Econometrics Unit 2
Suppose you are drawing 2) Suppose two seals are trying to get pregnant within 3 months. P(Pregnancy) = 1/3 each month; stays constant. What is the probability of getting pregnant within 3 months?
Tree: Total possibilities - 8 sodas - 2 beers --> All items = equally likely to be gotten --> Purchase of 2 items; probability for getting 0, 1, 2 beers? P(1st purchase): Coke = 8/10 Beer = 2/10 P(2nd purchase, if you got coke the first time): Coke = 7/9 (1 less coke) Beer = 2/9 P(2nd purchase, if you got beer the first time): Coke = 8/9 Beer = 1/9 P(1st purchase = coke; 2nd purchase = coke): (8/10)(7/9) = 56/90 (0 beers) P(1st purchase = coke; 2nd purchase = beer): (8/10)(2/9) = 16/90 P(1st purchase = beer; 2nd purchase = coke): (2/10)(8/9) = 16/90 P(1st purchase = beer; 2nd purchase = beer): (2/10)(1/9) = 2/90 Thus, probability for 0-2 beers--> # Beers: 0; 1; 2 P(# Beers): 56/90; 32/90; 2/90
Problem Set 3: Part 3- https://imgur.com/gallery/0p5eEwY shows the probability distribution for A and B. Complete the table, assuming that A and B are independent, with P[A] > P[B] , and answer the questions. 1) P[A∩B] = ? 2) P[A∩B bar] = ? 3) P[A∪B] = ? 4) P[A|B] = ? 5) P[A bar∪B bar] = ?
i. Remember that if A and B are independent: ==> Since independent, P(AnB) in table = P(A) x P(B)! 0.18 0.12 0.42 0.28 P(Abar)=0.4; P(Bbar)=0.7 --> P(A|B) = P(A) --> P(AnB) = P(A)*P(B) ii. Also keep in mind that P(A bar) = 1-P(A) iii. Given that P[A] > P[B] 1) P[A∩B] = P(A) x P(B) = 0.60 x 0.30 = 0.18 2) P[A∩B bar] = 0.60 x (1-0.30) = 0.42 3) P[A∪B] = [P(A)+P(B)]-(P(AnB)) = [0.60+0.30]-[0.18] = 0.72 4) P[A|B] = [P(AnB)/P(B)] = 0.18/0.30 = 0.60 5) P[A bar∪B bar] = [P(A bar)+P(B bar)]-(P(A barnB bar)) = [0.40+0.70]-[0.40 x 0.70] = 0.82
What are important things to know for statistics? What are the 2 types of estimates?
1. Expected values 2. Variance 2 types: 1. Point estimates 2. Interval estimates
1) Linear transformations: ☆ E[a + b⋅X] = ? ☆ Var(a + b⋅X) = ? ☆ Cov(a + b⋅X,Y) = ? ☆ Corr(a + b⋅X,Y) = ? 2) Linear combos: ☆ E[X +Y] = ? ☆ Var(X +Y) = ? ☆ Cov(X +Y,Z) = ? 3) Multiplying RVs: ☆ E[X ⋅Y] = ?
1) ☆ E[a + b⋅X] = a + b⋅E[X] ☆ Var(a + b⋅X) = b2 Var(X) ☆ Cov(a + b⋅X,Y) = b⋅Cov(X,Y) ☆ Corr(a + b⋅X,Y) = Corr(X,Y) 2) ☆ E[X +Y] = E[X]+ E[Y] ☆ Var(X +Y) = Var(X) + Var(Y) + 2⋅Cov(X,Y) ☆ Cov(X +Y,Z) = Cov(X,Z) + Cov(Y,Z) 3) ☆ E[X ⋅Y] = E[X]⋅E[Y], when X and Y are independent!
1) For a study involving one population and a sample size of 18 (assuming you have a t-distribution), what row of the t-table will you use to find the right-tail ("greater than") probability affiliated with the study results? 2) For a study involving a paired design with a total of 44 observations, with the results assuming a t-distribution, what row of the table will you use to find the probability affiliated with the study results? 3) A t-value of 2.35, from a t-distribution with 14 degrees of freedom, has an upper-tail ("greater than") probability between which two values on the t-table?
