Class 11 Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science Copyright 2017 by D.B. Rowe 1
Agenda: Recap Chapter 5.3 continued Lecture 6.1-6.2 Go over Eam 2. 2
5: Probability Distributions (Discrete Variables) 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 (1 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 )! 3
number of success in n trials page 713 prob of success on a trial number of trials. n n!!( n )! Figure from Johnson & Kuby, 2012. 4
5: Probability Distributions (Discrete Variables) 5.3 The Binomial Probability Distribution page page 713660 n=10, p=1/2 n! P( ) p (1 p)!( n )! n Figure from Johnson & Kuby, 2012. P ( ) 0 1 2 3 4 5 6 7 8 9 10 1 10 45 120 210 252 210 120 45 10 1 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 1024 5
5: Probability Distributions (Discrete Variables) 5.3 Mean and Standard Deviation of the Binomial Distribution The formula for the mean μ and variance σ 2 of Binomial is n 0 np n! p (1 p)!( n )! n 2 2 0 n n! ( ) p (1 p)!( n )! np(1 p) n np(1 p) (5.7) (5.8) 6
5: Probability Distributions (Discrete Variables) 5.3 Mean and Standard Deviation of the Binomial Distribution Eample: n Before, using, we found 1. Now using, we get. Before, using 2 1/ 2 found. [ P( )] 0 np (2) (1/ 2) 1 2 n 2 2 0 Now using, np(1 p) [( ) P( )] (2) (1/ 2) (1/ 2) 1/ 2 2, we we get. 0 1 2 P( ) 1 4 1 2 1 4 n=2 =1 p=1/2 7
5: Probability Distributions (Discrete Variables) Questions? Homework: Chapter 5 # 15, 17, 19, 29, 31, 43, 55a,b, 63, 77, 85, 89 8
Lecture Chapter 6.1-6.2 9
Chapter (Continuous Distribution) Daniel B. Rowe, Ph.D. Department of Mathematics, Statistics, and Computer Science 10
6.1 Normal Probability Distributions At the beginning of course we talked about types of data. Data Qualitative Quantitative Nominal Ordinal Discrete Continuous (names) (ordered) (gap) (continuum) Binomial Distribution (already covered) Normal Distribution (now covering) 11
6.1 Normal Probability Distributions We discussed discrete random variables and discrete probability functions, P(). 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. ={0,1,2} P ( ) 2! 1 1!(2 )! 2 2 2 12
6.1 Normal Probability Distributions The most important continuous distribution is the normal distribution (p 269). Insert and get f(). Probability distribution, continuous variable: the probability for a continuous random variable,, having values falling within a specified interval. Normal Probability Distribution Function: y f ( ) e 1 2 2 2 for all real (6.1) 13
6.1 Normal Probability Distributions The mathematical formula for the normal distribution is (p 269): f( ) e 1 2 2 where e = 2.718281828459046 π = 3.141592653589793 μ = population mean σ = population std. deviation 2 f(), 0 We will not use this formula. Figure from Johnson & Kuby, 2012. 14
6.1 Normal Probability Distributions Properties of Normal Distribution 1. Total Area under the normal curve is 1 2. Mound shaped, symmetric about mean, etends to ± 3. Has a mean of μ and standard deviation σ. 4. The mean divides area in half. 5. Nearly all area within 3σ of μ. f() f( ) e 1 2 2, 0 2.5 area.5 area Figure modified from Johnson & Kuby, 2012. 15
6.1 Normal Probability Distributions 1. Symmetric about the mean. 2. mean = median = mode. f() 3. Mean μ & variance σ 2 completely characterize. 4. P( ).68 P( 2 2 ).95 P( 3 3 ).99. 2% 14% 34% 34% 14% 2% μ-2σ μ-σ μ μ+σ μ+2σ 5. P( a b) area under curve from a to b. Figure modified from Johnson & Kuby, 2012. 16
6.1 Normal Probability Distributions When we discussed random eperiments such as flipping a coin or rolling a die, we described the outcomes and events. We then discussed the probabilities of these events which consisted of probabilities of the individual outcomes. With the discrete binomial distribution we were interested in events such as P(4 6)=P(4)+P(5)+P(6) 0.2 0.1 0 0 1 2 3 4 5 6 7 8 9 10 17
6.1 Normal Probability Distributions With the continuous normal distribution, we want areas. f() The probability is in the interval a to b is in red a b Shaded area: P(a b) Figure modified from Johnson & Kuby, 2012. 18
6.1 Normal Probability Distributions Areas of continuous functions are found with Calculus. f() a b We will not use Calculus in this class. Aside: Don t need to know. A b 1 e 2 a 1 2 f( ) This can not be done analytically and can only be done numerically with a computer. 2 d (6.2) Figure modified from Johnson & Kuby, 2012. 19
