Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Chapter Data and Distributions Section 1.2 (a) MINITAB generates the following stem-and-leaf display of this data: Stem-and-leaf of C1 Leaf Unit = 0.10 (11) 10 7 10 11 N = 27 33588 00234677889 127 077 368 The left most column in the MINITAB printout shows the cumulative numbers of observations from each stem to the nearest tail of the data For example, the in the second row indicates that there are a total of data points contained in stems and MINITAB uses parentheses around 11 in row three to indicate that the median (described in Chapter 2, Section 2.1) of the data is contained in this stem A value close to is representative of this data What constitutes large or small variation usually depends on the application at hand, but an often-used rule of thumb is: the variation tends to be large whenever the spread of the data (the difference between the largest and smallest observations) is large compared to a representative value Here, 'large' means that the percentage is closer to 100% than it is to 0% For this data, the spread is 11 - = 6, which constitutes 6/8 = 75, or, 75%, of the typical data value of Most researchers would call this a large amount of variation (b) The data display is not perfectly symmetric around some middle/representative value There tends to be some positive skewness in this data (c) In Chapter 1, outliers are data points that appear to be very different from the pack Looking at the stem-and-leaf display in part (a), there appear to be no outliers in this data (Chapter gives a more precise definition of what constitutes an outlier) (d) From the stem-and-leaf display in part (a), there are leaves associated with the stem of 11, which represent the data values that greater than or equal to 11 10.7, which is represented by the stem of 10 and the leaf of 7, also exceeds 10 Therefore, the proportion of data values that exceed 10 is 4/27 = 148, or, about 15% © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Data and Distributions 2 Chapter (a) Using the same stem and leaf units as in Exercise 1, a comparative stem-and-leaf display of this data is: data from Exercise (n=27) data from Exercise (n=20) p p 33588 00234677889 127 077 368 10 11 12 13 14 16 012488 13359 278 From this display, the cylinder data appears to be even more positively skewed than the data from Exercise The data value 14.1 appears to be an outlier From the stem-and-leaf display, there are values in the cylinder data that have stems of 11 or larger, so there the proportion of cylinder strengths that exceed 10 is 3/20 = 15, or, 15% (b) Both data sets have approximately the same representative value of about MPa and both stem-andleaf displays exhibit positive skewness The spread of the cylinder data is larger than that of the beam data and the cylinder data also appears to contain an outlier A MINITAB stem-and-leaf display of this data is: Stem-and-leaf of C1 Leaf Unit = 01 18 18 11 1 3 4 5 6 7 N = 36 56678 000112222234 5667888 144 58 6678 Another method of denoting the pairs of stems having equal values is to denote the first stem by L, for 'low', and the second stem by H, for 'high' Using this notation, the stem-and-leaf display would appear as follows: 3L 3H 4L 4H 5L 5H 6L 56678 000112222234 5667888 144 58 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Data and Distributions Chapter 6H 7L 7H 6678 The stem-and-leaf display shows that 45 is a good representative value for the data In addition, the display is not symmetric and appears to be positively skewed The spread of the data is 75 - 31 = 44, which is.44/.45 = 978, or about 98% of the typical value of 45 Using the same rule of thumb as in Exercise 1, this constitutes a reasonably large amount of variation in the data The data value 75 is a possible outlier (the definition of 'outlier' in Section 2.3, shows that 75 could be considered to be a 'mild' outlier) Because the stem-and-leaf display is nearly symmetric around 90, a representative value of about 90 is easy to discern from the diagram The most apparent features of the display are its approximate symmetry and the tendency for the data values to stack up around the representative value in a bell-shaped curve Also, the spread of the data, 100.3-83.4 = 16.9 is a relatively small percentage (16.9/90 | 18, or 18%) of the typical value of 90 (a) Stem-and-leaf of Tensile Leaf Unit = 1.0 11 17 28 53 (38) 62 40 22 12 12 12 12 12 13 13 13 13 13 14 14 14 14 N = 153 445 6667777 889999 00011111111 2222222222333333333333333 404444444444444444455555555555555555555 6666666666667777777777 888888888888999999 0000001111 2333333 444 77 The display is symmetric around the class with the stem of 13 and with the leaves of and This class is also the most peaked It is therefore easy to see that a representative value is about 134 or 135 ksi (b) The following histogram of the tensile ultimate strength values appears to form a bell shape around the value of 135 ksi © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Data and Distributions Chapter Frequency 30 20 10 120 130 140 150 Tensile Ultimate Strength (a) Two-digit stems would be best One-digit stems would create a display with only stems, and 7, which would give a display without much detail Three-digit stems would cause the display to be much too wide with many gaps (stems with no leaves) (b) The stem-and-leaf display below does not give up (truncate) the