期中考前回顧助教 : 王珊彗. Copyright 2009 Cengage Learning

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1 期中考前回顧助教 : 王珊彗

2 考前提醒考試時間 :11/17( 四 )9:10~12:10 考試地點 : 管二 104 ( 上課教室 ) 考試範圍 :C1-C9, 選擇 + 計算注意事項 : 考試請務必帶工程計算機可帶 A4 參考紙 ( 單面不能浮貼 ) 計算過程到第四位, 結果寫到小數點第二位不接受沒有公式, 也不接受沒算出最後答案考試只會附上 standard normal distribution 的表格 11/24( 四 ) 上課繳交 project 2

3 本課程回顧 3

4 Ch 1&2- 圖形技巧資料的型態有三種名目資料 (Nominal data)/ 屬質 (Qualitative) 資料的值是不同的類別長條圖圓餅圖雙變數長條圖順序資料 (Ordinal data) 看起來像名目資料, 但是它們的數值是有順序的長條圖圓餅圖區間資料 (Interval data) / 屬量 (Quantitative) 是真實的數字, 諸如身高體重所得和距離之類直方圖累積機率分配圖散佈圖 Sturges s rule: 1+3.3*log(n) 4

5 Ch1&2- 圖形技巧橫斷面資料 (Cross-Sectional) 時間序列資料 (Time-Series Data) 線形圖 (Line chart) 莖葉圖 (Stem-and-Leaf display) 克服直方圖無法看到每個組資料的情況 5

6 Ch 4 敘述性統計

7 Agenda 了解資料的分佈型態 Mean, Median, Mode Standard Deviation, Variance 了解資料的相對位子意涵 Percentiles Quartiles & Box plot 如何知道兩變數的線性關係 Covariance, Correlation 習題演練 7

8 一了解資料的相對位子意涵 Find the location of any percentile using the formula 算出在資料中的相對位子有些統計課本, 公式會不同 LP = (n + wherel P P 1) 100 is thelocation of thep th percentile 8

9 一了解資料的相對位子意涵 Quartiles: Commonly used percentiles First (lower) decile First (lower) quartile, Q 1, Second (middle)quartile,q 2, Third quartile, Q 3, Ninth (upper) decile = 10th percentile = 25th percentile = 50th percentile = 75th percentile = 90th percentile 9

10 一 Interquartile Range (IQR) This is a measure of the spread of the middle 50% of the observations Large value indicates a large spread of the observations Interquartile range = Q 3 Q 1 Outlier =1.5* IQR 10

11 ㄧ Box Plot ( 盒鬚圖 ) This is a pictorial display that provides the main descriptive measures of the data set: L - the largest observation Q 3 - The upper quartile Q 2 - The median Q 1 - The lower quartile S - The smallest observation IQR=Q3-Q1 鬍鬚長度 : 取數字比較大的鬍鬚長度 : 取數字比較小的 1.5(Q 3 Q 1 ) 1.5(Q 3 Q 1 ) Outlier Whisker Whisker S Q 1 Q 2 Q 3 L 11

12 練習題 :Box plot The score of baseball player s season scores are show as follow a. Determine the first, second, and third quartiles of the scores b. Draw the box plot 9.12

13 編號分數 Step 1: 排序後數列 Step2: 求出下列數值 Q1: L25=(13+1)*25/100=3.5 the first quartile is Q2: L50=(13+1)*50/100=7 the second quartile is 14.7 Q3: L75=(13+1)*75/100=10.5 the third quartile is 15.6 IQR= = *2.55= =9.225 ( 左 ) =19.425( 右 ) S:10 ( 左邊鬍鬚畫到 10 就好 ) L:18.5 ( 右邊鬍鬚畫到 18.5) 此資料顯示, 所有數值都介於 1.5*IQR( 沒有 outlier) 3 位子的數值 : 表示需要 3 和 4 位子間,0.5 位子的數值 :( )*0.5=0.85 Q1: = 位子的數值 : 表示需要 10 和 11 位子間,0.5 位子的數值 :( )*0.5=0.3 Q1: =

