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Math 2040 Chapter 5 Notes

by: Ang Judd

Math 2040 Chapter 5 Notes Math 2040

Marketplace > Southern Utah University > Math > Math 2040 > Math 2040 Chapter 5 Notes
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Pay attention to the formulas! (There aren't enough to bother making a chapter 5 formula page.)
Business Statistics
Said Bahi
Class Notes
Math, business, Statistics
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This 9 page Class Notes was uploaded by Ang Judd on Friday February 5, 2016. The Class Notes belongs to Math 2040 at Southern Utah University taught by Said Bahi in Winter 2016. Since its upload, it has received 24 views. For similar materials see Business Statistics in Math at Southern Utah University.


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Date Created: 02/05/16
Chapter 5 Probibility: Something that happens by chance; not governed by any rules Probibility Experiment: an experiment with 1) each trial having one single outcome 2) outcomes are uncertain Sample Space: The set of all possible outcomes for a given probability experiment S= set of all possible outcomes x= single outcome Event E= subset of S Tree Diagram Sample Space--> 1st toss either head or tail head HH The sample space S= HH,HT, TH, TT Event: Tossing exactly one head head E= HT, TH 1st toss tail HT head TH tail tail TT 5.2 Defining Probability Relative Frequency 1) in relation to relative frequency Relative Frequency A= event of a probability experiment that we carriA= a k number of times n ex) toss a coin Probability P (head)=0.5 Toss 100 times ex) toss coin 10 times event A= Head X 2 P(head)= 0.2 0 0 0 0 0 0 0 0 1 1 Classical (Theoritical) Probability *Probability is based on what it should be* P(A)= Count of outcomes i= A n(A) Total Count n(S) ex) Roll a die A= rolling an odd side S= 1, 2, 3, 4, 5, 6 n(A) 3 = 1 A= 1, 3, 5 n(S) 6 2 Experimental (empirical) Probability *Probability is based on an actual experiment in which we count the frequency of the event* P(event A= frequency of A n *n= number of trials* 5.3 Laws of Probability *5.3 Read on own, not much in book…* Most of the time, when working with samples, statisticians try to deduce from the samples the population parameters (means, proportions, variances, etc.) of certain variables. This process of making judgments about population parameters is called statistical inference. *Inferential Statistic: Infering about a population from a sample* 5.4 Laws of Probability Probability Law 1 A probability of zero means the event will not happen Probability Law 2 A probability of one means the event must happen Probability Law 3 0 ≤ P(A) ≤ 1 P(A) will always be between 0 and 1, and will never be negative P( empty set)= 0 P(S)=1 Probability Law 4 P   1  P A  2   P A   n 1. The sum of the probabilities of all outcomes in a sample space must equal 1 ex) S= HH, HT, TH, TT A1 = HH 1/4 A2 = HT, TH 2/4 A = TT 1/4 3 =1 Compound Events Union A, B two separete events A U B = outcomes that are in either A or B or both Intersection A ∩ B= all outcomes that are in both A AND B * sometimes called A and B Complement P(A)+ P(A )=1 Complement Law c P(A )= 1-P(A) P(A)= 1-P(A ) EX 5.5 DBS_5_4 slide 11 A = income ≥ %50,000 A = income ≤ $50,000 if P(A)=0.08 8% making 50K P(A )=1-P(A)=1-0.08=0.92 ex) toss a coin until the first head is tossed sample space 1,2,3,4,5,… *As you go further the probability of not getting a head decreases Define event A will need at least 2 tosses to get a head 2,3,14 A = getting head in one Head in 4 T T T H P(A)=n(A)/n(S) 0.5 Head in 5 T T T T H P(A)=P head in at least two tosses = 1- P(A )=1-1/2=1/2 Mutually Exclusive Events A and B have nothing in common No intersection FIND A PICTURE OF IT!!! Addition Rule P(A or B) P(A U B) *Count A then Count B Subtract one of them P(A U B)= P(A)+(PB)-P(A intersect B) EX) 5.4:12 pg 248 A mail order company *This type of table is called a contingency table- frequency table with two nominal variables* Male Female total Suburban 196 298 494 Urban 92 173 265 total 288 471 759 a) find P(customer is male) number of male 288/759 total b) find P(customer is female) number of females 471/759 total c) find P(customer is a suburban male) 196/759 P(customer is male or urban) P(male)+P(urban)-P(male and urban) *work out the whole before dividing* 288/759+265/759-92/759 = 288+265+92 = 645 P(female or suburban) 759 759 0.85 P(f)+P(sub)-P(f and sub) 471+494-248 759 *this example is mututally exclusive e) P(customer suburban male or urban female) nobody can be male and female at the SM UF same time, or urban and