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side of the vertical line. For example, for the first score we write the leaf 5 next to stem 7; for the second score we write the leaf 2 next to stem 5. The recording of these two scores in a stem-and-leaf display is shown in Figure 2.15. Now, we read all scores and write the leaves on the right side of the vertical line in the rows of corresponding stems. The complete stem-and-leaf display for scores is shown in Figure 2.16. Figure 2.16 Stem-and-leaf display of test scores. 5 2 0 7 6 5 9 1 8 4 7 5 9 1 2 6 9 7 1 2 8 0 7 1 6 3 4 7 9 6 3 5 2 2 8 By looking at the stem-and-leaf display of Figure 2.16, we can observe how the data values are distributed. For example, the stem 7 has the highest frequency, followed by stems 8, 9, 6, and 5. The leaves for each stem of the stem-and-leaf display of Figure 2.16 are ranked (in increas- ing order) and presented in Figure 2.17. Figure 2.17 Ranked stem-and-leaf display of test scores. 5 0 2 7 6 1 4 5 8 9 7 1 1 2 2 5 6 7 9 9 8 0 1 3 4 6 7 7 9 2 2 3 5 6 8 As already mentioned, one advantage of a stem-and-leaf display is that we do not lose infor- mation on individual observations. We can rewrite the individual scores of the 30 college students from the stem-and-leaf display of Figure 2.16 or Figure 2.17. By contrast, the information on individual observations is lost when data are grouped into a frequency table. ◼ 62 Chapter 2 Organizing and Graphing Data EXAMPLE 2–9 Monthly Rents Paid by Households The following data give the monthly rents paid by a sample of 30 households selected from a small town. 880 1081 721 1075 1023 775 1235 750 965 960 1210 985 1231 932 850 825 1000 915 1191 1035 1151 630 1175 952 1100 1140 750 1140 1370 1280 Construct a stem-and-leaf display for these data. Solution Each of the values in the data set contains either three or four digits. We will take the first digit for three-digit numbers and the first two digits for four-digit numbers as stems. Then we will use the last two digits of each number as a leaf. Thus for the first value, which is 880, the stem is 8 and the leaf is 80. The stems for the entire data set are 6, 7, 8, 9, 10, 11, 12, and 13. They are recorded on the left side of the vertical line in Figure 2.18. The leaves for the numbers are recorded on the right side. Figure 2.18 Stem-and-leaf display of rents. 6 30 7 21 75 50 50 8 80 50 25 9 65 60 85 32 15 52 10 81 75 23 00 35 11 91 51 75 00 40 40 12 35 10 31 80 13 70 Sometimes a data set may contain too many stems, with each stem containing only a few leaves. In such cases, we may want to condense the stem-and-leaf display by grouping the stems. Example 2–10 describes this procedure. EXAMPLE 2–10 Number of Hours Spent Working on Computers by Students The following stem-and-leaf display is prepared for the number of hours that 25 students spent working on computers during the past month. 0 6 1 1 7 9 2 2 6 3 2 4 7 8 4 1 5 6 9 9 5 3 6 8 6 2 4 4 5 7 7 8 5 6 Prepare a new stem-and-leaf display by grouping the stems. Solution To condense the given stem-and-leaf display, we can combine the first three rows, the middle three rows, and the last three rows, thus getting the stems 0–2, 3–5, and 6–8. The leaves for each stem of a group are separated by an asterisk (*), as shown in Figure 2.19. Thus, the leaf 6 in the first row corresponds to stem 0; the leaves 1, 7, and 9 correspond to stem 1; and leaves 2 and 6 belong to stem 2. Figure 2.19 Grouped stem-and- leaf display. 