14 二 Variance The variance of a population/ sample is: 速解法 14

15 二 Coefficient of Variation (CV) 線性關係量數共變異數, 相關係數母體共變異數 / 樣本共變異數 N i= 1 σ xy = ( x µ )( y µ ) i x i y N s xy = N i= 1 ( x x)( y y) i n 1 i 母體相關係數 / 樣本相關係數 ρ = σ x xy σσ y r = s x xy ss y 15

16 二 Coefficient of Variation (CV) The coefficient of variation of a set of observations is the standard deviation of the observations divided by their mean, that is: Population coefficient of variation = CV = σ / µ Sample coefficient of variation = cv = s/ x CV 是相對離勢量數 (measure of relative dispersion) 比較幾組資料單位不同的差異情形 ( 公斤與體重無法比 ) 比較幾組資料單位相同, 但平均數相差懸殊之差異情形 EX: 股票風險一樣時, 會選擇平均報酬較高的 16

17 練習題 Are the marks one receives in a course related to the amount of the time spent studying the subject? To analyze this mysterious possibility, a student took a random sample of 10 students who had enrolled in an accounting class last semester. She asked each to report his or her mark in the course and the total number of hours spent studying accounting. These data are listed here. Studying Marks a. Calculate the covariance. b. Calculate the coefficient of correlation. c. What do the statistics you have calculated tell you about the relationship between marks and study time? 1. 讀書時間與成績呈正相關 2. 因為讀書時間長, 所以可以獲得高成績哪個是對的? 9.17

18 9.18

19 Ch 6 機率

20 Agenda 機率相關名詞與關係 ( 如何判斷題目 ) 貝氏定理 Bayes Law Probability Trees 題目練習 20

Rule P(A B) 聯集 P (A or B) Either, union P( A or B) =P(A) +P(B)-P (A and B) Addition

21 ㄧ機率相關名詞與關係 ( 如何判斷題目 ) 語法翻譯關鍵字 / 公式圖形意涵 P(A B) 交集 P( A and B) Both, intersection, joint P( A and B) =P(A) *P(B) P(A and B)=P(A B)*P(B) P(A and B)= P(B A)*P(A) Multiplication Rule P(A B) 聯集 P (A or B) Either, union P( A or B) =P(A) +P(B)-P (A and B) Addition Rule P A B Conditional pro. ( 條件機率 ) The probability of A given B P(A B)= )(* +,-.) )(.) 21

22 一獨立與互斥 Ø 獨立 (independent) ü P(A B) = P(A) or P(B A) = P(B) ü P(A and B) = P(A)P(B) Ø 互斥 (Mutually exclusive) ü P(A or B) = P(A) + P(B) ü P(A and B)=0 22

23 二貝氏定理 ( 概念 ) The Graduate Management Admission Test (GMAT) is a requirement for all applicants of MBA programs. There are a variety of preparatory courses designed to help improve GMAT scores, which range from 200 to 800. Suppose that a survey of MBA students reveals that among GMAT scorers above 650, 52% took a preparatory course, whereas among GMAT scorers of less than 650 only 23% took a preparatory course. An applicant to an MBA program has determined that he needs a score of more than 650 to get into a certain MBA program, but he feels that his probability of getting that high a score is quite low--10%. He is considering taking a preparatory course that cost $500. He is willing to do so only if his probability of achieving 650 or more doubles. What should he do? 現在有幾個事件? (A: GMAT>650, B: 去補習 ) 23

24 二貝氏定理 ( 概念 ) A:GMAT > 650 (prior ) A c : GMAT <650 (prior) A A C A B 已知 : P(B A)=P ( 有補習 GMAT > 650) 反求 : P(A B) = P ( 多益 > GMAT 去補習 ) B: 有參加補習班 (after) 24

25 二貝氏定理 ( 概念 ) P(A B)= ) (* +,-.) )(.) = )( * +,-.) = ) * )(. *) ). +,- * 0)(. +,- * 1 ) ) * ) B A 0) * 1 )(. * 1 ) P(A and B)= P(A)*P(B A) 25

26 練習題 : 貝氏定理 9.26

27 二貝氏定理 ( 概念 ) A:DIJA increase (prior ) A c : DIJA not increase (prior) P(A): 04 P(A C )=0.6 B P(B A): 0.2 A P(B A c ): 0.05 B: NASDAQA(after) 已知 : P(B A)=P (NASDAQA DIJA increase) 反求 : P(A B) = P (DIJA increase NASDAQA ) 27

