A52 ODD ANSWERS 29. x is not possible; median is not possible; mode = “Junior”; The mean and median cannot be found because the data are at the nominal level of measurement. 31. x ≈ 29.2; median = 30.5; mode = 23, 34 33. x ≈ 19.5; median = 20; mode = 15 35. Cluster around 475–525, gap between 225 and 275 37. Mode, because the data are at the nominal level of measurement. 39. Mean, because the distribution is symmetric and there are no outliers. 41. 90.5 43. $612.73 45. 84 47. 87 49. 53.5 minutes 51. 42.3 years old 53. Class Frequency, f Midpoint 127 – 161 7 144 162 – 196 6 179 197 – 231 3 214 232 – 266 3 249 267 – 301 1 284 Frequency Number of beds 144 179 214 249 284 1 2 3 4 5 6 7 Hospital Beds Positively skewed 55. Class Frequency, f Midpoint 62 – 64 3 63 65 – 67 7 66 68 – 70 9 69 71 – 73 8 72 74 – 76 3 75 Frequency Height (in inches) 63 66 69 72 75 1 2 3 4 5 6 7 8 9 Heights of Males Symmetric 57. (a) x ≈ 1518.2, median = 1520.5 (b) x ≈ 1521.2, median = 1522.5 (c) Mean 59. The data are skewed right. A = mode, because it is the data entry that occurred most often. B = median, because the median is to the left of the mean in a skewed right distribution. C = mean, because the mean is to the right of the median in a skewed right distribution. 61. Increase one of the three-credit B classes to an A. The three-credit class is weighted more than the two-credit classes, so it will have a greater effect on the grade point average. 63. (a) Mean, because Car A has the highest mean of the three. (b) Median, because Car B has the highest median of the three. (c) Mode, because Car C has the highest mode of the three. 65. (a) x ≈ 49.2; median = 46.5 (b) Test Scores 1 1 3 Key:30 6 = 36 2 2 8 3 6 6 6 7 7 7 8 4 1 3 4 6 7 mean median 5 1 1 1 3 6 1 2 3 4 7 2 2 4 6 8 5 9 0 (c) Positively skewed Section 2.3 Activity (page 81) 1. The distribution is symmetric. The mean and median both decrease slightly. Over time, the median will decrease dramatically and the mean will also decrease, but to a lesser degree. 2. Neither the mean nor the median can be any of the points that were plotted. Because there are 10 points in each region, the mean will fall somewhere between the two regions. By the same logic, the median will be the average of the greatest point between 0 and 0.75 and the least point between 20 and 25. Section 2.4 (page 93) 1. The range is the difference between the maximum and minimum values of a data set. The advantage of the range is that it is easy to calculate. The disadvantage is that it uses only two entries from the data set. 3. The units of variance are squared. Its units are meaningless (example: dollars2). The units of standard deviation are the same as the data. 5. When calculating the population standard deviation, you divide the sum of the squared deviations by N, then take the square root of that value. When calculating the sample standard deviation, you divide the sum of the squared deviations by n - 1, then take the square root of that value. 7. Similarity: Both estimate proportions of the data contained within k standard deviations of the mean. Difference: The Empirical Rule assumes the distribution is approximately symmetric and bell-shaped and Chebychev’s Theorem makes no such assumption. 9. Approximately 50, or $50,000 11. (a) 17.8 (b) 39.8 13. Range = 1.1; m ≈ 10.33; s 2 ≈ 0.16; s ≈ 0.4 15. Range = 6; x = 19; s2 ≈ 3.5; s ≈ 1.9
RkJQdWJsaXNoZXIy NjM5ODQ=