Marquette University MATH 1700 Class 8 Copyright 2018 by D.B. Rowe
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1 Class 8 Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science Copyright 208 by D.B. Rowe
2 Agenda: Recap Chapter Lecture Chapter
3 Recap Chapter
4 4: Probability 4.3 Rules of Probability Eample: Pick Card, A=Heart, B=Ace P( A) P( A) P( A or B) P( A) P( B) P( A and B) P( A and B) P( A) P( B A) P( A B) P( A and B) P( B) Figure from Johnson & Kuby,
5 4: Probability 4.4 Mutually Eclusive Events Mutually eclusive events: Events that share no common elements In algebra: P( A and B) 0 (no overlap) In words:. If one event has occurred, the other cannot. 2. None of the elements in one is in other. 3. In Venn diagrams, no intersection. 4. Intersection of events has a probability of zero. 5
6 4: Probability 4.5 Independent Events Independent events: the occurrence or nonoccurence of one gives no information about occurrence for the other. P( A) P( A B) P( A not B) Dependent events: occurrence of one event does have an effect on the probability of occurrence of the other event. P( A) P( A B) Special multiplication rule: Let A and B independent events in a sample space S. P( A and B) P( A) P( B) 6
7 4: Probability Questions? Homework: Chapter 4 # 59, 63, 65, 69, 85, 89, 9, 97, 05, 07, 3 Read Chapter
8 Lecture Chapter
9 Chapter 5: Probability Distributions (Discrete Variables) Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science 9
10 5. Random Variables At beginning of course we talked about types of data. Data Qualitative Quantitative Nominal Ordinal Discrete Continuous (names) (ordered) (gap) (continuum) Binomial Distribution 0
11 5. Random Variables Random Variables: A variable that assumes a unique numerical value for each of the outcomes in the sample space of a probability eperiment. Eample: Let = the number of heads when we flip a coin twice. ={0,,2} {TT,TH,HT,HH} numerical values for outcomes in sample space
12 5. Random Variables Discrete Random Variables: A quantitative random variable that can assume a countable number of values. Continuous Random Variable: A quantitative random variable that can assume an uncountable number of values. Continuum of values. Eamples: Discrete: Number of heads when we flip a coin ten times. Continuous: Distance from earth center to sun center. 2
13 5.2 Probability Distributions of a Discrete Random Variable Probability Distribution: A distribution of the probabilities associated with each of the values of a random variable. The probability distribution is a theoretical distribution; it is used to represent populations. Flipping a Coin = # H Rolling a Die face = value Figures from Johnson & Kuby, 202. Recall:. each prob between 0 & and 2. sum of prob s = 3
14 5.2 Probability Distributions of a Discrete Random Variable Probability Function: A rule P() that assigns probabilities to the values of the random variables,. Eample: Let = # of heads when we flip a coin twice. ={0,,2} 2! P ( )!(2 )! P( ) Note:. 0 P() 2. ΣP()= 4
15 P() P() 5.2 Probability Distributions of a Discrete Random Variable Eample: Let = # of H when flip a coin twice. P ( ) 2 2!!(2 )! 2 2 0,,2 0 2 P( )
16 Marquette University Recall this slide MATH 700 : Statistics.2 What is Statistics? Data: The set of values collected from the variable from each of the elements that belong to the sample. Eperiment: A planned activity whose results yield a set of data. Sample: Subset of the population. Parameter: A numerical value summarizing all the data of an entire population. Statistic: A numerical value summarizing the sample data. 6
17 5.2 Mean and Variance of a Discrete Random Variable If we calculate a numerical summary from the sample of 2 data, it is called a statistic. i.e. and. s sample mean sample variance If we calculate a numerical summary from the population of 2 data, it is called a parameter. i.e. and. population mean population variance μ is the Greek letter lower case mu. σ is the Greek letter lower case sigma. 7
18 5.2 Mean and Variance of a Discrete Random Variable Mean of a discrete random variable (epected value): The mean, μ, of a discrete random variable is found by multiplying each possible value of by its own probability, P(), and then adding all of the products together: mean of : mu = sum of (each multiplied by its own probability) n [ ip( i)] i (5.) 8
