| Age | Spectacle Prescription |
Astigmatism | Tear Production Rate |
Recommended Lenses |
| young | yope | no | reduced | none |
| young | myope | no | normal | soft |
| young | myope | yes | reduced | none |
| young | myope | yes | normal | hard |
| young | hypermetrope | no | reduced | none |
| young | hypermetrope | no | normal | soft |
| young | hypermetrope | yes | reduced | none |
| young | hypermetrope | yes | normal | hard |
| pre-presbyopic | myope | no | reduced | none |
| pre-presbyopic | myope | no | normal | soft |
| pre-presbyopic | myope | yes | reduced | none |
| pre-presbyopic | myope | yes | normal | hard |
| pre-presbyopic | hypermetrope | no | reduced | none |
| pre-presbyopic | hypermetrope | no | normal | soft |
| pre-presbyopic | hypermetrope | yes | reduced | none |
| pre-presbyopic | hypermetrope | yes | normal | none |
| presbyopic | myope | no | reduced | none |
| presbyopic | myope | no | normal | none |
| presbyopic | myope | yes | reduced | none |
| presbyopic | myope | yes | normal | hard |
| presbyopic | hypermetrope | no | reduced | none |
| presbyopic | hypermetrope | no | normal | soft |
| presbyopic | hypermetrope | yes | reduced | none |
| presbyopic | hypermetrope | yes | normal | none |
2015年3月24日 星期二
Table 1-1 Contact Lens Data
weka.attributeSelection.SymmetricalUncertAttributeEval
(b)
(15 points) Apply Weka on this data set for attribute selection by the goodness
measure. Remember to append the Weka report.
=== Run information ===
Evaluator:
weka.attributeSelection.SymmetricalUncertAttributeEval
Search:weka.attributeSelection.Ranker -T
-1.7976931348623157E308 -N -1
Relation:
Table 1-1 Contact Lens Data
Instances:
24
Attributes: 5
Age
Spectacle Prescription
Astigmatism
Tear Production Rate
Recommended Lenses
Evaluation mode:evaluate on all training data
=== Attribute Selection on all input data ===
Search Method:
Attribute
ranking.
Attribute Evaluator (supervised, Class (nominal): 5
Recommended Lenses):
Symmetrical
Uncertainty Ranking Filter
Ranked attributes:
0.4719 4 Tear Production Rate
0.3242 3 Astigmatism
0.0583 2 Spectacle Prescription
0.0271 1 Age
Selected attributes: 4,3,2,1 : 4
Table-1-3
| Outlook | Temperature | Humidity | Windy | Play |
| Sunny | 85 | 85 | FALSE | no |
| Sunny | 80 | 90 | TRUE | no |
| Overcast | 83 | 86 | FALSE | yes |
| Rainy | 70 | 96 | FALSE | yes |
| Rainy | 68 | 80 | FALSE | yes |
| Rainy | 65 | 70 | TRUE | no |
| Overcast | 64 | 65 | TRUE | yes |
| Sunny | 72 | 95 | FALSE | no |
| Sunny | 69 | 70 | FALSE | yes |
| Rainy | 75 | 80 | FALSE | yes |
| Sunny | 75 | 70 | TRUE | yes |
| Overcast | 72 | 90 | TRUE | yes |
| Overcast | 81 | 75 | FALSE | yes |
| Rainy | 71 | 91 | TRUE | no |
2015年3月7日 星期六
case study-3-2_ blake electronics solution案例研究布雷克電子解決方案
case study-3-2_ blake electronics solution案例研究布雷克電子解決方案
5. Probability(機率)
5. Probability(機率)
A. Introduction簡介
B. Basic Concepts基本概念
C. Permutations and
Combinations排列組合
D. Poisson Distribution泊松分佈
E. Multinomial
Distribution多項分佈
F. Hypergeometric
Distribution超幾何分佈
G. Base Rates基準利率
H. Exercises練習
Case study 2-WTVX
https://prezi.com/g5arcchxx5w8/chapter-2/
Chapter 2-Probability concepts & Applications
Case study-Solution to Case study
WTVX was the only station that had a weatherperson who was the member of the Amarican Meteorological Society (AMS) - Joe Hummel - who was always trying to find innovative ways to make weather interesting by inviting questions during the actual broadcast
What the chances were of getting 15 days of rain in the next month (30 days), if there would be a 70% chance of rain everyday, and that what happens on one day (rain or shine) was not in any way dependent on what happened the day before?
