The class width for the given frequency distribution.

Question

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Answer 1

Question

Chapter 2.1, Problem 1E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 1E

Solution:

The class width of the given frequency distribution is 4.8 which is approximately equal to 5.

Explanation of Solution

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

The upper limit of the distribution is given as 39 and the lower limit is given as 15. The number of classes is 5.

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute 39 for upper limit and 15 for lower limit and 5 for the number of classes in the formula,

Class width=39−155=245=4.8≈5

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

(b)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Explanation of Solution

Formula used:

The formula to calculate the class boundary is,

Class boundary=Upper limit of one class+Lower limit of next class2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since the largest number is 19 and 20 is the smallest number in the next class, then, substitute 19 for upper limit and 20 for lower limit in the formula,

Class boundary=19+202=392=19.5

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Explanation of Solution

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute 19 for upper limit and 15 for lower limit in the formula,

Midpoint=15+192=342=17

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

(d)

Expert Solution

Answer to Problem 1E

Solution:

The relative frequency of each class for the given frequency distribution is shown in Table 3.

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Explanation of Solution

Formula used:

The formula to calculate the relative frequency of any class is,

Relative frequency=fN

Here N is the sum of the frequencies.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The sum of the frequencies is,

N=7+8+10+2+3=30

For the first class,

Substitute 7 for f and 30 for N in the formula,

Relative frequency=730=0.233=23%

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

(e)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Explanation of Solution

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

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Answer 2

Question

Chapter 2.1, Problem 1E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 1E

Solution:

The class width of the given frequency distribution is 4.8 which is approximately equal to 5.

Explanation of Solution

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

The upper limit of the distribution is given as 39 and the lower limit is given as 15. The number of classes is 5.

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute 39 for upper limit and 15 for lower limit and 5 for the number of classes in the formula,

Class width=39−155=245=4.8≈5

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

(b)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Explanation of Solution

Formula used:

The formula to calculate the class boundary is,

Class boundary=Upper limit of one class+Lower limit of next class2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since the largest number is 19 and 20 is the smallest number in the next class, then, substitute 19 for upper limit and 20 for lower limit in the formula,

Class boundary=19+202=392=19.5

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Explanation of Solution

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute 19 for upper limit and 15 for lower limit in the formula,

Midpoint=15+192=342=17

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

(d)

Expert Solution

Answer to Problem 1E

Solution:

The relative frequency of each class for the given frequency distribution is shown in Table 3.

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Explanation of Solution

Formula used:

The formula to calculate the relative frequency of any class is,

Relative frequency=fN

Here N is the sum of the frequencies.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The sum of the frequencies is,

N=7+8+10+2+3=30

For the first class,

Substitute 7 for f and 30 for N in the formula,

Relative frequency=730=0.233=23%

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

(e)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Explanation of Solution

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

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Answer 3

Question

Chapter 2.1, Problem 1E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 1E

Solution:

The class width of the given frequency distribution is 4.8 which is approximately equal to 5.

Explanation of Solution

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

The upper limit of the distribution is given as 39 and the lower limit is given as 15. The number of classes is 5.

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute 39 for upper limit and 15 for lower limit and 5 for the number of classes in the formula,

Class width=39−155=245=4.8≈5

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

(b)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Explanation of Solution

Formula used:

The formula to calculate the class boundary is,

Class boundary=Upper limit of one class+Lower limit of next class2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since the largest number is 19 and 20 is the smallest number in the next class, then, substitute 19 for upper limit and 20 for lower limit in the formula,

Class boundary=19+202=392=19.5

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Explanation of Solution

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute 19 for upper limit and 15 for lower limit in the formula,

Midpoint=15+192=342=17

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

(d)

Expert Solution

Answer to Problem 1E

Solution:

The relative frequency of each class for the given frequency distribution is shown in Table 3.