1) Degrees of freedom = the value to use on the sides of the t distribution df=n-1 n=18--> n-1=17; row 17 is the one you'd use to find the probability! 2) For a paired study, n=22 df=n-1=22-1=21 3) Reverse trace row 14 df--> 2.35 --> There is no exact 2.35 value; it falls between 2.145 and 2.624 --> Column headings for these values = 0.025 and 0.01 Thus, upper tail probability = between 0.025 and 0.01
Quiz 3: 1) The probability of A is 0.60, the probability of B is 0.45, and the probability of A or B is 0.80. The probability that both A and B occur is 2) The probability of A is 0.60, the probability of B is 0.45, and the probability of both is 0.30. --> True or False: A and B are independent. 3) Use this following probability distribution for X to solve the next four problems. A repairperson replaces broken windshield glass throughout the day. The number of windshields replaced is: X: 1; 2; 3; 5 P(X): 0.42; 0.26; 0.04; 0.28 4) The expected number of windshields replaced is µ = A. µ = 1 * 0.42 + 2 * 0.26 + 3 * 0.04 + 5 * 0.28 B. µ = (1 + 2 + 3 + 5) / 4 C. µ = (0.42 + 0.26 + 0.04 + 0.28) / 4 D. µ = (1 * 0.42 + 2 * 0.26 + 3 * 0.04 + 5 * 0.28) / 4 5) Using µ to represent the answer to the previous question, the variance in the number of windshields replaced is σ 2 = A. σ^2 = (1 - µ)2 * 0.42 + (2 - µ)2 * 0.26 + (3 - µ)2 * 0.04 + (5 - µ)^2 * 0.28 B. σ^2 = (12 * 0.42 + 22 * 0.26 + 32 * 0.04 + 52) / (4 - 1) C. σ^2 = ((1 - µ) * 0.42 + (2 - µ) * 0.26 + (3 - µ) * 0.04 + (5 - µ) * 0.28)^2 D. σ^2 = ((1 - µ) * 0.42 + (2 - µ) * 0.26 + (3 - µ) * 0.04 + (5 - µ)) * 0.28)2 / (4 - 1) E. σ^2 = ((1 - µ)2 + (2 - µ)2 + (3 - µ)2 + (5 - µ)2) / (4 -1) F. σ^2 = ((1 - µ)2 * 0.42 + (2 - µ)2 * 0.26 + (3 - µ)2 * 0.04 + (5 - µ)2 * 0.28) / (4 - 1) 6) The worker is paid a salary depending on his outcome: $S = 10 + 2*X. Let µ represent the expected value of X and σ 2 the variance in X, from Questions 2 and 3. --> E(x) = ? --> Var(x) = ? 7) A firm engages in two research projects. The returns from each project are random. In millions of dollars, the expected profit from Project X are E[X] = 3.7, and the variance is Var(X) = 9.9. The expected profit from Project Y, also in millions of dollars, are E[Y] = 8.3, and the variance is Var(Y) = 12.2. Because there is some spillover in the knowledge acquired during research between the two projects, Cov(X,Y) = 5.3. --> Expected total profit = ? --> Variance in total profits = ? --> Correlation between 2 profits = ? 8) Find cov(x,y) for the following distribution: x 2 8 y -3 | 0.43 0.22 | 5 | 0.19 0.16 |
1) Independence test: P(A|B) = P(A) --> P(AuB) = 0.80 = (0.6+0.45)-(P(AnB))--> P(AnB) = 0.25 2) P(A|B) = P(AnB)/P(B)--> (0.3)/0.45 = 0.66; P(A) = 0.6--> they're NOTindependent! --> False 3) 0.58 4) E(x) = sum of all (x*P(x) --> A. µ = 1 * 0.42 + 2 * 0.26 + 3 * 0.04 + 5 * 0.28 5) Var(x) = sum of all ((x-mean)^2)*(P(x)) --> A. σ^2 = (1 - µ)2 * 0.42 + (2 - µ)2 * 0.26 + (3 - µ)2 * 0.04 + (5 - µ)^2 * 0.28 6) Asking for linear transformations! Linear transformations: i. Mean of (a+bx) = a+b(mean of x) --> So if y=a+bx, mean of function y = a+b(mean of x) ii. Variance of (a+bx) = (b^2)(Var(x)) --> So if y=a+bx, variance of function y = (b^2)(Var(x)) iii. Covariance of (a+b(x,y)) = (b)(cov(x,y)) --> So if y=a+b(x,y), covariance of function y = (b)(cov(x,y)) mean(10+2x) = 10+2(mean of x) var(10+2x) = 4(var(x)) 7) Exp total profit = 3.7+8.3 = 12 Var(xprofits+yprofits) = var(x)+var(y)+(2 x cov(x,y)) = 9.9+12.2+10.6 = 32.7 Corr(2 profits) = Cov(x,y)/SDx*SDy = 5.3/(3.14643*3.4985) = 0.48 --> Remember that SD = (Var)^0.5 8) x 2 8 y -3 | 0.43 0.22 |--> P(y=-3) = 0.65 5 | 0.19 0.16 |--> P(y=5) = 0.35 --> P(x=2) = 0.62 --> P(x=8) = 0.38 Cov(x,y) for joint distr. table = sum of [(x-E(x))(y-E(y))*P(x,y)] i. E(x) = (2*0.62)+(8*0.38) = 4.28 ii. E(y) = (-3*0.65)+(5*0.35) = -0.2 --> Go through each sq and apply this to each sq as follows: [(2-4.28)(-3+0.2)(0.43)]+[(8-4.28)(-3+0.2)(0.22)]+[(2-4.28)(5+0.2)(0.19)]+[(8-4.28)(5+0.2)(0.16)] = *Cov(x,y) = 1.296*
List the formulas for the following: 1) Unions 2) Complements 3) Conditional Probability 4) Intersections 5) Bayes Rule 6) You're a doctor. Past data tells you 10% of patients entering clinic = have liver disease. 5% of clinic's patients = alcoholics. Among those diagnosed with liver disease, 7% = alcoholics. What is the probability of a patient having liver disease if they're an alcoholic?