rightmost digit in the data: 64 65 66 67 68 69 70 71 72 33 06 05 00 50 00 05 05 09 35 26 14 13 70 04 11 13 80 64 27 94 45 73 27 22 31 70 83 70 70 90 98 90 36 40 50 51 65 68 69 (c) A MINITAB stem-and-leaf display of this data appears below Note that MINITAB does truncate the rightmost digit in the data values 11 18 (4) 18 14 64 65 66 67 68 69 70 71 72 3367 0228 019 0147799 5779 0023 012455 013666 08 This display tends to be about as informative as the one in part (b) With larger sample sizes, the work involved in creating the display in part (c) would be much less than that required in part (b) In addition, for a larger sample size, the 'full' display in (b) would require a lot of room horizontally on the page to accommodate all the 2-digit leaves A MINITAB stem-and-leaf display in which each stem appears only once is: Stem-and-leaf of C1 N = 40 Leaf Unit = 1.0 034667899 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Data and Distributions Chapter 17 00122244 (19) 0011111223445557899 0358 A MINITAB stem-and-leaf display in which each stem appears twice is: Stem-and-leaf of C1 Leaf Unit = 1.0 N = 40 034 667899 17 00122244 17 (12) 001111122344 11 5557899 03 58 In the display with repeated stems it is apparent that there is a gap in the data at the second '7' stem This means that there are no exam scores between 75 and 79, which seems strange compared to the rest of the scores (a) Number Nonconforming Frequency 12 13 14 3 1 RelativeFrequency(Freq/60) 0.117 0.200 0.217 0.233 0.100 0.050 0.050 0.017 0.017 doesn't add exactly to because relative frequencies have been rounded o 1.001 (b) The number of batches with at most nonconforming items is 12 13 14 55, which is a proportion of 55/60 = 917 The proportion of batches with (strictly) fewer than nonconforming items is 52/60 = 867 Notice that these proportions could also have been computed by using the relative frequencies: e.g., proportion of batches with or fewer nonconforming items = (.05 017 107) 916 ; proportion of batches with fewer than nonconforming items = (.05 05 017 107) 866 (c) The following is a MINITAB histogram of this data The center of the histogram is somewhere around or 3, and it shows that there is some positive skewness in the data Using the rule of thumb in Exercise 1, the histogram also shows that there is a lot of spread/variation in this data © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Data and Distributions Chapter (a) The following histogram was constructed using MINITAB: The most interesting feature of the histogram is the heavy positive skewness of the data Note: One way to have MINITAB automatically construct a histogram from grouped data such as this is to use MINITAB's ability to enter multiple copies of the same number by typing, for example, 784(1) to enter 784 copies of the number The frequency data in this exercise was entered using the following MINITAB commands: MTB > DATA> DATA> DATA> set c1 784(1) 204(2) 127(3) 50(4) 33(5) 28(6) 19(7) 19(8) 6(9) 7(10) 6(11) 7(12) 4(13) 4(14) 5(15) 3(16) 3(17) end (b) From the frequency distribution (or from the histogram), the number of authors who published at least papers is 33+28+19+…+5+3+3 = 144, so the proportion who published or more papers is 144/1309 = 11, or 11% Similarly, by adding frequencies and dividing by n = 1309, the proportion who published 10 or more papers is 39/1309 = 0298, or about 3% The proportion who published more than 10 papers (i.e., 11 or more) is 32/1309 = 0245, or about 2.5% (c) No Strictly speaking, the class described by ' t15 ' has no upper boundary, so it is impossible to draw a rectangle above it having finite area (i.e., frequency) (d) The category 15-17 does have a finite width of 2, so the cumulated frequency of 11 can be plotted as a rectangle of height 6.5 over this interval The basic rule is to make the area of the bar equal to the class frequency, so area = 11 = (width)(height) = 2(height) yields a height of 6.5 (a) From this frequency distribution, the proportion of wafers that contained at least one particle is (1001)/100 = 99, or 99% Note that it is much easier to subtract (which is the number of wafers that contain particles) from 100 than it would be to add all the frequencies for 1, 2, 3,… particles In a similar fashion, the proportion containing at least particles is (100 - 1-2-3-12-11)/100 = 71/100 = 71, or, 71% © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Chapter Data and Distributions (b) The proportion containing between and 10 particles is (15+18+10+12+4+5)/100 = 64/100 = 64, or 64% The proportion that contain strictly between and 10 (meaning strictly more than and strictly less than 10) is (18+10+12+4)/100 = 44/100 = 44, or 44% (c) The following histogram was constructed using MINITAB The data was entered using the same technique mentioned in the answer to exercise 8(a) The histogram is almost symmetric and unimodal; however, it has a few relative maxima (i.e., modes) and has a very slight positive skew 10 The following Pareto chart was constructed using MINITAB: From this chart, the three most frequently occurring injury categories (A, B, and C) account for 90.8% of all injuries 11 (a) The following stem-and-leaf display was constructed using MINITAB: © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Data and Distributions Stem-and-leaf of C1 Leaf Unit = 100 12 23 (10) 14 5 Chapter N = 47 123334555599 00122234688 1112344477 0113338 37 23778 