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Solutions for Session 4, Part E
See solutions for Problems: E1 | E2 | E3 | E4 | E5 | E6| E7
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Problem E1 | |
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Here are some examples. If there were 15 raisins (n = 15), the median would be in position (8). If n = 16, the median would be in position (8.5). If n = 17, the median would be in position (9). A general mathematical rule is that the position of the median is (n + 1) / 2, where n is the number of items in the list.
<< back to Problem E1
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Problem E2 | |
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Since six is an even number, this is a case where you would need to draw a line to represent the position of Q1, the median of the six noodles to the left of Med. Using the formula (n + 1) / 2 from Problem E1 gives us (6 + 1) / 2 = 7 / 2 = 3.5. Therefore, position (3.5L), halfway between positions (3L) and (4L) from the low end of the noodles, is the position (though not the value) of Q1.
<< back to Problem E2
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Problem E3 | |
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Again, you'll need to draw a line to represent the position of Q3. As in Problem E2, the formula (n + 1) / 2 gives us (6 + 1) / 2 = 7 / 2 = 3.5. Therefore, position (3.5R), halfway between positions (3R) and (4R) from the noodles, is the position (though not the value) of Q3.
<< back to Problem E3
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Problem E4 | |
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Here is the Five-Number Summary:
• | Min is in position (1L); Min = 13. |
• | Max is in position (1R); Max = 127. |
• | Med is in position (13 + 1) / 2 = (7); Med = 74. |
• | There are six positions to the left of (7), so Q1 is in position (6 + 1) / 2 = (3.5L). The value of Q1 is (28 + 33) / 2; Q1 = 30.5. |
• | There are six positions to the right of (7), so Q3 is in position (6 + 1) / 2 = (3.5R). The value of Q3 is (102 + 118) / 2; Q3 = 110. |
<< back to Problem E4
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Problem E5 | |
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First, number the positions as you did in Problem E4. The center position will be marked with an (8). Here is the Five-Number Summary:
• | The minimum is in position (1L); Min = 10. |
• | The maximum is in position (1R); Max = 89. |
• | The median is in position (15 + 1) / 2 = (8); Med = 26. |
• | The first quartile is in position (7 + 1) / 2 = (4L); Q1 = 18. |
• | The third quartile is in position (4R); Q3 = 51. |
<< back to Problem E5
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Problem E6 | |
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Again, number the positions as you did in Problem E4. This time, there will be two values numbered (10) in the center of the ordered list. Here is the Five-Number Summary:
• | The minimum is in position (1L); Min = 1. |
• | The maximum is in position (1R); Max = 53. |
• | The median is in position (20 + 1) / 2 = (10.5), which means it is the average of the two values numbered (10), or (17 + 20) / 2; Med = 18.5. |
• | The first quartile is in position (10 + 1) / 2 = (5.5L), which is (4 + 4) / 2; Q1 = 4. |
• | The third quartile is in position (5.5R), which is (34 + 38) / 2; Q3 = 36. |
<< back to Problem E6
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Problem E7 | |
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The ordered list is as follows:
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Length of Needles (in millimeters) |
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37 |
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40 |
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43 |
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47 |
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48 |
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49 |
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56 |
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64 |
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68 |
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69 |
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71 |
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86 |
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93 |
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93 |
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109 |
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109 |
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110 |
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111 |
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117 |
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120 |
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a. Here is the Five-Number Summary:
Min = 37
Q1 = 48.5
Med = 70
Q3 = 109
Max = 120
b. Here is the box plot:
c. Based on these measurements:
• | All pine needles have lengths between 37 mm and 120 mm. |
• | Approximately half the pine needles have lengths less than 70 mm. |
• | Approximately half the pine needles have lengths greater than 70 mm. |
• | Approximately half the pine needles have lengths between 48.5 mm and 109 mm. |
• | The widest range of needle lengths seems to be in the third quartile, where 25% of the needles are between 70 mm and 109 mm. |
• | The longest and shortest needles fall in very tight ranges; the longest 25% of needles are between 109 and 120 mm, and the shortest 25% are between 37 mm and 48.5 mm. |
<< back to Problem E7
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