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Solution: identify the class width, class midpoints, and

Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola ISBN: 9780321836960 18

Solution for problem 10BSC Chapter 2.2

Elementary Statistics | 12th Edition

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Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola

Elementary Statistics | 12th Edition

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Problem 10BSC

In Exercises 5–10, identify the class width, class midpoints, and class boundaries for the given frequency distribution. The frequency distributions are based on real data from Appendix B.

Step-by-Step Solution:

Answer :

Step 1 of 1 :

Consider the data

White Blood Cell Count of females

Frequency

3.0-4.9

6

5.0-6.9

16

7.0-8.9

9

9.0-10.9

7

11.0-12.9

0

13.0-14.9

2

Now we have to identify the class width, class midpoints, and class boundaries.

Class Width :

Class width is found by finding the difference of the lower limits between two classes.

The class width should be uniform throughout the data set.

White Blood Cell Count of females

Frequency

Class Width

3.0-4.9

6

2

5.0-6.9

16

2

7.0-8.9

9

2

9.0-10.9

7

2

11.0-12.9

0

2

13.0-14.9

2

2

Therefore Class width: 2.

Class Midpoint :

Class midpoint is found by taking the average of the two limits within the class.

Class Midpoint = = 3.95

White Blood Cell Count of females

Frequency

Class Width

Class Midpoint

3.0-4.9

6

2

3.95

5.0-6.9

16

2

5.95

7.0-8.9

9

2

7.95

9.0-10.9

7

2

9.95

11.0-12.9

0

2

11.95

13.0-14.9

2

2

13.95

Therefore Class midpoints: 3.95, 5.95, 7.95, 9.95, 11.95,13.95

Class boundaries:

First find the gap value by subtracting the upper limit of the previous class from the lower limit of the next class.Then to find the lower boundary, subtract the gap value from the lower limit of the class. To find the upper boundary, add the gap value to the upper limit of the class.

White Blood Cell Count of females

Frequency

Class Width

Class Midpoint

Class Boundaries

3.0-4.9

6

2

3.95

(3-0.05) to (4.9+0.05) = 2.95 to 4.95

5.0-6.9

16

2

5.95

(5-0.05) to (6.9+0.05) = 4.95 to 6.95

7.0-8.9

9

Step 2 of 1

Chapter 2.2, Problem 10BSC is Solved
Textbook: Elementary Statistics
Edition: 12
Author: Mario F. Triola
ISBN: 9780321836960

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Solution: identify the class width, class midpoints, and