University:
Eastern Kentucky University
Course:
STA 270 | Applied Statistics
Academic year:

2020
Views:

257

Pages:

2
Author:

Hallie Randolph

Variable- categories Independence is to the multiplication law as mutually exclusive is to the addition law Individuals- numbers/data 1. For any A and B; P(AandB)= P(A|B) x P(B) multiplication Population- entire group of individuals interested in If A,B independent, P(AandB)= P(A) x P(B) Sample- part of your population we can directly 2. For any A and B; P(AUB)= P(A) + P(B) – P(AandB) Observe; where we get data If mutually exclusive, P(AUB)= P(A) + P(B) Parameter- a # describing a characteristic of a 3. What is the addition law for independent events? If A and B independent Population P(AUB)= P(A) + P(B) – P(A) x P(B) Statistic- a # describing a characteristic of a n!/(n-k!)k! Sample (nk)= (n n-k) We calculate a statistic from data to make a guess A random variable X associates a numerical value w/ each experimental outcome Out of a parameter (statistical inference) Discrete random variable- X has a finite # of values or infinitely many values that can bc Simple random sample- a sample where each arranged in a list. Eg: # of H’s in 3 tosses Individual is chosen randomly w/ each member Continuous random variable- X represents a measurement on a continuous scale and Of pop having equal chance of being selected can assume all values in an interval. Eg: height of randomly selected student in class And selecting one person doesn’t affect another’s Discrete- give distribution of X in terms of the probabilities f(x)= P(X=x) for each value x Chance of being chosen finitediscrete, the probability distribution must satisfy: 1. 01 like 0.3 and 0.8 A and B cannot be mutually exclusive because of General requirements for probability, not this specific P P(A) + P(B)=1.1 P(AUB)= 1.1- P(AandB) For P(AUB) <1, need P(AandB)>1 (n choose 1)= n (n choose n-1)=n (n choose n)=1 ** P(A)=0.36 P(B)=0.5 P(AandB)=0.18 P(B|A)= 0.18/0.36=0.5 P(Bˆc|A)= 1-0.5= 0.5 P(B|Aˆc)= P(AˆcandB)/P(Aˆc)= 0.32/1-0.36= 0.5 ** Carol makes 60% of repairs. Karl makes 40%. Carol’s incomplete 4% and Karl’s Incomplete 6% of time. P(I)= P(C)P(I|C)+ P(K)P(I|K)=0.6(0.04)+0.4(0.06)=0.048 P(K|I)= P(I|K)xP(K)/P(I)= 0.06x0.4/0.048= 0.5 ** 2 green, 3 red. 2 drawn at random w/o replacement P(A) and P(B) A= green on first B=green on second P(A)= 2/5 P(B)= P(A)P(B|A)+P(Aˆc)P(B|Aˆc)= 2/5x1/4+3/5x2/4=2/5 A and B are not independent because P(AandB)=2/5x1/4=1/10 and P(A)P(B)= 2/5ˆ2 ** P(A)=0.6 P(B)=0.22 P(AUB) if independent= P(AandB)=0.6x0.22=0.132= 0.6+0.22-0.132= 0.688 P(AUB) if mut ex= 0.6+0.22= 0.82 P(A|Bˆc) if mut ex= P(AandBˆc)=P(A)=0.6 P(AandBˆc)/P(Bˆc)= 0.6/0.78= 0.76 ** flip coin 8 times. Chance that at most 7 of tosses yield tails? P(At most 7 tails)=1-P(All 8 are tails)=1-(1/2)ˆ8=0.9961 ** 40% pop has type O blood. If 4 people selected random, find P(at least 1 has O) Prob having type O is 0.4. 1-P(None has type O)= 1-(0.6)ˆ4=0.8704 ** flip coin until you get H X=# of T’s you get before stopping X| 0 | 1 | 2 | 3 | infinity Y| ½ | ½ x ½ = ¼ | (1/2)ˆ3| (1/2)ˆ4| infinity Give distribution by probability mass function F(x)= P(X=x)= (1/2)ˆx+1 ** a roulette wheel has 18 red 18 black and 2 green slots, all equally likely. A bet on red or black is “even money”. What’s your expected payoff of $10 on black. X= payoff +10 or -10 E[X]= 10 x P(X=10) + (-10) x P(X=-10) = 10 x 18/38 + (-10) x 20/38 = -0.526. expect to lose 53 cents PRACTICE EXAM QUESTIONS: ** Any events A and B, P(A|B) + P(Aˆc|B)=1 true law of complement ** If events A and B are indep, then P(A|B)=P(B) false true if P(A|B)=P(A) *** If P(A)=0, then A is independent of any other event true ** If A and B are mut exclu, then P(A|B) = P(B|A) true ** If P(A)=0.3 P(B)=0.8 then A and B cannot be independent false ** If A is subset of B, then A and B are not mutually exclusive true ** If A is indep of B and B is indep of C, then A is indep of C false *** If A and B are mut excl and B and C are mut excl, then A and B are mut excl false ** If P(A|B)=0 then P(A)=0 false ** If P(B|A)=0 then A and B are mut excl true ** Hire 3 new employees from 12. 8 have college, 4 have high school. Picks at random, what’s the chance al l3 have same level of education? Sum of 3 from highschool or 3 from college/sum of 3 out of 12 (8 choose 3) + (4 choose 3)/ (12 choose 3) 56+4/220= 0.27 **or=add; and=multiply ** P(A)=0.2 P(B)=0.5 P(A|B)=0.2 A and B are independent because P(A)=P(A|B) P(AandB)= 0.2 x 0.5=0.1 P(B|A)= P(B)=0.5 P(AUB)= 0.2+0.5-0.1=0.6 P(Aˆc and Bˆc)= P(Aˆc)xP(Bˆc)= (1-0.2)x(1-0.5)=0.3 HOMEWORK PROBLEMS ** Inspect 4 bills chosen at random and passes batch if among the 4, none Are irregular. Find probability that batch will be passed if, 5% of bills are irregular: P(none are irregular) P(0.95)ˆ4=0.8145 20% of bills are irregular: P(0.8)ˆ4= 0.4096 ** Chance that a random sample of size 5 is selected from a pop of 20 1/(20 choose 5) = 1/15504 ** nine plots in square grid. Plots selected at random Find prob that all 3 are in same row: number of ways to choose plots all In same row is same as number of ways to choose a row, which is (3 choose 1) =3. There are (9 choose 3)=84 ways to choose 3 plots so the chance they are all in same row is: (3;1)/(9;3)= 3/84 Find prob that all 3 are in diff rows: (3 choose 1)ˆ3/(9 choose 3)= 27/84 ** P(A)=0.3 P(Bˆc)=0.6 P(AUB)=0.5 P(B)=0.4 P(Aˆc)= 1-0.3=0.7 P(AandB)= 0.3+0.4-0.5= 0.2 P(AandBˆc)= P(A)=P(AandB)+P(AandBˆc) P(AandBˆc)=P(A)-P(AandB)= 0.3-0.2=0.1 **Draw 2 cards from deck. One is spade, get $15. Two are red, lose $10. X=how much money you win Distribution of X

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