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Unformatted text preview: Ems 310,01 Rm 1:; SOLUTION. 2.4 The dot diagram of the paper strength data. 110 115 120 125 130 135 140 strength(pounds) 2.10 The “less than” distribution is given in the table: and the ogive below. o to
x No. less than 5
.c 14.95 0 :7) O 19.95 3 g V
24.95 18 : 29.95 42 .8 8
34.95 54 :E,
39.95 60 Z o 15 20 25 30 35 4O
Ounces
2.20 The difference between the histogram for this exercise and those for the preceed— ing exercises is that the classes are not of equal length. Thus, if the probability
distribution of the data were uniform over a interval covering the data, one
would expect that longer classes would have higher counts. To correct for this,
the area of each rectangle should be proportional to the proportion of obser—
vations in the class. Of course, when the class lengths are equal, plotting the
height equal to the class frequency results in rectangles with area proportional
to the corresponding class proportions. The incorrect histogram is given in Figure 2.7. The correct histogram is given in Figure 2.8. The height of the ﬁrst rectangle
is (relative frequency) /width = (3/80)/4 = 0.0094. 50 5‘
O
800
3
C
Q)
h
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O — Figure 2.7: Incorrect histogram using height=count. Exercise 2.20 0.04 Density
0.02 0.0 O 5 1O 20 30 Figure 2.8: Correct histogram using Density=count/(nxclasswidth). Exercise 2.20 2.21 The empirical cumulative distribution function is shown in Figure 2.9. 1.0 0.0 0.2 0.4 0.6 0.8 120 140 160 180 Degrees Figure 2.9: Empirical cumulative distribution for Exercise 2.21 2.22 The stem and leaf display is: 19* 8 20* 3 7 3 8 0 2 8 5
21* 4 7 2 22* 6 5 0 23* 24* 3...
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 Fall '03
 Mendell

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