57 lb.

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One convenient choice for the class interval size is 5 lb. Also, it is convenient to choose the class marks as 120, 125, 130, 135, . . . pounds. Therefore, the class intervals can be taken as 118 122, 123 127, 128 132, . . . . With this choice the class boundaries are 117.5, 122.5, 127.5, . . . , which do not coincide with observed data. The required frequency distribution is shown in Table 5-5. The center column, called a tally, or score, sheet, is used to tabulate the class frequencies from the raw data and is usually omitted in the final presentation of the frequency distribution.

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Another possibility

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Of course, other possible frequency distributions exist. Table 5-6, for example, shows a frequency distribution with only 7 classes, in which the class interval is 9 lb.

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CHAPTER 5 Sampling Theory

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Table 5-5

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Weight (lb) 118 122 123 127 128 132 133 137 138 142 143 147 148 152 153 157 158 162 163 167 168 172 173 177 > >> >> >>>> >>>> > >>>> >>> >>>> >>>> >> >>> > >> TOTAL Tally Frequency 1 2 2 4 6 8 5 4 2 3 1 2 40 Weight (lb)

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Table 5-6

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Tally Frequency

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118 126 127 135 136 144 145 153 154 162 163 171 172 180

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>>> >>>> >>>> >>>> >>>> >>>> >> >>>> >>>> >> TOTAL

3 5 9 12 5 4 2 40

5.29. Construct a histogram and a frequency polygon for the weight distribution in Problem 5.28.

The histogram and frequency polygon for each of the two cases considered in Problem 5.28 are given in Figs. 5-7 and 5-8. Note that the centers of the bases of the rectangles are located at the class marks.

Fig. 5-7

Fig. 5-8

5.30. Five pennies were simultaneously tossed 1000 times and at each toss the number of heads was observed. The numbers of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in Table 5-7. Graph the data.

The data can be shown graphically either as in Fig. 5-9 or Fig. 5-10. Figure 5-9 seems to be a more natural graph to use. One reason is that the number of heads cannot be 1.5 or 3.2. This graph is a form of bar graph where the bars have zero width, and it is sometimes called a rod graph. It is especially useful when the data are discrete. Figure 5-10 shows a histogram of the data. Note that the total area of the histogram is the total frequency 1000, as it should be.

CHAPTER 5 Sampling Theory

Table 5-7

Number of Heads 0 1 2 3 4 5 TOTAL Number of Tosses (frequency) 38 144 342 287 164 25 1000

Fig. 5-9

Fig. 5-10

Computation of mean, variance, and moments for samples 5.31. Find the arithmetic mean of the numbers 5, 3, 6, 5, 4, 5, 2, 8, 6, 5, 4, 8, 3, 4, 5, 4, 8, 2, 5, 4.

Method 1

x # ax n 5 96 20 3 4.8 6 5 4 5 2 8 6 5 20 4 8 3 4 5 4 8 2 5 4

Method 2

There are six 5s, two 3s, two 6s, five 4s, two 2s, and three 8s. Then a fx n (6)(5) (2)(3) 6 (2)(6) 2 2 (5)(4) 5 2 (2)(2) 3 (3)(8) 96 20

5.32. Four groups of students, consisting of 15, 20, 10, and 18 individuals, reported mean weights of 162, 148, 153, and 140 lb, respectively. Find the mean weight of all the students.

x # a fx n (15)(162) (20)(148) 15 20 (10)(153) 10 18 (18)(140) 150 lb

5.33. Use the frequency distribution of heights in Table 5-2, page 161, to find the mean height of the 100 male students at XYZ University.

CHAPTER 5 Sampling Theory

The work is outlined in Table 5-8. Note that all students having heights 60 62 inches, 63 65 inches, etc., are considered as having heights 61, 64, etc., inches. The problem then reduces to finding the mean height of 100 students if 5 students have height 61 inches, 18 have height 64 inches, etc. x # a fx af a fx n 6745 100 67.45 inches

Table 5-8

Height (inches) 60 62 63 65 66 68 69 71 72 74 Class Mark (x) 61 64 67 70 73 n gf Frequency ( f ) 5 18 42 27 8 100 g fx fx 305 1152 2814 1890 584 6745

The computations involved can become tedious, especially for cases in which the numbers are large and many classes are present. Short techniques are available for lessening the labor in such cases. See Problem 5.35, for example.