The class width for the given frequency distribution.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 2.1, Problem 8E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 8E

Solution:

The class width of the given frequency distribution is 0.99 which is approximately equal to 1.0.

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

The upper limit of the distribution is given as 15.49 and the lower limit is given as 10.50. The number of classes is 5.

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

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 15.49 for upper limit and 10.50 for lower limit and 5 for the number of classes in the formula,

Class width=15.49−10.505=4.995=0.998≈1.0

(b)

To determine

To find:

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

(b)

Expert Solution

Answer to Problem 8E

Solution:

The class boundary for each class of the given frequency distribution is,

Class	Frequency	Class boundaries
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the class boundary is,

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

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 11.49 and 11.50 is the smallest number in the next class, then, substitute 11.49 for upper limit and 11.50 for lower limit in the formula,

Class boundary=11.49+11.502=22.992=11.495

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
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 8E

Solution:

The midpoint of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

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 11.49 for upper limit and 10.50 for lower limit in the formula,

Midpoint=11.49+10.502=21.992=10.995

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
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

(d)

To determine

To find:

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

(d)

Expert Solution

Answer to Problem 8E

Solution:

The relative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

Relative frequency=fN

Here N is the sum of the frequencies.

Calculation:

The sum of the frequencies is,

N=92+78+68+45+34=317

For the first class,

Substitute 92 for f and 317 for N in the formula,

Relative frequency=92317=0.29=29%

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

(e)

To determine

To find:

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

(e)

Expert Solution

Answer to Problem 8E

Solution:

The cumulative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Want to see more full solutions like this?

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

Students have asked these similar questions

Problem 3. Pricing a multi-stock option the Margrabe formula The purpose of this problem is to price a swap option in a 2-stock model, similarly as what we did in the example in the lectures. We consider a two-dimensional Brownian motion given by W₁ = (W(¹), W(2)) on a probability space (Q, F,P). Two stock prices are modeled by the following equations: dX = dY₁ = X₁ (rdt+ rdt+0₁dW!) (²)), Y₁ (rdt+dW+0zdW!"), with Xo xo and Yo =yo. This corresponds to the multi-stock model studied in class, but with notation (X+, Y₁) instead of (S(1), S(2)). Given the model above, the measure P is already the risk-neutral measure (Both stocks have rate of return r). We write σ = 0₁+0%. We consider a swap option, which gives you the right, at time T, to exchange one share of X for one share of Y. That is, the option has payoff F=(Yr-XT). (a) We first assume that r = 0 (for questions (a)-(f)). Write an explicit expression for the process Xt. Reminder before proceeding to question (b): Girsanov's theorem…

Problem 1. Multi-stock model We consider a 2-stock model similar to the one studied in class. Namely, we consider = S(1) S(2) = S(¹) exp (σ1B(1) + (M1 - 0/1 ) S(²) exp (02B(2) + (H₂- M2 where (B(¹) ) +20 and (B(2) ) +≥o are two Brownian motions, with t≥0 Cov (B(¹), B(2)) = p min{t, s}. " The purpose of this problem is to prove that there indeed exists a 2-dimensional Brownian motion (W+)+20 (W(1), W(2))+20 such that = S(1) S(2) = = S(¹) exp (011W(¹) + (μ₁ - 01/1) t) 롱) S(²) exp (021W (1) + 022W(2) + (112 - 03/01/12) t). where σ11, 21, 22 are constants to be determined (as functions of σ1, σ2, p). Hint: The constants will follow the formulas developed in the lectures. (a) To show existence of (Ŵ+), first write the expression for both W. (¹) and W (2) functions of (B(1), B(²)). as (b) Using the formulas obtained in (a), show that the process (WA) is actually a 2- dimensional standard Brownian motion (i.e. show that each component is normal, with mean 0, variance t, and that their…

The scores of 8 students on the midterm exam and final exam were as follows. Student Midterm Final Anderson 98 89 Bailey 88 74 Cruz 87 97 DeSana 85 79 Erickson 85 94 Francis 83 71 Gray 74 98 Harris 70 91 Find the value of the (Spearman's) rank correlation coefficient test statistic that would be used to test the claim of no correlation between midterm score and final exam score. Round your answer to 3 places after the decimal point, if necessary. Test statistic: rs =

Answer 2

Question

Chapter 2.1, Problem 8E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 8E

Solution:

The class width of the given frequency distribution is 0.99 which is approximately equal to 1.0.