28 Let A=DJIA increase and B= NASDAQ increase P(A B)= ) (* +,-.) = )(.) )( * +,-.) = ). +,- * 0)(. +,- * 1 ) ) * )(. *) ) * ) B A 0) * 1 )(. * 1 ) 9.28

29 Ch 7 隨機變數與機率分配

30 題目公式化不管是一個隨機變數兩個隨機變數或是特定分配 The population mean The population variance or standard deviation 特定值的機率 ex: P(X=3) 將題目公式化 The number of students who seek assistance with their statistics assignments is Poisson distributed with a mean of three per day. What is the probability that no student seek assistance tomorrow? P(X=0) What is the probability that two student seek assistance tomorrow? P(X=2) What is the probability two or more than two student seek assistance tomorrow? P(X>=2) What is the probability at least two student seek assistance tomorrow? P(X>=2) What is the probability less than two student seek assistance tomorrow? P(X<2) Shan Huei, Wang 30

31 Agenda 隨機變數的特性兩個隨機變數 ( 以上 ) 時 Bivariate Distributions (Joint distributions) Portfolio concept 特殊分配 Binomial distribution Poisson Distribution Shan Huei, Wang 31

32 ㄧ隨機變數和離散機率分配隨機變數為了降低分析的複雜性, 將所有可能結果加以數值化短期不知道是什麼, 長期下來會呈現某種分配離散隨機分配圖表 (table), 公式 (formula), or 圖形 (graph) Write down the probability distribution P(X = x) = p(x) = C n x n x x p (1 p) 期望值 E ( X ) = µ = xi p( x i ) all x i 32

33 ㄧ隨機變數和離散機率分配變異數 / 標準差 V (X ) = σ 2 2 = E[ (X µ ) ]= i l V( X) = σ = E( X ) µ = x p( x ) µ i i all x i σ = 2 σ 期望值法則 / 變異數法則 E(c) = c V(c) = 0 E(X + c) = E(X) + c V(X + c) = V(X) E(cX) = ce(x) V(cX) = c 2 V(X) 33

34 ㄧ隨機變數和離散機率分配 X+ Y 的隨機變數 E(X+Y) =E(X) + E(Y); V(X+Y) = V(X) +V(Y) +2COV(X,Y) When X and Y are independent COV(X,Y) = 0, and V(X+Y) = V(X)+V(Y) 34

35 練習題 :portfolio Two investments, X and Y, have the following characteristics: If the weight of portfolio assets assigned to investment X and Y are equal, compute the E X = 50, E Y = 100, V X = 9000, V Y = 15000,Cov X, Y = 7500 a. E 0.5X + 0.5Y = = 75 b. σ 0.5X + 0.5Y = 0.5 J J (0.5)(0.5)(7500) 9.35

36 二 Bivariate Distributions (Joint distributions) Bivariate (or joint) distribution The 共變異數 Ø COV(X,Y) = Σ(X µ x )(Y- µ y )p(x,y)=e(xy)-e(x)e(y) 相關係數 ρ NOP QRS(T,U) V W V X joint probability function satisfies thefollowingconditions 0 p(x,y) 1 p(x,y) = 1 all x all y : 36

37 練習題 :Joint probability The following table lists the probabilities of unemployed females and males and their educational attainment. Female male Less than highschool Highschool graduate Somecollege/ university-no degree College/University graduate a. If one unemployed person is selected at random, what is the probability that he or she did notfinish high school? b. If one unemployed female is selected at random, what is the probability that she has a college or university degree? c. If an unemployed high school graduate is selected at random, what is the probability that he is a male? 9.37

38 練習題 :Joint probability Female male P(A) Less than highschool Highschool graduate Some college/ university-no.132 degree College/University graduate P(B) a. If one unemployed person is selected at random, what is the probability that he or she did not finish high school? b. If one unemployed female is selected at random, what is the probability that she has a collegeor university degree? c. If an unemployed high school graduate is selected at random, what is the probability that he is a male? 9.38

39 三 Binomial distribution 二項分配 ( 成功 or 失敗 ) P(X = x) = p(x) = C n x n x x p (1 p) E(X) = µ = np V(X) = s 2 = np(1- p) 知道圖形長相知道平均數與標準差 39