Joe's mistake
Joe's quick caculation:
(70%) * (15 days/ 30 days)
= (70%) * (1/2)
= 35%
Binomial distribution
=> Binomial formula
Probability of r successes in n trials =
n = number of trials
r =number of successes
p = the probability of success on any single trial
q = 1 – p = the probability of a failure
1. What are the chances of getting 15 days of rain during the next 30 days?
We want to find the probability of getting 15 days of rain during the next 30 days.
Thus, the chance of rain every day is 70%, and that what happens on one days (rain or shine) was not in any way dependent on what happened the day before.
In this case;
n =number of days = 30
r= number of rainy days = 15
p = probability of rain = 70% = 0.70
q = 1-p = 1- 0.70 = probability of sun = 30% = 0.30
2. What do you think about Joe’s assumptions concerning the weather for the next 30 days?
Sajeena Vivatbutsiri _ G1022241301
Mayy Nguyen _ G1022241009
Probability of r successes in n trials
Thus, the probability of getting 15 days of rain in the next 30days is 1.06%.
- Joe’s assumption that the chance of getting 15 days of rain in the next 30 days (35%) is wrong. He uses the rule of the rule of three. This method is to find the fourth term of a mathematical proportion when three terms are known. It is based on the principle that the product of the first and fourth terms (called the extremes) is equal to the product of the second and third terms (called the means).
The two variables are exactvalue, for example given that one apple costs 10 NTD, so 2 apples cost 20 NTD. But this case is talking about chance which is not exact value. It means the probability of rain today is perhaps 80% as opposed to 70%. Thus, Joe should concern that this is the probability. He should think about the probability methods and 70% chance of rain every day doesn’t depend on the day before.
第2章概率的概念及應用案例研究，解決方案案例研究WTVX的是，有一個weatherperson誰是Amarican氣象學會（AMS）的成員中唯一站 - 喬胡默爾 - 誰總是想尋找創新的方法，使天氣有趣的實際廣播期間邀請的問題什麼機會是越來越雨15天下個月（30天），如果會有雨每天70％的機會，在一天（晴雨）什麼情況是不以任何方式依賴於前一天發生了什麼事？喬的錯誤喬的快速計算研究：（70％）*（15天/ 30天）=（70％）*（1/2）= 35％二項分佈=>二項式公式ř成功在n次試驗的概率=
試驗N =數量R =一些成功P =任何單一試驗成功的概率Q = 1 - P =失敗的概率1.什麼是讓雨15天在未來30天內的機會？我們希望找到讓雨15天在未來30天內的概率。因此，雨每天的機率為70％，並且會發生什麼在一個工作日（晴雨）不以任何方式依賴於前一天發生的事情。在這種情況下;N =天數= 30R =陰雨天數= 15p值=下雨的概率= 70％= 0.70Q = 1-P =太陽1- 0.70 =概率= 30％= 0.302.你認為對有關天氣在未來30天喬的假設？Sajeena Vivatbutsiri _ G1022241301Mayy阮_ G1022241009ř成功在n次試驗的概率因此，得到的雨水15天下一30天的概率是1.06％。 - 喬的假設，讓雨15天未來30天（35％）的機率是錯誤的。他用三個規則的規則。此方法是找到的數學比例第四術語時三個項是已知的。它是基於這樣的原則，第一和第四方面（稱為極值）的乘積等於所述第二項和第三項的產物（稱為單元）。
這兩個變量是exactvalue，例如因為一個蘋果花費10 NTD，所以2個蘋果花20台幣。但是，這種情況下談論機會這是不準確的值。這意味著當今下雨的概率是可能的80％，而不是70％。因此，喬應該關注，這是概率。他應該想想概率方法和70％的機率下雨，每天不依賴於前一天。
Chapter 2-Probability concepts & Applications
Case study-Solution to Case study
WTVX was the only station that had a weatherperson who was the member of the Amarican Meteorological Society (AMS) - Joe Hummel - who was always trying to find innovative ways to make weather interesting by inviting questions during the actual broadcast
What the chances were of getting 15 days of rain in the next month (30 days), if there would be a 70% chance of rain everyday, and that what happens on one day (rain or shine) was not in any way dependent on what happened the day before?