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

The upper limit of the distribution is given as 15.49 and the lower limit is given as 10.50. The number of classes is 5.

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

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 15.49 for upper limit and 10.50 for lower limit and 5 for the number of classes in the formula,

Class width=15.49−10.505=4.995=0.998≈1.0

(b)

To determine

To find:

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

(b)

Expert Solution

Answer to Problem 8E

Solution:

The class boundary for each class of the given frequency distribution is,

Class	Frequency	Class boundaries
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the class boundary is,

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

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 11.49 and 11.50 is the smallest number in the next class, then, substitute 11.49 for upper limit and 11.50 for lower limit in the formula,

Class boundary=11.49+11.502=22.992=11.495

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
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 8E

Solution:

The midpoint of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

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 11.49 for upper limit and 10.50 for lower limit in the formula,

Midpoint=11.49+10.502=21.992=10.995

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
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

(d)

To determine

To find:

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

(d)

Expert Solution

Answer to Problem 8E

Solution:

The relative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

Relative frequency=fN

Here N is the sum of the frequencies.

Calculation:

The sum of the frequencies is,

N=92+78+68+45+34=317

For the first class,

Substitute 92 for f and 317 for N in the formula,

Relative frequency=92317=0.29=29%

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

(e)

To determine

To find:

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

(e)

Expert Solution

Answer to Problem 8E

Solution:

The cumulative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Want to see more full solutions like this?

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

Answer 3

Question

Chapter 2.1, Problem 8E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 8E

Solution:

The class width of the given frequency distribution is 0.99 which is approximately equal to 1.0.

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

The upper limit of the distribution is given as 15.49 and the lower limit is given as 10.50. The number of classes is 5.

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

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 15.49 for upper limit and 10.50 for lower limit and 5 for the number of classes in the formula,

Class width=15.49−10.505=4.995=0.998≈1.0

(b)

To determine

To find:

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

(b)

Expert Solution

Answer to Problem 8E

Solution:

The class boundary for each class of the given frequency distribution is,

Class	Frequency	Class boundaries
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the class boundary is,

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

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 11.49 and 11.50 is the smallest number in the next class, then, substitute 11.49 for upper limit and 11.50 for lower limit in the formula,

Class boundary=11.49+11.502=22.992=11.495

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
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 8E

Solution:

The midpoint of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

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 11.49 for upper limit and 10.50 for lower limit in the formula,

Midpoint=11.49+10.502=21.992=10.995

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
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

(d)

To determine

To find:

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

(d)

Expert Solution

Answer to Problem 8E

Solution:

The relative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

Relative frequency=fN

Here N is the sum of the frequencies.

Calculation:

The sum of the frequencies is,

N=92+78+68+45+34=317

For the first class,

Substitute 92 for f and 317 for N in the formula,

Relative frequency=92317=0.29=29%

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

(e)

To determine

To find:

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

(e)

Expert Solution

Answer to Problem 8E

Solution:

The cumulative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Want to see more full solutions like this?

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

Answer 4

Question

Chapter 2.1, Problem 8E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 8E

Solution:

The class width of the given frequency distribution is 0.99 which is approximately equal to 1.0.

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

The upper limit of the distribution is given as 15.49 and the lower limit is given as 10.50. The number of classes is 5.

