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Solutions for Session 4, Part C
See solutions for Problems: C1 | C2 | C3 | C4 | C6| C7 | C8| C9
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Problem C1 | |
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For six noodles, the median is located between the third and fourth noodle. For the six noodles to the left of Q2, this median will be Q1. Similarly, the median of the six noodles to the right of Q2 will be Q3.
<< back to Problem C1
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Problem C2 | |
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You would know that all of the lengths are between Min and Max, and that Med (Q2) divides the ordered data into two equal-sized groups. Six noodles will be shorter than Med, and six will be longer. The quartiles then divide the ordered data into four equal-sized groups. The first group contains three noodles shorter than Q1; these three noodles must have lengths the size of or larger than Min and smaller than Q1. The second group contains three noodles that are longer than Q1 but shorter than Med. The third group contains three noodles that are longer than Med but shorter than Q3. The final group contains three noodles that are longer than Q3 and the size of or smaller than Max. (For example, the ninth noodle is longer than Med and shorter than Q3.) You still don't know how the three noodles in each group are distributed -- only the ends of each interval. (For example, you don't know whether the ninth noodle is closer to Q3 or to Med.)
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Problem C3 | |
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You would know that N4 is larger than Q1, the first quartile, and that it is shorter than Med, the median.
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Problem C4 | |
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No, Ralph's reasoning is not necessarily valid. Here is a sample data set of noodle lengths, measured to the nearest millimeter: 30, 35, 38, 60, 61, 62, 64, 67, 70, 75, 90, 96. The fourth noodle, N4, has a length of 60 mm. The first quartile, Q1, is (38 + 60) / 2 = 49 mm. The median is (62 + 64) / 2 = 63 mm. In this set, N4 is closer to Med than to Q1. Remind Ralph that the information in the Five-Number Summary, while valuable, does not tell us anything about the actual values in each interval. Ralph's claim would only be valid if the data are equally spaced, for example if each length was a multiple of 10.
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Problem C6 | |
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First, find the median, between the seventh and eighth noodles. The first quartile is the median of the seven shortest noodles, which is the fourth noodle. The third quartile is the median of the seven longest noodles, which is the 11th noodle. There will be three noodles in each of the four groups.
<< back to Problem C6
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Problem C7 | |
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First, find the median, which will be the eighth noodle. The first quartile is the median of the seven shortest noodles, which is the fourth noodle. The third quartile is the median of the seven longest noodles, which is the 12th noodle. There will be three noodles in each of the four groups.
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Problem C8 | |
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If you started with 57 noodles, there would be 14 noodles in each group (the median is the 29th noodle). If you started with 112 noodles, there would be 28 noodles in each group (the median is between the 56th and 57th noodles). One possible rule is to take the number of noodles, divide by four, and then round down if you have a fractional result.
<< back to Problem C8
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Problem C9 | |
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The interquartile range contains the center 50% of the data. This is a useful interval for describing variation; if the interquartile range is small compared to the overall range (from Min to Max), it suggests that there are a lot of extreme values in the data. If the interquartile range is wide compared to the overall range, it suggests that there are few extreme values and that the data are pretty tightly grouped.
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