Joe's mistake
Joe's quick caculation:
(70%) * (15 days/ 30 days)
= (70%) * (1/2)
= 35%
Binomial distribution
=> Binomial formula
Probability of r successes in n trials =
n = number of trials
r =number of successes
p = the probability of success on any single trial
q = 1 – p = the probability of a failure
1. What are the chances of getting 15 days of rain during the next 30 days?
We want to find the probability of getting 15 days of rain during the next 30 days.
Thus, the chance of rain every day is 70%, and that what happens on one days (rain or shine) was not in any way dependent on what happened the day before.
In this case;
n =number of days = 30
r= number of rainy days = 15
p = probability of rain = 70% = 0.70
q = 1-p = 1- 0.70 = probability of sun = 30% = 0.30
2. What do you think about Joe’s assumptions concerning the weather for the next 30 days?
Sajeena Vivatbutsiri _ G1022241301
Mayy Nguyen _ G1022241009
Probability of r successes in n trials
Thus, the probability of getting 15 days of rain in the next 30days is 1.06%.
- Joe’s assumption that the chance of getting 15 days of rain in the next 30 days (35%) is wrong. He uses the rule of the rule of three. This method is to find the fourth term of a mathematical proportion when three terms are known. It is based on the principle that the product of the first and fourth terms (called the extremes) is equal to the product of the second and third terms (called the means).
The two variables are exactvalue, for example given that one apple costs 10 NTD, so 2 apples cost 20 NTD. But this case is talking about chance which is not exact value. It means the probability of rain today is perhaps 80% as opposed to 70%. Thus, Joe should concern that this is the probability. He should think about the probability methods and 70% chance of rain every day doesn’t depend on the day before.
第2章概率的概念及應用案例研究，解決方案案例研究WTVX的是，有一個weatherperson誰是Amarican氣象學會（AMS）的成員中唯一站 - 喬胡默爾 - 誰總是想尋找創新的方法，使天氣有趣的實際廣播期間邀請的問題什麼機會是越來越雨15天下個月（30天），如果會有雨每天70％的機會，在一天（晴雨）什麼情況是不以任何方式依賴於前一天發生了什麼事？喬的錯誤喬的快速計算研究：（70％）*（15天/ 30天）=（70％）*（1/2）= 35％二項分佈=>二項式公式ř成功在n次試驗的概率=
試驗N =數量R =一些成功P =任何單一試驗成功的概率Q = 1 - P =失敗的概率1.什麼是讓雨15天在未來30天內的機會？我們希望找到讓雨15天在未來30天內的概率。因此，雨每天的機率為70％，並且會發生什麼在一個工作日（晴雨）不以任何方式依賴於前一天發生的事情。在這種情況下;N =天數= 30R =陰雨天數= 15p值=下雨的概率= 70％= 0.70Q = 1-P =太陽1- 0.70 =概率= 30％= 0.302.你認為對有關天氣在未來30天喬的假設？Sajeena Vivatbutsiri _ G1022241301Mayy阮_ G1022241009ř成功在n次試驗的概率因此，得到的雨水15天下一30天的概率是1.06％。 - 喬的假設，讓雨15天未來30天（35％）的機率是錯誤的。他用三個規則的規則。此方法是找到的數學比例第四術語時三個項是已知的。它是基於這樣的原則，第一和第四方面（稱為極值）的乘積等於所述第二項和第三項的產物（稱為單元）。
這兩個變量是exactvalue，例如因為一個蘋果花費10 NTD，所以2個蘋果花20台幣。但是，這種情況下談論機會這是不準確的值。這意味著當今下雨的概率是可能的80％，而不是70％。因此，喬應該關注，這是概率。他應該想想概率方法和70％的機率下雨，每天不依賴於前一天。
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