When you are asked to do a calculation, you do not need to compute the decimal equivalent of a fraction or radical (square root). Fractions should be reduced to lowest terms for convenience in grading. Radicals do not need to be reduced.
計算を行うときには分数または根数のままを書いてもよい。少数にする必要は ない。ただし、分数の分母と分子は互いに素にすること。
For Problems 1 to 6, use Data Set A. (Each student receives a different data set. Make sure your Data Set ID is correctly entered in the space at the top of the page.) Data Set A is a data set of rice yield (productivity) in metric tonnes per hectare in Japan. Classify productivity according to the following subjective yield scale: x <= 5 is “poor,” 5 < x <= 6.5 is “average,” 6.5 < x <= 8 is “good,” and x > 8 is “excellent.”
問題1〜6にデータセットAを利用してください。(注意:皆に別のデータ を用意する。必ずデータセットIDを確認すること。)データセットAは日本の 米の生産性(t/ha)である。 サイズを「主観的生産性」に以下の表によって区別する:x <= 5は「わる い」、5 < x <= 6.5は「平均」、6.5 < x <= 8は「よい」、そして x > 8は「優秀」。
Copy your data set in the space below:
ここにデータを写ってください:
[7.4, 7.0, 7.5, 7.4, 6.8, 6.5, 6.7, 7.5, 6.4, 6.5]
Sort your raw data, and write the sorted data here.
データを順序に並べてここに書くこと。
[6.4, 6.5, 6.5, 6.7, 6.8, 7.0, 7.4, 7.4, 7.5, 7.5]
Label Up to Value Frequency Relative CDF poor 5 4.25 0 0.0 0.0 average 6.5 5.75 3 0.3 0.3 good 8 7.25 7 0.7 1.0 very fast infinity 8.75 0 0.0 1.0The median is "['good']."
The values are entered in the table above. Reason: For the middle ranges, the middle of the range is chosen. For the end ranges (which are actually open-ended), the distance from the middle values is chosen to make the respresentative values evenly space (all differences are the same).
Label x f(x) xf(x) x - mu (x-mu)^2 (x-mu)^2f(x) poor 4.25 0.00 0.0000 -2.55 6.50 0.0000 average 5.75 0.30 1.7250 -1.05 1.10 0.3307 good 7.25 0.70 5.0750 0.45 0.20 0.1418 very fast 8.75 0.00 0.0000 1.95 3.80 0.0000 moments 6.8000 0.4725The mean is μ=6.8, the variance is σ^2=0.47250000000000003, and the standard deviation is σ=0.687386354243376.
Note that your answer to this kind of question should not depend on the data! This case is relatively difficult. One likely "hidden" factor is that the farmer knows about the productivity of the different fields, so even though he uses the same seed, he may put different amounts of effort into cultivating each field according to productivity. Another one is that the fields probably have different levels of exposure to animals, disease, and weather, so the losses from these causes will vary. Both of these factors suggest that the distribution of yields differs for each field.
In this case, although the problem states the farms are representative for each prefecture, 2010 might not be a representative year. For major weather disasters such as typhoons, even though the farms are in ten different prefectures it seems likely that losses from weather conditions would be correlated (the observations are not independent).
In this case, we don't know that the farm was representative of other farms, or what aspects of management may have changed. If the farm is unrepresentative, its time series data might not be useful to assess trends at other farms. If management has changed, such as an older farmer retiring and passing on management to a successor, that would likely change productivity according to the skill of the new manager in ways that would not apply to other farms.
Sample answers are provided above with each variant.For Problems 6 to 7, use Data Set B. (Each student receives a different data set.) Data Set B is a data set of examination scores on a 0-100 scale. 問題6〜7にデータセットBを利用して ください。(注意:皆に別のデータを用意する。必ずデータセットIDを確認す ること。)データセットBはある試験の点数データで、0〜100の範囲である。 Copy your data set here: ここにデータを写ってください:
[67, 66, 74, 73, 92, 62, 60, 47, 65, 81]
The median is the 50-percentile value, that is, the value such the 50% of the distribution is larger than this value and 50% is smaller. Sorting the data set gives [47, 60, 62, 65, 66, 67, 73, 74, 81, 92], so the median is some number between 66 and 67. Some statisticians report both (the median values are 66 and 67), others take the average (the median value is 66.5). The letter grade distribution isLabel Up to Value Frequency Relative CDF D 60 55 2 0.2 0.2 C 70 65 4 0.4 0.6 B 80 75 2 0.2 0.8 A 100 85 2 0.2 1.0The median of the letter grades is ['C']. This does not correspond accurately to the letter grade corresponding to the median point score.
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This is purely an observational study. An experiment would require telling some people to smoke and others to stop smoking. That was not done.
Probably because these variables modify the effect that smoking has on health. By studying each group separately, the study was able to compare similar people who differed only in their smoking history. (Note that this does not help control for unrecorded factors that might be correlated with health risks of smoking, such as obesity.)
That is not a valid inference from this data. The problem is that it is very hard to stop smoking; one needs a strong incentive. Being told by your doctor that smoking has destroyed your health and you must stop or you will die is much more motivating than if you are healthy and your advises you to stop. Probably many of those who recently stopped had such motivation, and therefore were less healthy than those who "happily" continue to smoke.
(i) H T T H T H (ii) H H H H H H(The coin must land H or T in the order given; H = heads, T = tails.) Which of the following statements is correct? Explain briefly. 硬貨を6回なげる。数多くの結果の順序の中で
(i) H T T H T H (ii) H H H H H Hの2つがある。(硬貨は書いた通りHまたはTにならなければならない。H =表、T=裏。)以下のケースの中から正しいのはどれであるか。その理由 を簡単に説明せよ。
The correct answer is "both sequences are equally likely" because the tosses are independent. Since both heads and tails have the same probability of 1/2, each sequence has probability 1/(2*2*2*2*2*2) = 1/64.
The precise answer is, "We don't know, because it depends on the school." Let A = the event "the child is in 2d grade" and B = the event "the child is female". The two events are independent if P(A and B) = P(A)P(B). Since the child is picked at random, P(A) = (number of 2d grade students) / (number of all students), and P(B) = (number of female students) / (number of all students). Now consider the case (unusual, but I know of such a school!) where all of the second grade students are male (but there are some girls in other grades). Then P(A and B) = 0, but P(A)P(B) > 0. So they are not independent. On the other hand, in this case A and B are mutually exclusive. Of course normally we expect male and female to be half and half in all grades, so A and B should be "close" to independent, and similarly they would not be mutually exclusive. But the example shows that we can't be sure.
Pr({}) <= Pr(A) <= Pr(A ∪ B) <= Pr(Ω)
Pr({} <= Pr(B) <= Pr(A ∪ B) <= Pr(Ω)
Pr(A ∩ B) <= Pr(A) <= Pr(Ω)
Pr(A ∩ B) <= Pr(B) <= Pr(Ω)