1) Unions: P(AuB) = P(A)+P(B)-P(AnB) 2) Complements: P(A)=1-P(A) 3) Conditional Probability: P(A|B) = P(AnB)/P(B) 4) Intersections: P(AnB) = P(A|B)*P(B) 5) Bayes Rule: used for conditional probability Bayes formula = P(A|B) = P(B|A) * P(A) / P(B) - Conversely, P(B|A) = [P(A|B)P(B)]/[[P(A|B)*P(B)]+[P(A|B)*P(B)]] --> Patient does this, GIVEN a positive result for the disease; working backwards almost) --> A = patient has liver disease --> B = patient is alcoholic --> P(A) = 0.10 --> P(B) = 0.05 --> P(B|A) or P(patient is alcoholic, if they have/given that they have liver disease) = 0.07 --> P(A|B) = (0.07 * 0.1)/0.05 = 0.14 1/1000 has disease; if disease, P(positive) = 0.99 Test: if no disease, P(positive) = 0.005 P(positive|disease) = "sensitivity"
Problem Set 3: Part 4- This question uses Bayes' Rule. (It can be solved with other techniques, as well). P[A|B] = 0.55 , P[A|B bar ] = 0.72 , and P[B] = 0.09 . What is P[B|A]? Part 5- This question uses Bayes' Rule. (It can also be solved with other techniques). P[A|B] = 0.80 , P[A] = 0.76 , and P[B] = 0.60 . What is P[B|A]?
4) Recall that P[A|B] = [P(B|A)P(A)]/P(B) = [(P(AnB)/P(B)] --> P[B|A]= [(P(A|B))(P(B))]/[[P(A|B)P(B)]+[P(A|Bbar)P(Bbar)]] P(B|A)=(0.55x0.09)/((0.55x0.09)+(0.72x0.91))=0.07 5) P[A|B] = [P(B|A)P(A)]/P(B); x = P(A|B) --> 0.80 = (x*0.76)/0.60--> P(B|A) = 0.632
When is it better to use complementary probability (finding the inverse, and subtracting from inverse)? 1) Weather forecast: 0.10 chance of rain each day; this is independent (rain on tue doesn't change probability of rain on wed). P(rains within 7 days?)
When it's sequential; asking for AT LEAST ONE of... 1) 1-P(no rain, 7 times in a row) = 1-(1-0.10)^7 = 0.5217
1) What are the 2 requirements for random sampling?
1) i. Probability of inclusion is equal for all the population ii. Whether individual is included = independent of whether anyone else is included
1. Expected value! 2. Linear transformations! Knowing for x its E(x) and var(x) Want to know: for (a+bx), what is E(a+bx), and var(a+bx) --> a,b = constant
1. E(x) = x*f(x)*dx Var(x) = all x [(x-E(x))^2]f(x)dx discrete: sum of all x; P(x) continuous: S(x); f(x) 2. E(a+bx_= a+b(E(x)) Avg temp < 72 degrees F avg in C? va(a+bx) = b^2(var(x) --> LMAO I ALREADY MEMORIZED THIS SHIT
What is the standard error of the: 1) Sample mean (x)? 2) Sample proportion (p)? 3) x1-x2? 4) p1-p2? What is proportion? --> ie- you're interested how many red m&ms there exist out of every m&m that exists. there are 170,000 m&ms total, 17,000 red m&ms. what is the population proportion?
--> SE = sample SD 1) SEx=s/(n^0.5) --> s=sample estimate of population SD 2) SEp=[p(1-p)/n]^0.5 3) SEx1-x2=[(s1^2/n1)+(s2^2/n2)]^0.5 4) SEp1-p2=([(p1(1-p1))/n1]+[(p2(1-p2))/n2])^0.5 Proportion = number of subjects that possess the traits you are investigating/interested in --> the population proportion = 17,000/170,000=0.1
Problem Set 3: Part 1- https://imgur.com/gallery/qaI0vxB shows the probability distribution for A and B. 1) P[A∩B] = ? 2) P[A∩B bar] = ? 3) P[A∪B] = ? 4) P[A|B] = ? 5) P[A bar∪B bar] = ?
--> Since joint distribution is given, AnB = 0.19; AuB = P(A)+P(B)-P(AnB), etc.--> n = in table!!!! WOW 1) P(AnB) = 0.19 2) P[A∩B bar] = 0.46 3) P[A∪B] = P(A)+P(B)-P(AnB) = (0.65+0.4)-0.19 = 0.86 4) P[A|B] = P(AnB)/P(B) = 0.19/0.4 = 0.475 5) P[A bar∪B bar] = (0.35+0.6)-(0.14) = 0.81
--> what is p hat? --> what is parameter? --> what is statistic? 1) What is the E(p hat)? 2) What is var(p hat)? 3) What are the 3 requirements for p hat? 4) What is bayne's formula? 5) What is a "space" in regards to probability? --> event=? 6) Simple vs. complex events
-->p hat=sample proportion; proportion of interest within sample --> parameter = descriptor of population --> statistic = descriptor of sample 1) E(p hat)=P 2) Var(p hat)=[p(1-p)]/N 3) unbiased; accurate; consistent 4) PA|B=(A|BxB)/[(A|BxB)+(A|BbarxBbar)] 5) "Space"- describes all possible outcomes --> Event=subset/part of the space 6) --> Simple = 1 desired outcome --> Complex = multiple/intersecting desired outcomes (drawing a 6 of spades)
Problem Set 3: Part 6- State whether each of the following statements is true or false, and explain. 1) The complement of the union of two events is the intersection of their complements. 2) If A and B are mutually exclusive, then they are independent. 3) If P[A|B] = P[B|A], then P[A] = P[B]. 4) If an event and its complement are equally likely to occur, the probability of that event must be 0.5. 5) If A and B are independent, then A bar and B bar are independent. 6) If A and B are mutually exclusive, then A bar and B bar are mutually exclusive 7) P[A∪B] is greater than or equal to P[A∩B]. 8) P[A∪B] is greater than or equal to P[A]+ P[B]. 9) P[A∩B] is less than or equal to both P[A] or P[B] . 10) P[A]+ P[B] is less than or equal to 1. 11) P[A|B] is greater than or equal to P[A∩B] . 12) P[A|B] is greater than or equal to P[A].
