Frequency Distributions Minutes Spent on the Phone 102 124 108 86 103 82 71 104 112 118 87 95 103 116 85 122 87 100 105 97 107 67 78 125 109 99 105 99 101 92 This data will be used for several examples. You might want to duplicate it for use with slides that follow. Make a frequency distribution table with five classes. Minimum value = Maximum value = 67 Key values: 125

Steps to Construct a Frequency Distribution 1. Choose the number of classes Should be between 5 and 15. (For this problem use 5) 2. Calculate the Class Width Find the range = maximum value – minimum. Then divide this by the number of classes. Finally, round up to a convenient number. (125 - 67) / 5 = 11.6 Round up to 12. 3. Determine Class Limits The lower class limit is the lowest data value that belongs in a class and the upper class limit is the highest. Use the minimum value as the lower class limit in the first class. (67) 4. Mark a tally | in appropriate class for each data value. After all data values are tallied, count the tallies in each class for the class frequencies.

Construct a Frequency Distribution Minimum = 67, Maximum = 125 Number of classes = 5 Class width = 12 3 5 8 9 Class Limits Tally 78 90 102 114 126 67 79 91 103 115 Do all lower class limits first.

Frequency Histogram Boundaries Class 3 66.5 - 78.5 67 - 78 78.5 - 90.5 90.5 - 102.5 102.5 -114.5 114.5 -126.5 Class 67 - 78 79 - 90 91 - 102 103 -114 115 -126 3 5 8 9 Time on Phone 9 9 8 8 7 6 5 5 5 A histogram is a bar graph for which the bars touch. To form boundaries, find the distance between consecutive classes. Add half that distance to the lower limits and half to the upper limits. In this case the distance is 1 unit so add .5 to all upper limits and subtract .5 from all lower ones. The data must be quantitative. This histogram is labeled at the class boundaries. Explain that midpoints could have been labeled instead. 4 3 3 2 1 6 6 . 5 7 8 . 5 9 . 5 1 2 . 5 1 1 4 . 5 1 2 6 . 5 minutes

Frequency Polygon Time on Phone minutes Class 3 67 - 78 79 - 90 5 91 - 102 103 -114 115 -126 3 5 8 9 Time on Phone 9 9 8 8 7 6 5 5 5 4 3 3 2 1 The frequency polygon is labeled at midpoints. Explain that midpoints could have been labeled instead. 72.5 84.5 96.5 108.5 120.5 minutes Mark the midpoint at the top of each bar. Connect consecutive midpoints. Extend the frequency polygon to the axis.

Other Information Midpoint: (lower limit + upper limit) / 2 Relative frequency: class frequency/total frequency Cumulative frequency: number of values in that class or in lower Class Midpoint Relative Frequency Cumulative Frequency (67 + 78)/2 3/30 67 - 78 79 - 90 91 - 102 103 - 114 115 - 126 3 5 8 9 72.5 84.5 96.5 108.5 120.5 0.10 0.17 0.27 0.30 3 8 16 25 30 The first two columns reflect the work done in previous slides. Once the first midpoint is calculated, the others can be found by adding the class width to the previous midpoint. Notice the last entry in the cumulative frequency column is equal to the total frequency.

Relative Frequency Histogram Time on Phone Relative frequency A relative frequency histogram will have the same shape as a frequency histogram. minutes Relative frequency on vertical scale

Ogive An ogive reports the number of values in the data set that are less than or equal to the given value, x. Minutes on Phone 3 8 16 25 30 66.5 78.5 90.5 102.5 114.5 126.5 10 20 30 Cumulative Frequency Label each boundary on the horizontal axis. Start with 0 for the lower boundary of the first class. Then mark points corresponding to cumulative frequencies at the upper boundaries. Connect with line segments. minutes