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

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 15.49 for upper limit and 10.50 for lower limit and 5 for the number of classes in the formula,

Class width=15.49−10.505=4.995=0.998≈1.0

(b)

To determine

To find:

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

(b)

Expert Solution

Answer to Problem 8E

Solution:

The class boundary for each class of the given frequency distribution is,

Class	Frequency	Class boundaries
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the class boundary is,

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

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 11.49 and 11.50 is the smallest number in the next class, then, substitute 11.49 for upper limit and 11.50 for lower limit in the formula,

Class boundary=11.49+11.502=22.992=11.495

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
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 8E

Solution:

The midpoint of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

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 11.49 for upper limit and 10.50 for lower limit in the formula,

Midpoint=11.49+10.502=21.992=10.995

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
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

(d)

To determine

To find:

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

(d)

Expert Solution

Answer to Problem 8E

Solution:

The relative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

Relative frequency=fN

Here N is the sum of the frequencies.

Calculation:

The sum of the frequencies is,

N=92+78+68+45+34=317

For the first class,

Substitute 92 for f and 317 for N in the formula,

Relative frequency=92317=0.29=29%

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

(e)

To determine

To find:

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

(e)

Expert Solution

Answer to Problem 8E

Solution:

The cumulative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Want to see more full solutions like this?

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

Answer 5

Chapter 2.1, Problem 8E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 8E

Solution:

The class width of the given frequency distribution is 0.99 which is approximately equal to 1.0.

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

The upper limit of the distribution is given as 15.49 and the lower limit is given as 10.50. The number of classes is 5.

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

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 15.49 for upper limit and 10.50 for lower limit and 5 for the number of classes in the formula,

Class width=15.49−10.505=4.995=0.998≈1.0

(b)

To determine

To find:

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

(b)

Expert Solution

Answer to Problem 8E

Solution:

The class boundary for each class of the given frequency distribution is,

Class	Frequency	Class boundaries
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the class boundary is,

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

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 11.49 and 11.50 is the smallest number in the next class, then, substitute 11.49 for upper limit and 11.50 for lower limit in the formula,

Class boundary=11.49+11.502=22.992=11.495

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
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 8E

Solution:

The midpoint of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

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 11.49 for upper limit and 10.50 for lower limit in the formula,

Midpoint=11.49+10.502=21.992=10.995

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
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

(d)

To determine

To find:

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

(d)

Expert Solution

Answer to Problem 8E

Solution:

The relative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

Relative frequency=fN

Here N is the sum of the frequencies.

Calculation:

The sum of the frequencies is,

N=92+78+68+45+34=317

For the first class,

Substitute 92 for f and 317 for N in the formula,

Relative frequency=92317=0.29=29%

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

(e)

To determine

To find:

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

(e)

Expert Solution

Answer to Problem 8E

Solution:

The cumulative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Answer 6

Chapter 2.1, Problem 8E

(a)

To determine

To find:

The class width for the given frequency distribution.

(a)

Expert Solution

Answer to Problem 8E

Solution:

The class width of the given frequency distribution is 0.99 which is approximately equal to 1.0.

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

The upper limit of the distribution is given as 15.49 and the lower limit is given as 10.50. The number of classes is 5.

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

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 15.49 for upper limit and 10.50 for lower limit and 5 for the number of classes in the formula,

Class width=15.49−10.505=4.995=0.998≈1.0

Answer 7

Chapter 2.1, Problem 8E

(a)

To determine

To find:

The class width for the given frequency distribution.

Answer 8

(a)

Expert Solution

Answer to Problem 8E

Solution:

The class width of the given frequency distribution is 0.99 which is approximately equal to 1.0.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

The upper limit of the distribution is given as 15.49 and the lower limit is given as 10.50. The number of classes is 5.