1) The complement of the union of two events is the intersection of their complements. --> True; *complement of P(AuB) = P(Abar n Bbar)* 2) If A and B are mutually exclusive, then they are independent. --> False 3) If P[A|B] = P[B|A], then P[A] = P[B]. --> True, (5*5/5 = 5) 4) If an event and its complement are equally likely to occur, the probability of that event must be 0.5. --> True, 1-0.5 = 0.5 5) If A and B are independent, then A bar and B bar are independent. --> True; independence transfers over to complements—remember dat! 6) If A and B are mutually exclusive, then A bar and B bar are mutually exclusive --> FALSE; mutual exclusivity doesn't necessarily transfer over to complements. 7) P[A∪B] is greater than or equal to P[A∩B]. --> True apparently lmao wtf 8) P[A∪B] is greater than or equal to P[A]+ P[B]. --> False, it's less than since PAuB = (PA+PB)-PAnB 9) P[A∩B] is less than or equal to both P[A] or P[B] . --> P(AnB) = P(A|B)*P(B) --> TRUE! bc P(A) and P(B) are always going to be decimals; so decimal*decimal = smaller decimal :) 10) P[A]+ P[B] is less than or equal to 1. --> False, A and B aren't necessarily related. If P(A)=0.9 and P(B)=0.9, then obviously P(A)+P(B)=/=1, since the events are necessarily related! 11) P[A|B] is greater than or equal to P[A∩B] . --> True, PA|B = PAnB/PB--> divided by decimal = greater or equal :) 12) P[A|B] is greater than or equal to P[A]. --> False, this varies depending on B
1. Random variables! 2. What is discrete rv? 3. What is continuous rv? 4. What are expected values? 5. What is the variance in relation to probability? 6. What is a binomial distribution? 7. What is the covariance/joint distribution (discrete)? 8. What is correlation (discrete)? --> What if x and y are statistically independent?
1. Specifically, outcomes = quantitative (often x/y/z) 2. Discrete random variables = limited set of values --> Described with probability distribution or table 3. Continuous = any value in some range 4. Expected value = average mean = E(x) = SUM of all (x*P(X)) --> so P(x) = (x)*(probability of x) 5. Variance = sum of all (x-E(x))^2(P(x)) --> ux = mean of population = mu 6. Binomial distribution: Way to measure probability of specific successes/failures --> N attempts P = probability of success x = number of successes --> P (x successes) = N!/[X!(N-X)!]*(P^x)(1-P)^(N-X) 7. cov(x,y) = sum of all [(X-E(x))(y-E(y))*P(x,y)] --> x and y= the values; P(x,y) = probability of both or where the cells join in joint distribution 8. corr(x,y) = [cov(x,y)]/[SDx*SDy] = Pxy --> range = from -1 to 1 --> If x and y are independent, then corr = 0. (basically means they're mutually exclusive)
Problem Set 3: Part 10- A firm receives widgets from two sources. The first manufacturer is more reliable, producing defective widgets 5% of the time. This manufacturer provides 2/3 of the firm's widgets. The second manufacturer produces defective widgets 15% of the time. This manufacturer provides the remaining 1/3 of widgets. Given that a randomly selected widget is defective, what is the probability that it came from the more reliable firm?
10) P(defective from firm 1)/(P(defective from firm 1)+P(defective from firm 2)) --> P(defective from firm 1) = 0.05*0.667 --> P(defective from firm 2) = 0.15*0.33 (0.05*0.667)/(( 0.05*0.667)+(0.15*0.33)) = 0.402 --> great value Baynes; just use the number of widgets provided as B) --> *Essentially, (# defected from reliable firm)/(total number of defected widgets!)* uhh OR Bayne's Rule: A=from manufacturer 1 Abar=from manufacturer 2 B=defective Bb=not defective (0.8) P(A|B)=0.05 P(A|Bbar)=0.95 P(B|A)=(0.05x0.2)/(0.05x0.2)+(0.95x0.8)
Problem Set 3: Part 2- https://imgur.com/gallery/DpoozJz shows the probability distribution for A and B. Complete the table and answer the questions. 1) P[A∩B] = ? 2) P[A∩B bar] = ? 3) P[A∪B] = ? 4) P[A|B] = ? 5) P[A bar∪B bar] = ?