40 練習題 : Binomial distribution When a customer places an order with Alibaba, their IT department automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is Suppose that, on a given day, 20 customers place orders. Assume that the number of customers that the IT department detects as having exceeded their credit limit is distributed as a binomial random variable. a. What are the mean and standard deviation of the number of customers exceeding their credit limits? b. What is the probability that zero customers will exceed their limits? c. What is the probability that one customer will exceed his or her limit? 9.40

41 練習題 : Binomial distribution When a customer places an order with Alibaba, their IT department automatically checks to see if the customer has exceeded his or her credit limit. Past records indicate that the probability of customers exceeding their credit limit is Suppose that, on a given day, 20 customers place orders. Assume that the number of customers that the IT department detects as having exceeded their credit limit is distributed as a binomial random variable. a. What are the mean and standard deviation of the number of customers exceeding their credit limits? mean=1, standard deviation = b. What is the probability that zero customers will exceed their limits? P(X=0) = C Y JY (0.05) Y (0.95) JY = c. What is the probability that one customer will exceed his or her limit? P(X=1) = C Z JY (0.05) Z (0.95) Z[

42 四 Poisson distribution Poisson 分配 ( 單位時間的來客數 ) P(X = x) = p(x) = e µ x! µ x x = 0,1,2... E(X) = V(X) = µ 42

43 練習題 :Poisson distribution Airline passenger arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. a) Compute the probability of no arrivals in a one-minute period. b) Compute the probability that three or fewer passenger arrive in a one-minute period. c) Compute the probability of no arrivals in a 15-second period. d) Compute the probability of at least one arrivals in a 15-second period. 9.43

44 練習題 :Poisson distribution Airline passenger arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. a) Compute the probability of no arrivals in a one-minute period. P(X=0) a) Compute the probability that three or fewer passenger arrive in a one-minute period. P(X<=3) 9.44

練習題 :Poisson distribution Airline passenger arrive randomly and independently at the passenger-screening facility at a major international airport.

45 練習題 :Poisson distribution Airline passenger arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. C. Compute the probability of no arrivals in a 15-second period. P(X=0) Compute the probability of at least one arrivals in a 15-second period. P(X>=1) 9.45

46 Ch 8 連續機率分配

47 Agenda Uniform distribution Normal Distribution Exponential Distribution Poisson & Exponential distribution 查表 ( 會給標準常態分配表 ) Shan Huei Wang 47

48 ㄧ Uniform distribution 均勻分配 (Uniform distribution) 1 f ( x) = a x b. b a a + b ( b a) E(X) = V (X) = F(x)= 1/(b-a) a b 48

二 Normal distribution 常態分配 x 1 (1/ 2) f ( x) = e σ 2π where π = 3.14159.

49 二 Normal distribution 常態分配 x 1 (1/ 2) f ( x) = e σ 2π where π = µ σ and e x = Normalization ( 正規化 ) X µ Z = σ 可以查表 x x 49

50 練習題 : Normal distribution The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions. a. What is the probability of completing the exam in one hour or less? b. What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes? c. Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time? 9.50

51 練習題 : Normal distribution The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions. a. What is the probability of completing the exam in one hour or less? P(X<=60) b. What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes? P(60<X<75) 9.51

52 練習題 : Normal distribution The time needed to complete a final examination in a particular college course is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes. Answer the following questions. c. Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time? 9.52

53 練習題 : Normal distribution ( 求 X) The height of 2 years old children are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches. Pediatricians regularly measure the height of toddlers to determine whether there is a problem. There may be a problem when a child is in the top or bottom 5% of heights. Determine the heights of 2-year-old children that could be a problem. 9.53

54 練習題 : Normal distribution ( 求 X) The height of 2 years old children are normally distributed with a mean of 32 inches and a standard deviation of 1.5 inches. Pediatricians regularly measure the height of toddlers to determine whether there is a problem. There may be a problem when a child is in the top or bottom 5% of heights. Determine the heights of 2-year-old children that could be a problem 這裡 95%, Z 是 1.96 嗎? 9.54