suburban at P(uf)+P(sm)-P(uf and sm) the same time* 173+92-0 = 265 0.34914361 759 759 *when they're mutually exclusive you P(ᶲ)=0 can ignore the last part of the addition rule* General Addition Rule P(A or B)=P(A)+P(B)-(A and B) Special Rule for Addition A, B are mutually exclusive then P(A or B)=P(A)+P(B) g) P(customer is not suburban female) *Use the complement* 1-P(suburban female) 1-298/759=759-298/759 *This is experimental probability* *Any time you have a table of data you will use experimental probability* 5.5-5.6 Multiplication Rule and Conditional Probability Conditonal Probibility Conditional Probibility: Probibility of an event given that another event has occurred. P(A given B happened) *Sometimes B doesn't have and affect on A, sometimes it does ex) set of cards P(king)= 4/52 If two cards are drawn without replacement P(2nd card is King given the 1st was a King)= 3/51 Notation P(A l B)= P(A given B) P(A l BP(A intersect B) P(B) ex) Speeding and gender A (men) B(women) total C (speeding) 10 14 24 D(not speeding) 5 6 11 total 15 20 35 P(speeding given it was a man) P(C l A) = P(C ∩ A)= 10/35 P(A) 15/35 10 x 35 = 10 35 15 15 P(A l C) P(A U C) P(C ) P(man given speeding) 10 x 35 = 10 = 5 35 24 24 12 P(A)=15/35=3/7 does not equal P(A l C) = 5/12 P(A l C) does not equal P (A) = 1 A and C are not independent Independence A and B are independent if P( A l B)= P(A) and P(B l A) = P(B) Multiplication Rule *Multiplication formula for probibility * "or" addition "and" multiplication P(A ∩ B)= P(A l B) P(B) or P(A U B)=P(B l A)x P(A) ex) drawing two cards from a 52 deck without replacement P(king and queen) P(queen l king) x P(King) queen king 4 x 4 = 4 x 1 51 52 51 13 ex) drawing two cards from a 52 deck with replacement P(king and queen) P(queen l king) x P(King) queen king 4 x 4 *Events are independent 52 52 If A and B are independent then P(A and B)=P(A)x P(B) P(A intersect B)= P(A)xP(B) EX) pg 252:3 mail order company classified customers according to gender and location Male Female total Suburban (a)196 (b)298 494 Urban (c)92 (d)173 265 total 288 471 759 a) given a customer is a male what is the probibility he is urban? P(A given C) P(C l A)= P( C and A) P(A) 92 x 759 = 92 = 0.3194 759 288 288 b) given urban, find the probibility of male? P(A l C)= P(C l A) P(C ) 92 x 759 = 92 = 0.35 759 265 265 P(C l A) 0.3471698 not equal P(A) 0.3794466 Gender and location of customers are dependent d) given suburban chance of being woman? P(B l D) 298 x 759 = 298 0.6 P(B) 759 494 494 *Useful in marketing- if in this age range what is the chance of her liking the product? 5.8 Counting Recall P(A)= n(A)/n(S) ex) toss a coin 4 times event A= obtaining 3 heads in 4 tosses P(A)=n(A)/n(S)=number of ways we get three heads/all outcomes from 4 tosses *We could use a tree diagram, but it is time consuming to draw it out and it is very easy to get lost! Principle of Counting If x can occur in n ways, and y can occur in m ways. The number of ways (X,Y) will occur is nm. ex) Buying a car color: 4 colors size: 3 sizes make: 5 makes *In how many ways can you select a car from that data? Number of choices of car C-3ways S-3ways M-5ways Totals 4x3x5=60 ways ex) License plates in UT Digits 3 10 choices 10 choices 10 choices Letters 3 23 choices 23 choices 23 choices 3 3 Total (10) (23) = 12167000 Permutations The number of permutation of n objects is n! nx(n-1)x(n-2)x(n-3)x(n-4)x(n-…..) Calculator n --> math-->prob--> 4 0!=1 ex) n=5 how many ways can we arrange 5 objects? 5 4 3 2 1 total 5x4x3x2x1 possible arrangement of 5 objects 5! Permutation: Selection with order ex) how many ways can they picked for 9 positions on a baseball team? n=11 kids 11 10 9 8 7 6 5 4 3 P 11 9= 11! = 11! (11-9)! 2! calculator 11-->math -- =19958400 possible permutations of 11 kids taken 9 >prob-->2-->9 at a time 1)number of permutations of n objects is n! 2)number of permutations of n objects taken k at a time nPk=n!/(n-k)! ex)5 5 5! 5! 5! 5x4x3x2x1 120 (5-5)! 0! 1 1 n n=n! Combinations Selection without order ex) 10 individuals Want a committee of 3 How many ways can we select 3 out of 10? n k n! (n-k)!k! 10 3 10! *10! 3!7! 3! (10-3)! 10x9x8x7x6x5x4x3x2x1) (3x2x1)(7x6x5x4x3x2x1) 10x3x4 10x3x4= 120 1x1x1 calculator 10-->math-->3-->3 ex) DBS_5_8 slide 11 a) what is the number of possible winning combinations? megaball 56 56 56 56 56 46 56 5 Total number of winning tickets is (456 5C ) = (46)(3819816) = 175711536 possible winning tickets b) what is the probability of winning? P(1 winning ticket)=1/175,711,536 ex)5 5 5! 5x4x3x2x1 1! 4! 4x3x2x1 n n=1


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