First, complete the table: --> Remember that P(x) = sum of probability distributions in row of x! --> Pd = probability distribution in each specific cell and also intercepts! A A_ B | 0.48 0.25 |--> P(B) = 0.73 B_ | 0.09 0.18 |--> P(B_) = 0.27 --> P(A) = 0.57 --> P(A_) = 0.43 1) P(AnB) = 0.48 2) P[A∩B bar] = 0.09 3) P[A∪B] = P(A)+P(B)-P(AnB) = (0.57+0.73)-(0.48) = 0.82 4) P[A|B] = P(AnB)/P(B) = 0.48/0.73 = 0.658 5) P[A bar∪B bar] = (0.43+0.27)-(0.18) = 0.52
How do you control for variables in regression? 7. Dr. Jones gives two exams in his Introduction to Archaeology course. We know the chance of getting a grade of A on his exams: P[A on the midterm] = 0.40 P[A on the final] = 0.35 P[A on the final A on the midterm] = 0.80 What is the probability of getting an A on at least one of his exams? 17. Ashton needs to purchase gadgets. Each store that Ashton visits has a probability p of selling a gadget; a probability 1 − p if not. Ashton continues visiting stores until he or she has all the gadgets needed. [Note: this question does not use the binomial, Poisson, or Bernoulli distribution. You should use more general techniques to solve it.] a) Ashton needs to purchase 1 gadget. What is the probability that Ashton needs to visit exactly 4 stores to get them? b) Ashton needs to buy 1 gadget. What is the chance that he or she finds it in 3 visits or fewer? c) Ashton needs to purchase 2 gadgets. What is the probability that Ashton needs to visit exactly 4 stores to get them?
Just include them in the regression as independent variables (x) lmao 7) P(A on at least one of his exams)=P(A on the midterm OR final)=P(AuB) --> P(AuB)=P(A)+P(B)-P(AnB) i. P(AnB)=P(A|B)xP(B)--> 0.8x0.4=0.32 ii. 0.35+0.4-0.32=0.43 17. This is just a sequence probability problem! a) P(success on 4th)=no/no/no/yes -->(1-p)^3(p) b) P(success on 1st)+P(success on 2nd)+P(success on 3rd)=(1-p)^0(p)+(1-p)^1(p)+(1-p)^2(p) =D c) In order to get 2 gadgets within N=4, this ~space~ will contain: 2 yeses, 2 no's (this is P(AnB) Possible sequencing (remember that it has to be 4 visits, so cannot be yes/yes/no/no as that would only entail 2 visits): no/no/yes/yes yes/no/no/yes no/yes/no/yes --> Formula for each visit = (1-p)^2(p^2) --> 4 attempts = (1-p)^2(p^2)+(1-p)^2(p^2)+(1-p)^2(p^2)+(1-p)^2(p^2)=(1-p)^2(3p^2)! =D
Problem Set 3: Part 7- During the semester, a student takes N = 6 quizzes. The probability of failing any individual quiz is 0.05. What is the probability that the student fails at least one quiz? Part 8- This question is the same as before, except that the professor drops the lowest quiz. What is the probability that the student still fails at least one of the remaining five quizzes? Part 9- Ginkgo trees are dioecious, meaning that they are either male or female(sounds fake but okay lol), with equal probability. Producing seeds requires at least one of each sex. When purchasing 5 trees, what is the probability of at least one male and one female?
P = probability of success x = number of successes n = number of trials --> P (x successes) = N!/[X!(N-X)!]*(P^x)(1-P)^(N-X) 7) N=6; P(failing individual quiz)=0.05 P(AT LEAST 1 quiz)--> find P(0); 1-P(0) = P(at least 1) --> P(fail >= 1 quiz) = 1-P(fail=0 quizzes); plug into formula: 1 - [6!/[0!(6-0)!]*(0.05^0)(1-0.05)^(6-0)] = 1-0.735 = *0.26491* 8) P(fail >= 2 quizzes) = 0.032773 --> P(fail>=2 quizzes) = 1-[(P(fail=1 quiz))+(P(fail=0 quizzes))] P(fail=0 quizzes) = 0.735 from (7) P(fail=1 quiz) = [6!/[1!(6-1)!]*(0.05^1)(1-0.05)^(6-1)] --> 1-(0.735+0.23213) = *0.03277* 9) P(F) = 0.5; P(M) = 0.5; N = 5 --> Producing seeds = AT LEAST 1 M + 1 F --> Purchasing 5 trees; P(n>=1 M)nP(n=1 F) = ? n=5; P(either "gender") = 0.5--> P(f>=1) = ?; P(m>=1) = ? P(either "gender">=1) = 1-(P(either "gender" = 0) --> 1 - [5!/[0!(5-0)!]x(0.05^0)(1-0.5)^(5-0)] P(either "gender">=1) = 0.96875 P(FnM) = 0.968x0.968 = *0.937* 4+11+12+9+13+8
Continuous distribution! 1. What is the probability of density function (PDF)? --> What is the f(x) that gives you the bell curve? 2. What is the area under the curve? 3. What is the cumulative density function (CDF)?