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Explanation of Solution

Formula used:

The formula to calculate the relative frequency of any class is,

Relative frequency=fN

Here N is the sum of the frequencies.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The sum of the frequencies is,

N=7+8+10+2+3=30

For the first class,

Substitute 7 for f and 30 for N in the formula,

Relative frequency=730=0.233=23%

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

(e)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Explanation of Solution

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

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Answer 4

Question

Chapter 2.1, Problem 1E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 1E

Solution:

The class width of the given frequency distribution is 4.8 which is approximately equal to 5.

Explanation of Solution

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

The upper limit of the distribution is given as 39 and the lower limit is given as 15. The number of classes is 5.

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute 39 for upper limit and 15 for lower limit and 5 for the number of classes in the formula,

Class width=39−155=245=4.8≈5

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

(b)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Explanation of Solution

Formula used:

The formula to calculate the class boundary is,

Class boundary=Upper limit of one class+Lower limit of next class2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since the largest number is 19 and 20 is the smallest number in the next class, then, substitute 19 for upper limit and 20 for lower limit in the formula,

Class boundary=19+202=392=19.5

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Explanation of Solution

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute 19 for upper limit and 15 for lower limit in the formula,

Midpoint=15+192=342=17

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

(d)

Expert Solution

Answer to Problem 1E

Solution:

The relative frequency of each class for the given frequency distribution is shown in Table 3.

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Explanation of Solution

Formula used:

The formula to calculate the relative frequency of any class is,

Relative frequency=fN

Here N is the sum of the frequencies.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The sum of the frequencies is,

N=7+8+10+2+3=30

For the first class,

Substitute 7 for f and 30 for N in the formula,

Relative frequency=730=0.233=23%

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

(e)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Explanation of Solution

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

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Answer 5

Chapter 2.1, Problem 1E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 1E

Solution:

The class width of the given frequency distribution is 4.8 which is approximately equal to 5.

Explanation of Solution

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

The upper limit of the distribution is given as 39 and the lower limit is given as 15. The number of classes is 5.

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute 39 for upper limit and 15 for lower limit and 5 for the number of classes in the formula,

Class width=39−155=245=4.8≈5

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

(b)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Explanation of Solution

Formula used:

The formula to calculate the class boundary is,

Class boundary=Upper limit of one class+Lower limit of next class2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since the largest number is 19 and 20 is the smallest number in the next class, then, substitute 19 for upper limit and 20 for lower limit in the formula,

Class boundary=19+202=392=19.5

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Explanation of Solution

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute 19 for upper limit and 15 for lower limit in the formula,

Midpoint=15+192=342=17

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

(d)

Expert Solution

Answer to Problem 1E

Solution:

The relative frequency of each class for the given frequency distribution is shown in Table 3.

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Explanation of Solution

Formula used:

The formula to calculate the relative frequency of any class is,

Relative frequency=fN

Here N is the sum of the frequencies.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The sum of the frequencies is,

N=7+8+10+2+3=30

For the first class,

Substitute 7 for f and 30 for N in the formula,

Relative frequency=730=0.233=23%

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

(e)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Explanation of Solution

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Answer 6

Chapter 2.1, Problem 1E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 1E

Solution:

The class width of the given frequency distribution is 4.8 which is approximately equal to 5.

Explanation of Solution

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

The upper limit of the distribution is given as 39 and the lower limit is given as 15. The number of classes is 5.

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute 39 for upper limit and 15 for lower limit and 5 for the number of classes in the formula,

Class width=39−155=245=4.8≈5

Answer 7

Chapter 2.1, Problem 1E

(a)

To determine

To find:

The class width for the given frequency distribution.

Answer 8

(a)

Expert Solution

Answer to Problem 1E

Solution:

The class width of the given frequency distribution is 4.8 which is approximately equal to 5.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

The upper limit of the distribution is given as 39 and the lower limit is given as 15. The number of classes is 5.

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

Substitute 39 for upper limit and 15 for lower limit and 5 for the number of classes in the formula,

Class width=39−155=245=4.8≈5

Answer 13

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

(b)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Explanation of Solution

Formula used:

The formula to calculate the class boundary is,

Class boundary=Upper limit of one class+Lower limit of next class2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since the largest number is 19 and 20 is the smallest number in the next class, then, substitute 19 for upper limit and 20 for lower limit in the formula,

Class boundary=19+202=392=19.5

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Answer 14

(b)

To determine

To find:

The class boundary for each class of the given frequency distribution.

