest class limit in the first class to find the lowest class limit in the second class. There are 7 desired classes. Hence, the class limits are 16-22, 23-29, 30-36, 37-43, 44-50, 51-57, and 58-64. Now, to find the class boundaries subtract 0.5 from lower limit of every class and add 0.5 to the upper limit of the every class interval. Hence, the class boundaries are 15.5-22.5, 22.5-29.5, 29.5-36.5, 36.5-43.5, 43.5-50.5, 50.5-57.5, and 57.5-64.5.

Next to find the midpoint of the class is calculated by using formula,

Midpoint=Lower class limit + Upper class limit2

Midpoint of first class is calculated as:

Midpoint=15.5+ 22.52=19

The frequencies for respective classes are 8, 11, 11, 7, 6, 4 and 3.

Relative frequency is calculated by using the formula,

Relative frequency=fn=frequencyTotal of all frequencies

The frequency for 1^st class is 8 and total frequencies are 50 so the relative frequency is 850=0.16(Same calculation for other class).

The calculated frequency table is as follows:

Class limits	Class boundaries	midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Interpretation: Hence, the complete frequency table is as:

Class limits	Class boundaries	Midpoints	Frequency	Relative Frequency	Cumulative Frequency
16-22	15.5-22.5	19	8	0.16	8
23-29	22.5-29.5	26	11	0.22	8+11=19
30-36	29.5-36.5	33	11	0.22	19+11=30
37-43	36.5-43.5	40	7	0.14	30+7=37
44-50	43.5-50.5	47	6	0.12	37+6=43
51-57	50.5-57.5	54	4	0.08	43+4=47
58-64	57.5-64.5	61	3	0.06	47+3=50

Answer 19

(c)

To determine

To graph: A histogram for the data of the age distribution.

(c)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct the histogram by using the MINITAB, the steps are as follows:

Step 1: Enter the class boundaries in C1 and frequency in C2.

Step 2: Go to Graph > Histogram > Simple.

Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.

Step 4: Click on OK.

The obtained histogram is

UNDERSTANDING BASIC STAT LL BUND >A< F, Chapter 2, Problem 8CR

Answer 20

(c)

To determine

To graph: A histogram for the data of the age distribution.

Answer 21

(c)

Expert Solution

Explanation of Solution

The data shows the ages of 50 drivers arrested while driver under the influence of alcohol.

Graph: To construct the histogram by using the MINITAB, the steps are as follows:

Step 1: Enter the class boundaries in C1 and frequency in C2.

Step 2: Go to Graph > Histogram > Simple.

Step 3: Enter C1 in Graph variable then go to Data options > Frequency > C2.

Step 4: Click on OK.

The obtained histogram is

UNDERSTANDING BASIC STAT LL BUND >A< F, Chapter 2, Problem 8CR

Answer 22

(c)

Expert Solution

Answer 23

(c)

Expert Solution

Answer 24

Expert Solution

Answer 25

(d)

To determine

The shape of the histogram of age distribution..

(d)

Expert Solution

Answer to Problem 8CR

Solution: The shape of histogram of age distribution is skewed to the right.

Explanation of Solution

A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions. That’s because there is a long tail in the positive direction on the number line.

From above histogram, there are two class boundaries (22.5-29.5 and 29.5-36) have higher frequencies 11 on left side and most of the data values fall on the left side of the graph. The data to lean towards right side of the graph and also there is tail on right side.

Hence, the shape of histogram of age distribution is skewed to the right.

Answer 26

(d)

To determine

The shape of the histogram of age distribution..

Answer 27

(d)

Expert Solution

Answer to Problem 8CR

Solution: The shape of histogram of age distribution is skewed to the right.

Answer 28

(d)

Expert Solution

Answer 29

(d)

Expert Solution

Answer 30

Expert Solution

Answer 31

Explanation of Solution

A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions. That’s because there is a long tail in the positive direction on the number line.

From above histogram, there are two class boundaries (22.5-29.5 and 29.5-36) have higher frequencies 11 on left side and most of the data values fall on the left side of the graph. The data to lean towards right side of the graph and also there is tail on right side.

Hence, the shape of histogram of age distribution is skewed to the right.