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Chapter 2 example questions

# Chapter 2 example questions - Student Grady Simonton...

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Unformatted text preview: Student: Grady Simonton Instructor: Shawn Parvini Date: 2/18/10 Course: Mathll9: Elementary Statistics - Spring 2.010 - CRN: 49239 - 16 weeks Book: Triola: Elementary Statistics, lle Time: 10:57Al\t[ Identify the class width, class Daily Low Frequency Daily Low Frequency midpoints, and class boundaries for the Temperature (°F) Temperature (°F) given frequency distribution. 5 5- 59 1 7 5-19 7 60-64 3 80-84 7 65-69 5 85-89 1 70-74 1 1 Class width is the difference between two consecutive lower class limits or two consecutive lower class boundaries. The difference between two consecutive lower class limits is S. 60 — 55 = 5 Class midpoints are the values in the middle of the classes. Each class midpoint can be found by adding the lower class limit to the upper class limit and dividing the sum by 2. To calculate the class midpoints, find the average of the lower and upper class limits. For example, the ﬁrst midpoint is 55 + 59 equal to = 57. Therefore, the class midpoints are 57,62,67,72,77,82,87. Class boundaries are the numbers used to separate classes, but without the gaps created by class limits. You can simplify the process of ﬁnding class boundaries by understanding that they basically split the difference between the end of one class and the beginning of the next class. To calculate the class boundaries, ﬁnd the average of the upper class limit and the consecutive lower class limit. For 59 + 60 example, one of the boundaries is equal to = 59.5. First determine the middle boundaries. Note, we will determine the end boundaries from the other boundaries. ?,59.5,64.5,69.5,74.5,79.5,84.S,? Notice that the difference between each boundary is 5. Therefore, to ﬁnd the ﬁrst boundary, subtract 5 from 59.5. Then to ﬁnd the last boundary, add 5 to 84.5. Thus, the class boundaries are 54.5595,64.5,69.5,74.5,79.5,84.5,89.5. Page 1 ...
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