Answer 15

(b)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Formula used:

The formula to calculate the class boundary is,

Class boundary=Upper limit of one class+Lower limit of next class2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

It is known that the smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class, since the largest number is 19 and 20 is the smallest number in the next class, then, substitute 19 for upper limit and 20 for lower limit in the formula,

Class boundary=19+202=392=19.5

Continuing the same way, the class boundaries of all the five classes are shown in the table given below,

Table 1

Class	Frequency	Class boundaries
15−19	7	14.5−19.5
20−24	8	19.5−24.5
25−29	10	24.5−29.5
30−34	2	29.5−34.5
35−39	3	34.5−39.5

Answer 20

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Explanation of Solution

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute 19 for upper limit and 15 for lower limit in the formula,

Midpoint=15+192=342=17

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Answer 21

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

Answer 22

(c)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The smallest number that can belong to a particular class is termed as lower class limit and the largest number that can belong to a particular class is termed as upper class limit.

For the first class,

Substitute 19 for upper limit and 15 for lower limit in the formula,

Midpoint=15+192=342=17

Continuing the same way, the midpoint of all the five classes are shown in the table given below,

Table 2

Class	Frequency	Class boundaries	Midpoint
15−19	7	14.5−19.5	17
20−24	8	19.5−24.5	22
25−29	10	24.5−29.5	27
30−34	2	29.5−34.5	32
35−39	3	34.5−39.5	37

Answer 27

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

(d)

Expert Solution

Answer to Problem 1E

Solution:

The relative frequency of each class for the given frequency distribution is shown in Table 3.

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Explanation of Solution

Formula used:

The formula to calculate the relative frequency of any class is,

Relative frequency=fN

Here N is the sum of the frequencies.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The sum of the frequencies is,

N=7+8+10+2+3=30

For the first class,

Substitute 7 for f and 30 for N in the formula,

Relative frequency=730=0.233=23%

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Answer 28

(d)

To determine

To find:

The relative frequency of each class for the given frequency distribution.

Answer 29

(d)

Expert Solution

Answer to Problem 1E

Solution:

The relative frequency of each class for the given frequency distribution is shown in Table 3.

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Formula used:

The formula to calculate the relative frequency of any class is,

Relative frequency=fN

Here N is the sum of the frequencies.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The sum of the frequencies is,

N=7+8+10+2+3=30

For the first class,

Substitute 7 for f and 30 for N in the formula,

Relative frequency=730=0.233=23%

Continuing the same way, the relative frequency of all the five classes are shown in the table given below,

Table 3

Class	Frequency	Class boundaries	Midpoint	Relative frequency
15−19	7	14.5−19.5	17	730=23%
20−24	8	19.5−24.5	22	830=27%
25−29	10	24.5−29.5	27	1030=33%
30−34	2	29.5−34.5	32	230=7%
35−39	3	34.5−39.5	37	330=10%

Answer 34

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

(e)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Explanation of Solution

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Answer 35

(e)

To determine

To find:

The cumulative frequency of each class for the given frequency distribution.

Answer 36

(e)

Expert Solution

Answer to Problem 1E

Solution:

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Formula used:

The cumulative frequency of any class is calculated by adding the frequency of that class and all the previous classes.

Given:

The table is given as,

Ages of Taste-Test Participants (in Years)
Class	Frequency
15−19	7
20−24	8
25−29	10
30−34	2
35−39	3

Calculation:

The cumulative frequency of all the five classes are shown in the table given below,

Table 4

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
15−19	7	14.5−19.5	17	730=23%	7
20−24	8	19.5−24.5	22	830=27%	7+8=15
25−29	10	24.5−29.5	27	1030=33%	15+10=25
30−34	2	29.5−34.5	32	230=7%	25+2=27
35−39	3	34.5−39.5	37	330=10%	27+3=30