6.1 Normal Probability Distributions How are we going to find areas in this class? We find areas of the normal distribution by using the standard normal distribution and tables in the back of the book. When μ=0 and σ=1, the curve is called the standard normal distribution. f() f( ) e 1 2 2 2 I will describe the standard normal, then discuss finding areas. Figure from Johnson & Kuby, 2012. 20
6.2 The Standard Normal Distribution Properties of the Standard Normal Distribution: 1. Total area under the normal curve is 1. 2. The distribution is mounded and symmetric, it etends indefinitely in both directions; approaching but never touching the horizontal ais. 3. The distribution has a mean of 0 and a standard deviation of 1. 4. The mean divides the area in half,.5 on each side. 5. Nearly all the area is between z =-3.00 and z=3.00. 21
6.2 The Standard Normal Probability Distributions Normal distribution with population mean μ and variance σ 2. f() 1 μ 2 We want to know the (red) area under the normal distribution between 1 and 2. Note: Similar to discrete probabilities adding to 1. The total area under the normal distribution is 1. 22
6.2 The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() σ=2 Let s say we want to know the red area under the normal distribution between 1 = 2.28 and 2 = 9.28. μ= What is the area under the normal distribution between these two values? 23
6.2 The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() Aside: Don t need to know. σ=2 A 9.28 2.28 1 e 2 2 1 5 2 2 2 d f( ) μ= We would normally do this numerically with a computer. 24
6.2 The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() μ= σ=2 But we can t do calculus in this class. Someone had the idea to convert normal distribution to the standard normal. Subtract μ and divide this by σ for every value of. z = (- μ)/σ. Area between 1 and 2 is the same as area between z 1 and z 2. 25
6.2 The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() f(z) σ=2 σ=1 μ= z 1 z 2 Area between 1 and 2 is the same as area between z 1 and z 2. μ= z 26
6.2 The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() f(z) σ=2 σ=1 μ= z 1 z 2 If 1 = 2.28 and 2 = 9.28 then z 1 = ( 1 - μ)/σ and z 2 = ( 2 - μ)/σ are? μ= z 27
6.2 The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() f(z) σ=2 σ=1 μ= z 1 z 2 If 1 = 2.28 and 2 = 9.28 then z 1 = ( 1 - μ)/σ and z 2 = ( 2 - μ)/σ are? μ= z 2.28 5 2 9.28 5 2 28
6.2 The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() f(z) σ=2 σ=1 μ= We find z 1 = -1.36 and z 2 = 2.14? Do we agree with my z's? μ= z 29
6.2 The Standard Normal Probability Distributions Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. z f() f(z) 1 = 2.28 2 = 9.28 σ=2 1 z1 2 z2 σ=1 z 1 = -1.36 z 2 = 2.14 1 = μ= 2 = z 1 = μ= z 2 = We find z 1 = -1.36 and z 2 = 2.14? Do we agree with my z's? z 30
6.2 The Standard Normal Probability Distributions z 1 = -1.36 z 2 = 2.14 Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() f(z) σ=2 σ=1 μ= Area between 1 and 2 is same as the area between z 1 and z 2. μ= z 31
6.2 The Standard Normal Probability Distributions z 1 = -1.36 z 2 = 2.14 Standard normal curve μ = 0 and σ 2 = 1. Now we can simply look up the z areas in a table. f(z) Appendi B Table 3 Page 716. σ=1 μ= z 32
First Decimal Place in z Appendi B Table 3 Page 716 Second Decimal Place in z z 1 = -1.36 z 2 = 2.14 33
Appendi B, Table 3, Page 716 z 1 = -1.36 z 2 = 2.14 This table gives us the area less than a z value. P(z<z 1 )=Area less than z 1. We get this from Table 3. 34
Appendi B, Table 3, Page 716 z 1 = -1.36 z 2 = 2.14 P(z<-1.36)=Area less than -1.36. We get this from Table 3. Row labeled -1.3 over to column Labeled.06. z 35
Appendi B, Table 3, Page 717 z 1 = -1.36 z 2 = 2.14 (continued) P(z<2.14)=Area less than 2.14. We get this from Table 3. Row labeled 2.1 over to column Labeled.04. z 36
Appendi B, Table 3, Page 716-717 z 1 = -1.36 z 2 = 2.14 P(-1.36<z<2.14) = P(z<2.14) - P(z<-1.36) = - z z 37 z
6.2 The Standard Normal Probability Distributions z 1 = -1.36 z 2 = 2.14 Eample: Here is a normal distribution with μ = 5 and σ 2 = 4. f() f(z) 0.8969 0.8969 Area between 1 and 2 is same as the area between z 1 and z 2. z 39
Questions? Homework: Chapter 6 # 7, 9, 13, 17, 19, 29, 31, 33, 41 40