0–2 6 * 1 7 9 * 2 6 3–5 2 4 7 8 * 1 5 6 9 9 * 3 6 8 6–8 2 4 4 5 7 * * 5 6 Mark Harmel/Stone/Getty Images Constructing a stem-and-leaf display for three- and four-digit numbers. Preparing a grouped stem-and-leaf display. ◼ 2.3 Stem-and-Leaf Displays 63 If a stem does not contain a leaf, this is indicated in the grouped stem-and-leaf display by two consecutive asterisks. For example, in the stem-and-leaf display of Figure 2.19, there is no leaf for 7; that is, there is no number in the 70s. Hence, in Figure 2.19, we have two asterisks after the leaves for 6 and before the leaves for 8. ◼ Some data sets produce stem-and-leaf displays that have a small number of stems relative to the number of observations in the data set and have too many leaves for each stem. In such cases, it is very difficult to determine if the distribution is symmetric or skewed, as well as other char- acteristics of the distribution that will be introduced in later chapters. In such a situation, we can create a stem-and-leaf display with split stems. To do this, each stem is split into two or five parts. Whenever the stems are split into two parts, any observation having a leaf with a value of 0, 1, 2, 3, or 4 is placed in the first split stem, while the leaves 5, 6, 7, 8, and 9 are placed in the second split stem. Sometimes we can split a stem into five parts if there are too many leaves for one stem. Whenever a stem is split into five parts, leaves with values of 0 and 1 are placed next to the first part of the split stem, leaves with values of 2 and 3 are placed next to the second part of the split stem, and so on. The stem-and-leaf display of Example 2–11 shows this procedure. EXAMPLE 2–11 Consider the following stem-and-leaf display, which has only two stems. Using the split stem procedure, rewrite this stem-and-leaf display. 3 1 1 2 3 3 3 4 4 7 8 9 9 9 4 0 0 0 1 1 1 1 1 1 2 2 2 2 2 3 3 6 6 7 Solution To prepare a split stem-and-leaf display, let us split the two stems, 3 and 4, into two parts each as shown in Figure 2.20. The first part of each stem contains leaves from 0 to 4, and the second part of each stem contains leaves from 5 to 9. Figure 2.20 Split stem-and-leaf display. 3 1 1 2 3 3 3 4 4 3 7 8 9 9 9 4 0 0 0 1 1 1 1 1 1 2 2 2 2 2 3 3 4 6 6 7 Figure 2.21 Split stem-and-leaf display. 3 1 1 3 2 3 3 3 3 4 4 3 7 3 8 9 9 9 4 0 0 0 1 1 1 1 1 1 4 2 2 2 2 2 3 3 4 4 6 6 7 In the stem-and-leaf display of Figure 2.20, the first part of stem 4 has a substantial number of leaves. So, if we decide to split stems into five parts, the new stem-and-leaf display will look as shown in Figure 2.21. There are two important properties to note in the split stem-and-leaf display of Figure 2.21. The third part of split stem 4 does not have any leaves. This implies that there are no observations in the data set having a value of 44 or 45. Since there are observations with values larger than 45, we need to leave an empty part of split stem 4 that corresponds to 44 and 45. Also, there are no observations with values of 48 or 49. However, since there are no values larger than 47 in the data, we do not have to write an empty split stem 4 after the largest value. ◼ Stem-and-leaf display with split stems. 64 Chapter 2 Organizing and Graphing Data EXERCISES 1, 2, 3, and 4, and the second part should contains the leaves 5, 6, 7, 8, and 9. c. Which display (the one in part a or the one in part b) provides a better representation of the features of the distribution? Explain why you believe this. 2.28 The following data give the taxes paid (rounded to thousand dollars) in 2014 by a random sample of 30 families. 11 17 35 3 15 9 21 13 5 19 5 12 8 16 10 8 12 6 14 18 8 12 5 3 14 28 38 18 22 15 a. Prepare a stem-and-leaf display for these data. Arrange the leaves for each stem in increasing order. b. Prepare a split stem-and-leaf display for these data. Split each stem into two parts. The first part should contain the leaves 0 through 4, and the second part should contain the leaves 5 through 9. 2.29 The following data give the one-way commuting times (in minutes) from home to work for a random sample of 50 workers. 23 17 34 26 18 33 46 42 12 37 44 15 22 19 28 32 18 39 40 48 16 11 9 24 18 26 31 7 30 15 18 22 29 32 30 21