A typical data value is somewhere in the low 2000's The display almost unimodal (the stem at would be considered a mode, the stem at another) and has a positive skew (b) A histogram of this data, using classes of width 1000 separated at 0, 1000, 2000, and 6000 is shown below The proportion of subdivisions with total length less than 2000 is (12+11)/47 = 489, or 48.9% Between 2000 and 4000, the proportion is (10 + 7)/47 = 362, or 36.21% The histogram shows the same general shape as depicted by the stem-and-leaf display in part (a) 12 (a) A histogram of the y data appears below From this histogram, the number of subdivisions having no cul-de-sacs (i.e., y = 0) is 17/47 = 362, or 36.2% The proportion having at least one cul-de-sac (y t 1) is (47-17)/47 = 30/47 = 638, or 63.8% Note that subtracting the number of cul-de-sacs with y = from the total, 47, is an easy way to find the number of subdivisions with y t (b) A histogram of the z data appears above From this histogram, the number of subdivisions with at most intersections (i.e., z d 5) is 42/47 = 894, or 89.4% The proportion having fewer than intersections (z < 5) is 39/47 = 830, or 83.0% © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Chapter 13 Data and Distributions (a) Proportion of herds with only one giraffe = 589/1570 = 0.3752 (b) Proportion of herds with six or more giraffes = (89+57+…+ + 1)/1570 or – (589 + 190 + 176 + 157 + 115)/1570 = 0.2185 (c) Proportion of herds that had between and 10 giraffes, inclusive = (115+89+57+55+33+31)/1570 = 0.242 (d) The distribution of herd size is skewed to the right, with very few large herds, and majority of herds being smaller than to in size 14 Note: since the class intervals have unequal length, we must use a density scale The distribution of tantrum durations is unimodal and heavily positively skewed Most tantrums last between and 11 minutes, but a few last more than half an hour! With such heavy skewness, it’s difficult to give a representative value © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Data and Distributions 10 Chapter 15 (a) Yes: the proportion of sampled angles smaller than 15° is 177 + 166 + 175 = 518 (b) The proportion of sampled angles at least 30° is 078 + 044 + 030 = 152 (c) The proportion of angles between 10° and 25° is roughly 175 + 136 + (.194)/2 = 408 (d) The distribution of misorientation angles is heavily positively skewed Though angles can range from 0° to 90°, nearly 85% of all angles are less than 30° Without more precise information, we cannot tell if the data contain outliers Histogram of Angle 0.04 Density 0.03 0.02 0.01 0.00 10 20 40 90 Angle 16 A histogram of the raw data appears below: After transforming the data by taking logarithms (base 10), a histogram of the log10 data is shown below The shape of this histogram is much less skewed than the histogram of the original data © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Solution Manual for Applied Statistics for Engineers and Scientists 3rd Edition by Devore Data and Distributions Chapter 17 The histogram of this data appears below A typical value of the shear strength is around 5000 lb The histogram is almost symmetric and approximately bell-shaped 18 (a) The classes overlap For example, the classes 0-50 and 50-100 both contain the number 50, which happens to coincide with one of the data values, so it would not be clear which class to put this observation in (b) The lifetime distribution is positively skewed A representative value is around 100 There is a great deal of variability in lifetimes and several possible candidates for outliers Class Interval 0–< 50 50– -.80) = - Proportion(z d -.80) = - 2119 = 7881 (d) Proportion(.21 d z d 1.21) = Proportion(z d 1.21) - Proportion(z d 21) = 8869 - 5832 = 3037 (e) Proportion(z d -2.00 or z t 2.00) = Proportion (zd -2.00) + [1- Proportion(z < 2.00)] = 0228 + [1 - 9772] = 0456 Alternatively, using the fact that the z density is symmetric around z = 0, Proportion(z d -2.00) = Proportion(z t 2.00), so the answer is simply Proportion(z d -2.00) = 2(.0228) = 0456 (f) Proportion(z d -4.2) = 0000 (g) Proportion(z > 4.33) = 0000 32 (a) Proportion(z d z*) = 9082 when z* = 1.33 (Table I) (b) Proportion(z d 1.33) = 9082 and Proportion(z d 1.32) = 9066; using linear interpolation, 9080 is (.9080-.9066)/(.9082-.9066) = 0014/.0016 = 875 (or, 87.5%) of the way from 9066 to 9082, so z* is 875 of the way from 1.32 to 1.33; i.e., z* = 1.32 + 875(1.33-1.32) = 1.32875, or, about 1.329 (c) Proportion(z > z*) = 1210 is a right-tail area Converting to a left-tail area (so that we can use table I), Proportion(z d z*) = - 1210 = 8790 From Table I, z* = 1.17 has a left tail area of 8790 and, therefore, has a right-tail area of 1210 (d) Proportion(-z* d z d z*) = 754 Because the z density is symmetric the two tail areas associated with z < - z* and z > z* must be equal and they also account for all the remaining area under the z density curve That is, 2uProportion(z d - z*) = - 754 = 246, which means that Proportion(z d - z*) = 1230 From Table I, Proportion(z d -1.16) = 1230, so - z* = -1.16 and z* = 1.16 (e) Proportion(z > z*) = 002 is equivalent to saying that Proportion(z d z*) = 998 From Table I, Proportion(z d 2.88) = 9980, so z* = 2.88 Similarly, you would have to go a distance of -2.88 (i.e., 2.88 units to the left of 0) to capture a left-tail area of 002 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part