19 5.2 Mean and Variance of a Discrete Random Variable n [ i P( i )] P( ) 2P( 2)... np( n) i For the # of H when we flip a coin twice discrete distribution: μ = ( ) P( ) + ( 2 ) P( 2 ) + ( 3 ) P( 3 ) 0 2 P( )
20 5.2 Mean and Variance of a Discrete Random Variable Variance of a discrete random variable: The variance, σ 2, of a discrete random variable is found by multiplying each possible value of the squared deviation, ( μ) 2, by its own probability, P(), and then adding all of the products together: variance of : sigma squared = sum of (squared deviation times probability) n 2 2 i i equivalent formula [( ) P( )] i n [ i P( i)] i (5.2) (5.3b) 2
21 5.2 Mean and Variance of a Discrete Random Variable n i P i P 2 P 2 n P n i [( ) ( )] ( ) ( ) ( ) ( )... ( ) ( ) For the # of H when we flip a coin twice discrete distribution: σ 2 = ( -μ) 2 P( ) + ( 2 -μ) 2 P( 2 ) + ( 3 -μ) 2 μ = P( 3 ) P( ) P ( ) P ( 2) P ( 3) 22
22 5.2 Mean and Variance of a Discrete Random Variable Standard deviation of a discrete variable: The positive square root of the variance. 2 (5.4) σ 2 = σ = n 2 [( i ) P( i)] i 24
23 5.3 The Binomial Probability Distribution Let s assume we have two independent events E and E 2. We know that P( E and E ) P( E ) P( E ). Page More generally, if we have n independent events E,,E n. We know that P( E and E and E ) P( E ) P( E ) P( E ). 2 n 2 n 26
24 5.3 The Binomial Probability Distribution Let s assume we are flipping a coin twice. E =Head on first flip, E 2 =Tail on second flip. The probability of heads on any given flip is p = P(H). The probability of tails (not heads) on any given flip is q = (-p). Then P(HT)=P(H)P(T) Similarly P(TH) = P(T)P(H) =p(-p). = (-p)p. Let = # of heads in two flips of a coin. P(=) = P(HT)+P(TH) = p(-p)+(-p)p = 2p(-p). 2 ways to get one H and one T 2 ways to get = heads 27
25 5.3 The Binomial Probability Distribution An eperiment with only two outcomes is called a Binomial ep. Call one outcome Success and the other Failure. Each performance of ept. is called a trial and are independent. Prob of eactly successes n! P( ) p ( p)!( n )! n = number of trials or times we repeat the eperiment. = the number of successes out of n trials. p = the probability of success on an individual trial. n num( successes) P( successes and n- failures) Bi means two like bicycle 0,..., n (5.5) n n!!( n )! 28
26 5.3 The Binomial Probability Distribution Flip coin once. num flips num succ. prob succ O H T n= = p=/2 P( O) / 2 / 2 P(0) = # of Heads n()= ways to get Heads P( ) n( ) 0 / 2 0 / 2 n! P( ) p ( p)!( n )! P() n 29
27 5.3 The Binomial Probability Distribution Flip coin twice. O HH HT TH TT num flips num succ. prob succ P( O) / 4 / 4 / 4 / 4 = # of Heads n()= ways to get Heads n=2 = p=/2 P( ) 0 / 4 2 / 4 2 / 4 n( ) n! P( ) p ( p)!( n )! P() n 3
28 5.3 The Binomial Probability Distribution Flip coin three times. O P( O) HHH / 8 HHT / 8 HTH / 8 HTT / 8 THH / 8 THT / 8 TTH / 8 TTT / 8 = # of Heads n=3 = p=/2 P( ) 0 / 8 3 / / 8 3 / 8 n()= ways to get Heads n( ) n! P( ) p ( p)!( n )! P() n 33
29 5.3 The Binomial Probability Distribution Flip coin ten times. = # of Heads n()= ways to get Heads n ( ) P ( ) n=0 =0,,0 p=/2 n ( ) n!!( n )! n p ( p) /024 n! P( ) p ( p)!( n )! n Note:. 0 P() 2. ΣP()= 35
30 5.3 The Binomial Probability Distribution Flip coin once. O H T p=2/3 P( O) 2 / 3 / 3 = # of Heads n()= ways to get Heads P( ) n( ) 0 / / 3 num flips num succ. prob succ n= = p=2/3 n! P( ) p ( p)!( n )! n! P() 2 / 3 ( 2 / 3)!( )! 37
31 5.3 The Binomial Probability Distribution Flip coin twice. O HH HT TH TT num flips num succ. prob succ p=2/3 P( O) 4 / 9 2 / 9 2 / 9 / 9 n=2 = p=2/3 = # of Heads P( ) 0 / 9 4 / / 9 n()= ways to get Heads n( ) n! P( ) p ( p)!( n )! n 2! P() 2 / 3 ( 2 / 3)!(2 )! 2 39
32 5.3 The Binomial Probability Distribution Flip coin three times. O P( O) HHH HHT HTH HTT THH THT TTH TTT 8 / 27 4 / 27 4 / 27 2 / 27 4 / 27 2 / 27 2 / 27 / 27 p=2/3 = # of Heads n=3 = p=2/3 P( ) 0 / 27 6 / / / 27 n()= ways to get Heads n( ) n! P( ) p ( p)!( n )! n 3! P() 2 / 3 ( 2 / 3)!(3 )! 3 4
33 Questions? Homework: Chapter 5 # 5, 7, 9, 29, 3, 43 Read Chapter
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