Formula used:

The formula to calculate the class width is,

Class width=Upper limit−Lower limitNumber of class

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 15.49 for upper limit and 10.50 for lower limit and 5 for the number of classes in the formula,

Class width=15.49−10.505=4.995=0.998≈1.0

Answer 13

(b)

To determine

To find:

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

(b)

Expert Solution

Answer to Problem 8E

Solution:

The class boundary for each class of the given frequency distribution is,

Class	Frequency	Class boundaries
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the class boundary is,

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

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 11.49 and 11.50 is the smallest number in the next class, then, substitute 11.49 for upper limit and 11.50 for lower limit in the formula,

Class boundary=11.49+11.502=22.992=11.495

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
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

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 8E

Solution:

The class boundary for each class of the given frequency distribution is,

Class	Frequency	Class boundaries
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the class boundary is,

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

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 11.49 and 11.50 is the smallest number in the next class, then, substitute 11.49 for upper limit and 11.50 for lower limit in the formula,

Class boundary=11.49+11.502=22.992=11.495

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
10.50−11.49	92	10.495−11.495
11.50−12.49	78	11.495−12.495
12.50−13.49	68	12.495−13.495
13.50−14.49	45	13.495−14.495
14.50−15.49	34	14.495−15.495

Answer 20

(c)

To determine

To find:

The midpoint of each class for the given frequency distribution.

(c)

Expert Solution

Answer to Problem 8E

Solution:

The midpoint of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

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 11.49 for upper limit and 10.50 for lower limit in the formula,

Midpoint=11.49+10.502=21.992=10.995

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
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

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 8E

Solution:

The midpoint of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

The formula to calculate the midpoint of any class is,

Midpoint=Upper limit+Lower limit2

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 11.49 for upper limit and 10.50 for lower limit in the formula,

Midpoint=11.49+10.502=21.992=10.995

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
10.50−11.49	92	10.495−11.495	10.995
11.50−12.49	78	11.495−12.495	11.995
12.50−13.49	68	12.495−13.495	12.995
13.50−14.49	45	13.495−14.495	13.995
14.50−15.49	34	14.495−15.495	14.995

Answer 27

(d)

To determine

To find:

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

(d)

Expert Solution

Answer to Problem 8E

Solution:

The relative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

Relative frequency=fN

Here N is the sum of the frequencies.

Calculation:

The sum of the frequencies is,

N=92+78+68+45+34=317

For the first class,

Substitute 92 for f and 317 for N in the formula,

Relative frequency=92317=0.29=29%

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

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 8E

Solution:

The relative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

Relative frequency=fN

Here N is the sum of the frequencies.

Calculation:

The sum of the frequencies is,

N=92+78+68+45+34=317

For the first class,

Substitute 92 for f and 317 for N in the formula,

Relative frequency=92317=0.29=29%

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%
11.50−12.49	78	11.495−12.495	11.995	78317=25%
12.50−13.49	68	12.495−13.495	12.995	68317=21%
13.50−14.49	45	13.495−14.495	13.995	45317=14%
14.50−15.49	34	14.495−15.495	14.995	34317=11%

Answer 34

(e)

To determine

To find:

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

(e)

Expert Solution

Answer to Problem 8E

Solution:

The cumulative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

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 8E

Solution:

The cumulative frequency of each class for the given frequency distribution is,

Class	Frequency	Class boundaries	Midpoint	Relative frequency	Cumulative frequency
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given:

The table is given as,

Hourly Wages of Surveillance Operators (in Dollars)
Class	Frequency
10.50−11.49	92
11.50−12.49	78
12.50−13.49	68
13.50−14.49	45
14.50−15.49	34

Formula used:

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

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
10.50−11.49	92	10.495−11.495	10.995	92317=29%	92
11.50−12.49	78	11.495−12.495	11.995	78317=25%	92+78=170
12.50−13.49	68	12.495−13.495	12.995	68317=21%	170+68=238
13.50−14.49	45	13.495−14.495	13.995	45317=14%	238+45=283
14.50−15.49	34	14.495−15.495	14.995	34317=11%	283+34=317