55 Shan Huei Wang 55

56 9.56

57 三 Exponential distribution Ø 指數分配 (Exponential 分配 ) 兩個事件發生所需的時間 f λx ( x) = λe, x 0 E(X) = Z \ ; V(X) = Z \ P(X > x) = e λx. P(X < x) = 1 e λx P(x 1 < X < x 2 ) = e λ(x1) e λ(x2) 57

58 練習題 :Exponential distribution 9.58

59 練習題 :Exponential distribution 9.59

60 四 Poisson vs Exponential Distribution EX: 一小時有 6 個客人 (a) u Poisson = 6 person( 單位時間的來客數 ) (b) u exponential = 1/6 hr ( 客人與客人間來的時間 ) λ = 1/u= 6 ( 帶公式要用 6) 今天打工一小時, 要服務 6 個客人 (Poisson) 大概 1/6 hr (10 分鐘 ) 後, 下一個客人會來 (exponential) 60 Shan Huei Wang

61 練習題 :Poisson vs Exponential Distribution a) Show the probability distribution for the time between interruptions. b) What is the probability a businessperson will have no interruptions during a 15- minute period? c) What is the probability that the next interruption will occur within 10 minutes for a particular businessperson? 9.61

62 練習題 :Poisson vs Exponential Distribution a) Show the probability distribution for the time between interruptions. exponential 9.62

63 練習題 :Poisson vs Exponential Distribution What is the probability a businessperson will have no interruptions during a 15- minute period? Poisson è 4 =

64 練習題 :Poisson vs Exponential Distribution What is the probability that the next interruption will occur within 10 minutes for a particular businessperson? exponential 9.64

65 Ch 9 抽樣分配

66 學習目標 1. Sampling distribution & 中央極限定理 (CLT) Example: 1. 一般狀況 2. Proposition 3. Difference (X1-X2) 9.66

67 1. Sampling distribution and Central Limit Theorem Sampling distribution 特色 : 1. 母體為常態分配 : If the population is normal, then X` is normally distributed for all values of n. 2. 母體不是常態分配 ( 依據中央極限定理 ): If the population is non-normal, then X` is approximately normal only for larger values of n (n>30). 9.67

68 9.68

69 類型二 : Proportion(Proportion, Binomial) The parameter of interest for nominal data is the proportion of times a particular outcome (success) occurs. To estimate the population proportion p we use the sample proportion. 母體參數樣本統計量 The number of successes The estimate of p = p^ = X n

70 類型二 : Proportion Using the laws of expected value and variance, we can determine the mean, variance, and standard deviation of. (The standard deviation of the proportion.) is called the standard error of Sample proportions can be standardized to a standard normal distribution using this formulation: 9.70

71 練習題 : Proportion In the last election a state representative received 52% of the votes cast. One year after the election the representative organized a survey that asked a random sample of 300 people whether they would vote for him in the next election. If we assume that his popularity has not changed what is the probability that more than half of the sample would vote for him? P= p = 9.71

72 類型三 :Difference of two means Since the distribution of is normal and has a mean of and a standard deviation of We can compute Z (standard normal random variable) in this way: 9.72

73 練習題 :Difference of two means Suppose that we have the first exam score of two classes, the first class follow normal populations with the means 75 and standard deviations 20. The second class follow normal populations with the means 65 and standard deviations 21. If random samples of size 5 students are drawn from each population, what is the probability that the the class 1 is outperformance than the class 2? 9.73

74 u1=75 u2=65 σ dz =20 σ dz =21 n1=n2=5 u Z u J = σ gh igj = σ Z J J + σ J = n Z n J 20J J 5 = P(X Z - X J >0) =p(z > YiZY ) = P(Z > 0.77)=0.78 ZJ.[m[ 9.74

0 0 = 1 0 = 0 1 = = 1 1 = 0 0 = 1

0 0 = 1 0 = 0 1 = = 1 1 = 0 0 = 1 0 0 = 1 0 = 0 1 = 0 1 1 = 1 1 = 0 0 = 1 : = {0, 1} : 3 (,, ) = + (,, ) = + + (, ) = + (,,, ) = ( + )( + ) + ( + )( + ) + = + = = + + = + = ( + ) + = + ( + ) () = () ( + ) = + + = ( + )( + ) + = = + 0

期中考前回顧 助教 : 王珊彗. Copyright 2009 Cengage Learning

期中考前回顧助教 : 王珊彗. Copyright 2009 Cengage Learning