P(x=a) = 0~generic value--> P(a<=x<=b) 1. PDF = probability that x>some value (remember cumpdf on the calc? lmao) PDF of P(a<=x<=b) = area under f(x) from a to b Example of PDF: f(x) = 1/(2*pi*r^2)^0.5 --> This function gives you the bell curve! 2. Area under curve = surplus 3. CDF = probability that x<some value --> f(x) = 1-e --> antiderivative of f(X) = x^3 antiderivative = 1/4(x^4) --> f(x) = e^(x^2) has no antiderivative
Continuous Random Variables What are the PDFs of the following? 1) x 2) E(x) 3) Var(x) 4) Mean = 100; variance = 225 --> P(110<=n<=120) = ? 5) What is the value of z, given that there is an area of 0.20 that falls to the right of the z score? 6) Rules to remember for normal distribution? (3)
PDF = area under bell curve; PDF for x P(a<=x<=b) = S f(x)dx E(x) = S (all poss x) x*f(x)dx Var(x) = S ((x-E(x))^2)*f(x)dx Uniform distribution = places equal likelihood on all values inside some range; usually described from lower limit (l) to upper limit (u) 1) PDF of x: If we want to find out probability of P(a<=x<=b) = integrate from a to b--> S f(X)dx--> S 1/(u-1)dx--> find antiderivative with respect to x--> x/(u-l) --> Upper limit = b; lower limit = a --> Plug in for x!--> (b/(u-l))-(a/(u-l)) = *PDF of x = ((b-a)/(u-l))* --> (u-l) = entire range; (b-a) = range we're interested in Uniform PDF: f(x) = 1/(u-l)--> if l<=x<=u, otherwise, = 0 2) PDF of E(x): E(x) = S all poss x (x*f(x)dx) = S (x*(1/(u-l))dx)--> antiderviative with regards to x = 1/(u-l)(0.5x^2) --> Plug in upper limit = u; lower limit = l (plug into x) 1/(u-l)*0.5(u^2)-(l^2) = 0.5[(u^2-l^2)/(u-l)] = 0.5[((u+l)(u-l))/u-l)] = *PDF of E(x) = 0.5(u+l)--> the average of u and l!* 3) PDF of Var(x): Var(x) = S all poss x (x-E(x))^2(f(x)dx) = S (x-0.5(u+l))^2(1/(u-l))*dx = ? --> Will end up being a HW problem where we have to derive the actual problem! --- In theory: probabilities come from integration In practice: We use numerical approximations; come up with them from tables; or use software! i. Standard normal distr: mean=0; variance=SD=1 4) P(110<=n<=120)--> z = (x-mean)/SD --> Mean = 100; variance = 225 --> Standardizing = transitioning from P(a<=x<=b)--> P(st. value of a<=z<=st. value of b) --> Find probabilities for st. values of distribution iii. Standardize values of 110 and 120! --> P[((110-100)/15)<=z<=((120-100)/15)] (use Standard Norm Distr Table, which gives area to the right of the value) --> normal value = P(below value) --> P[(0.251)<=z<=(0.092)]--> these are the z scores! find the corresponding normal distr. decimals in the table--> area falls to right --> 0.251-0.092=0.159 5) z = ?--> normal value = 0.20; find 0.20 on the graph and then find corresponding *z score = 0.84* --> This was basically asking for the z score of 0.2! 6) Rules to remember(TMS): i. TOTAL AREA: All area under curve = 1 ii. MEAN: Mean sits in middle of distribution (area above 0 = 1/2; area below 0 = 1/2) iii. SYMMETRY: bell curve is symmetric, so P(z>=c) = P(z<=c) --> Bell curve is perfectly symmetric, that's literally all he's saying lmao he sucks at teaching
Problem Set 3.5 Part 1: 1) Calculate E(x) and E(y) 2) Calculate Var(x) and Var(y) 3) Calculate Cov(x,y) Part 2: 4) Calculate E(x) and E(y) 5) Calculate Var(x) and Var(y) 6) Calculate Cov(x,y)
Part 1: --> P(x) = 0 = 0.35 --> P(x) = 1 = 0.65 --> P(y) = 0 = 0.65 --> P(y) = 1 = 0.35 1) E(x) = sum of all [(x)(P(x))] = [(0)(0.35)]+[(1)(0.65)] = *0.65* E(y) = sum of all [(y)(P(y))] = [(0)(0.65)]+[(1)(0.35)] = *0.35* 2) Var(x) = sum of all[(x-E(x))^2(P(x))] = [(0-0.65)^2(0.35)]+[(1-0.65)^2(0.65)] = *0.2275* Var(y) = sum of all[(y-E(y))^2(P(y))] = [(0-0.35)^2(0.65)]+[(1-0.35)^2(0.35)] = *0.2275* 3) Cov(x,y) = sum of all[(x-E(x))(y-E(y))(P(x,y))] = [(0-0.65)(0-0.35)(0.20)]+[(0-0.65)(1-0.35)(0.15)]+[(1-0.65)(0-0.35)(0.45)]+[(1-0.65)(1-0.35)(0.20)] = *-0.0275 (inversely related!)* Part 2: --> P(x) = -1 = 0.50 --> P(x) = 1 = 0.50 --> P(y) = 3 = 0.85 --> P(y) = 5 = 0.15 4) E(x) = sum of all [(x)(P(x))] = [(-1)(0.55)]+[(1)(0.50)] = *-0.05* E(y) = sum of all [(y)(P(y))] = [(3)(0.85)]+[(5)(0.15)] = *3.3* 5) Var(x) = sum of all[(x-E(x))^2(P(x))] = [(-1+0.05)^2(0.50)]+[(1+0.05)^2(0.50)] = *1.0025* Var(y) = sum of all[(y-E(y))^2(P(y))] = [(3-3.3)^2(0.85)]+[(5-3.3)^2(0.15)] = *0.51* 6) Cov(x,y) = sum of all[(x-E(x))(y-E(y))(P(x,y))] = [(-1+0.05)(3-3.3)(0.40)]+[(-1+0.05)(5-3.3)(0.10)]+[(1+0.05)(3-3.3)(0.45)]+[(1+0.05)(5-3.3)(0.05)] = *-0.1 (inversely related!)*
Problem Set 3.5 Part 3: Z ~ N(0,1) 7) P[Z > 1.44] = ? 8) P[Z > 2.31] = ? 9) P[Z < 1.59] = ? 10) P[0.13 < Z < 0.34] = ? 11) P[−2.02 < Z < −1.44] = ? Part 4: X ~ N(50,100)??? 12) P[X > 64] = ? 13) P[X < 73] = ? 14) P[58 < X < 67] = ? Part 5: Z ~ N(0,1) . 15. P[Z < ??] = 0.6844 16. P[Z < ??] = 0.7939 17. P[Z > ??] = 0.0934 Part 6: Z ~ N(0,1) . Find the symmetric intervals that have these probabilities. (In other words, the upper and lower limits are the same value with different signs.) 18. P[?? < Z < ??] = 0.5407 19. P[?? < Z < ??] = 0.8638 20. P[?? < Z < ??] = 0.9815
Part 3: Z ~ N(0,1)--> mean=0; SD=1 (no need for conversion bc this is the normal distribution) --> Use normal distribution to find Z-score, then find the area under the curve for probability (all areas fall to the *right* of the z score) 7) P[Z > 1.44] = 0.075 8) P[Z > 2.31] = 0.01 9) P[Z < 1.59] = 1-0.056 = 0.944 10) P[0.13 < Z < 0.34] = P(0.13)-P(0.34) = 0.448-0.367 = 0.081 11) P[−2.02 < Z < −1.44] = P(1.44)-P(2.02) = 0.075-0.022 = 0.053 --> Remember that normal distribution = symmetric, so just find the area between 1.44-2.02 Part 4: X ~ N(50,100)--> mean=50; SD=100 12) P[X > 64] = 0.081 --> Convert X to z=(64-50)/100 = 1.4--> find on normal distribution 13) P[X < 73] = 0.989 --> (73-50)/100 = 2.3 = 0.011--> find complement for Z<2.3 = 1-0.011=0.989 14) P[58 < X < 67] = 0.167 --> (58-50)/100=0.8; (67-50)/100=1.7 --> P(0.8)=0.212; P(1.7)=0.045--> area=0.212-0.045=0.167 Part 5: Z ~ N(0,1) . 15. P[Z < ??] = 0.6844 --> 1-0.6844 = 0.3156--> Z=0.374; P(Z>0.364)=0.3156 --> ??=0.364 16. P[Z < ??] = 0.7939 --> 1-0.7939 = 0.2061--> Z=0.82; P(Z>0.82)=0.2061 --> ??=0.82 17. P[Z > ??] = 0.0934 --> ??=1.32 Part 6: Z ~ N(0,1) . Find the symmetric intervals that have these probabilities. (In other words, the upper and lower limits are the same value with different signs.)?? 18) P[?? < Z < ??] = 0.5407 --> 0.5407/2 = 0.27035--> this is area for Z>0.27035; find its complement--> 0.50-0.27035 = 0.22965--> now trace this value to original Z value in chart = 0.74 --> ?? = 0.74; -0.74 19) P[?? < Z < ??] = 0.8638 --> 0.8638/2 = 0.4319--> this is area for Z>0.8638; find its complement--> 0.50-0.4319 = 0.0681--> now trace this value to original Z value in chart = 1.49 --> ?? = 1.49; -1.49 20) P[?? < Z < ??] = 0.9815 --> 0.9815/2 = 0.49075--> this is area for Z>0.49075; find its complement--> 0.50-0.49075 = 0.00925--> now trace this value to original Z value in chart = 2.38 --> ?? = 2.38; -2.38
Problem Set 3.5 Part 7: Calculate the standard error (that is, the square root of the variance) in the following estimates. 21) Standard error in X, in a sample of N = 100 with sX^2 = 25 22) Standard error in X, in a sample of N = 20 with sX^2 = 18 23) Standard error in pˆ, in a sample of N = 400 with pˆ= 0.5 24) Standard error in pˆ, in a sample of N = 188 with pˆ = 0.62 Part 8: Calculate the upper and lower limits of these confidence intervals. 25) The 95% confidence interval for µ, in a (large) sample of N = 100 with X = 75 and sX^2 = 25 26) The 95% confidence interval for µ, in a (small) sample of N = 20 with X = 8.3 and sX^2 = 18 27) The 85% confidence interval for p, in a (large) sample of N = 400 with p^ = 0.5
Part 7: Calculate the standard error (that is, the square root of the variance) in the following estimates. 21) Standard error in X, in a sample of N = 100 with sX^2 = 25 --> SE=s/(n^0.5)=5/10=0.5 22) Standard error in X, in a sample of N = 20 with sX^2 = 18 --> SE=s/(n^0.5)=18/(20^0.5)=4.472 23) Standard error in pˆ, in a sample of N = 400 with pˆ= 0.5 --> SEp=[p(1-p)/n]^0.5=[(0.5(1-0.5))/400]^0.5=0.025 24) Standard error in pˆ, in a sample of N = 188 with pˆ = 0.62 --> SEp=[(0.62(1-0.62))/188]^0.5=0.0354 Part 8: Calculate the upper and lower limits of these confidence intervals. --> Formula for CIx=mean +/- z(SD/n^0.5)=mean +/- (z x standard error of mean) --> CIp=proportion +/- [z x [(p(1-p))/n]^0.5] --> z=z value for the confidence interval; reverse trace from the tail area of one side not included in the confidence interval! --> ie- if 95% confidence interval, end tail of one side not included = (0.05/2) = 0.025--> corresponding z=1.96 =D 25) The 95% confidence interval for µ, in a (large) sample of N = 100 with mean = 75 and sX^2 = 25 -: 75-1.96(5/10)=74.02 +: 75+1.96(5/10)=75.98 --> CIx=75±0.98--> (74.02, 75.98) 26) The 95% confidence interval for µ, in a (small) sample of N = 20 with mean = 8.3 and sX^2 = 18 -: 8.3-1.96(4.2426/4.472)=6.4405 +: 8.3+1.96(4.2426/4.472)=10.1595 --> CIx=8.3±0.9487--> 27) The 85% confidence interval for p, in a (large) sample of N = 400 with p^ = 0.5 --> individual tail area=0.15/2=0.075--> z=1.44 --> CI of sample p=p±[z x ((p(1-p)/n)^0.5] -: 0.5-[1.44(0.5(1-0.5))/400)^0.5] +: 0.5+[1.44(0.5(1-0.5))/400)^0.5] --> CIp=0.5±0.03--> (0.47, 0.53)
Sample statistics: E[x bar] = mean of population Var(x bar) = variance of x/N Type of distribution: -->
Sample statistics: E[x bar] = mean of population Var(x bar) = variance of x/N Type of distribution: --> IF X~=N; x bar ~= N --> If N is large, x bar ~=N Sample proportion: p hat = # observations with characteristics/# total observations E[p hat]=p Var(p hat) = p(1-p)/N Type? --> P hat ~=N, when large sample P=population P hat=sample SD in x bar = SD/(N^0.5) Sd in p hat = (p(1-p)/N)^0.5 St. error in X bar = Sx/(N^0.5) St. error in p hat = [(p hat(1-p hat))/N]^0.5
Problem Set 3: XPart 11: Sue calls classmates to ask for help with homework. Each classmate has an independent 0.3 chance of answering their question. Sue continues calling classmates until they get an answer. Then they stop. 1. What is the probability that Sue receives an answer from the fourth classmate they call? 2. What is the probability that they must call at least three classmates? 3. What is the probability that they receive an answer from the N-th classmate they call? A. (0.7)^N B. (0.7)^(N-1) * (0.3) C. 1 - (0.3)^N D. 1 - (0.7)^(N-1) Part 12. A firm hires 3 workers from an applicant pool of 5 men and 5 women. What is the probability that all three employees are male? Part 13. & 14. For these questions, calculate: 1. The probability distribution for the random variable. (For example, P[X = 0], P[X = 1], P[X = 2].) 2. The expected value of the random variable. 3. The variance in the random variable. XPart 13. The Acme Manufacturing Company produces a large batch of widgets, of which 20% are purple. Two widgets are chosen at random from the batch. Let X denote the number of purple widgets. XPart 14. The Acme Manufacturing Company produces a batch of N = 20 widgets, of which 4 are purple. Two widgets are chosen at random from the batch. Let X denote the number of purple widgets.
X11: 1) REMEMBER THAT SUE KEEPS CALLING UNTIL THEY GET AN ANSWER; SO FOR 4TH CLASSMATE TO ANSWER MEANS PREVIOUS THREE DID NOT ANSWER! 4th classmate answering: 0.7x0.7x0.7x0.3 = 0.1029 2) Probability of calling AT LEAST 3 students = AT LEAST/3 OR MORE 3 students do NOT answer the question (3 or more) = P(at least 3 do not answer the question)= 1-(P(call 1 student)+P(call 2 students) --> DOES NOT INCLUDE P(CALL 3 STUDENTS) BC THEY'RE ASKING FOR PROBABILITY OF SUE HAVING TO CALL THREE OR MORE (that's what at least means, remember!) --> P(call 1 student)=0.7^0x0.3=0.3 --> P(call 2 students)=0.7^1x0.3=0.21 1-(0.3+0.21)=0.49 0.2401 3) B. (0.7)^(N-1) * (0.3) --> Probability of student NOT answering (necessitating Sue to continue calling on students)^(N-1)*Probability of success on the Nth time --> ie- P(n=3) = 0.7*0.7*0.3 = 0.7^(3-1)*0.3 :-) 12: P(women) = 0.5; P(men) = 0.5 First pick = man = 5/10 Second pick = man = 4/9 Third pick = 3/8 --> 0.5*(4/9)*(3/8) = 0.083 X13: 1. P[X = 0]=0.8(complement of all purple widgets); P[X = 1]=0.2; P[X = 2]=((2/10)(1/9))=0.02 2. The expected value of the random variable. --> (0*0.8)+(1*0.2)+(2*0.02)=0.24 3. The variance in the random variable. --> (((0-0.24)^2)*0.8)+(((1-0.24)^2)*0.2)+(((2-0.24)^2)*0.02) = 0.2236 X14: 1. The probability distribution for the random variable. (For example, P[X = 0], P[X = 1], P[X = 2].) --> P(x=0)=0.8; P( 2. The expected value of the random variable. 3. The variance in the random variable.
