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# Notes MATH 220 220

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This 6 page Class Notes was uploaded by Jay Ty on Tuesday October 4, 2016. The Class Notes belongs to 220 at James Madison University taught by Colleen Watson in Fall 2016. Since its upload, it has received 5 views. For similar materials see Elementary Statistics in Math at James Madison University.

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Date Created: 10/04/16

9/27/2016 MATH 220 Watson Notes I. Correlation coefficient = r a. Coefficient of determination i. % of the change in y can be explained by the linear relationship between‘x’ and ‘y’ 1. Tells you weak, moderate strong 2. Tells you how much you know 3. How much you don’t know ii. REMEMBER WORD FOR WORD II. Interpret Slope a. M= y2- y1/x2-x1 i. Change in y over change inx 1. What if we let x change 1 unit? a. Then slope is reallyhow y is changing ii. As x increases or decreasesone unit, y increases or decreasesslope units! 1. Positive slope: both go up and both go down 2. Negative slope: one go up and one go down III. R and R^2 a. Cannot pick up every data point or residual plot b. Can lead you astray occasionally IV. Calculate Linear Regressionwith your calculator a. Stat fill in L1, L2, L3 Stat Calculate LinReg L2-L1 L3-L1 Calculate V. Weak moderate strong number line a. The further you are from zero, the better i. -.73 is better than .6 because it is furthest from zero (we use absolute value) VI. Probability a. Systematic study of uncertainty i. Subjective Probability- personal to you, no one can calculate it 1. Ex: parking ii. Classical/Theoretical-games of chance 1. Ex: dice,coins, etc. 2. Law of Large Numbers (picture on pg. 149 of book) a. Do “experiment” over and over and over and it willsettle downto what we call the true probability iii. Empirical- accumulation of past results 1. Ex: number of times a Dixie cup willland on its bottom in a kindergarten class 9/29/2016 MATH 220 Watson Notes Chapter 4.1 I. My focus for probability: Apply rules to word problems (hard to do) and “reading off of data charts” (easierto do, on video) roll 2 dice where x=sum 0 ≤ P ≤ 1 0 ≤ P ≤ 100% a. There are multiple ways to find the sum of five, because (2,3) is different from (3,2). This method is called Probability Distribution. 1. Practice: Probability(roll sum of at least three). This means from the number 3 to the number 12. We do not want “2”. Chapter 4.2 II. My focus for probability: Addition Rule a. The Addition Rule states that P(A or B) = P(A) + P(B)- P(A and B) where the “or” in P(A or B) is a union and the P(A and B) is an intersection. b. Disjoint: mutually exclusive, or there is no intersection, it can fall out [P(A and B)]. a. “Or” means addition III. Complement Rule: “1 – what I don’t want= what I do want” a. P(A not occur) – P(A occur) b. Dependent events: knowing 1 thing first affects the probabilityof the other i. Most of our wordproblems willbe Independent Chapter 4.3 IV. Multiplication Rule: IF A, B are INDEPENDENT, P(A and B)= P(A) ・P(B). a. My focus for probabilityexample: A fire department knows that 85% of its calls are for medical assistance. i. This does not imply that the next 20 calls willbe for medical assistance ii. P(MA)= .85 1. MA= Medical Assistance iii. There’s only ever two choices!!! iv. This phone call willeither be for medical assistance or not v. Phone calls are generally independent of one another vi. What is the probabilitythat callswillnot be for medical assistance? 1. P(a call is not for MA)= .15 = P(Not MA) vii. For the next two calls,what is the probabilitythat both are for medical assistance? 1. P(both are MA) a. It would be .85/MA ・.85/MA b. Multiplying them because it is providedthat these situations are independent of each other viii. 2 Calls st nd 1. P(1 is for MA, 2 is not MA) a. Multiply .85/MA with.15/Not MA to get .1275 because they are both independents and should be multiplied b. This is easy c. Not vague 2. P(only 1 of them is for MA) a. Two different scenarios that satisfy this question b. Did not specify, so you must do both problems i. (.85/MA times .15/Not MA) + (.15/Not MA times .85/MA) to get .255 ii. Could alsojust do 2(.85)(.15) V. More practice problems a. There’s this girl named Jeannie, and she’s a bit forgetful. The probabilitythat she forgets any chore is 10% (.1) and she has 3 chores to do today. i. What’s the probabilitythat Jeannie forgets all 3? 1. P(forgets all 3) = fff = (.1)(.1)(.1) ii. Probability that she remembers at least 1 1. Long version: What does at least 1 mean? a. It means she remembers 1 OR remembers 2 OR remembers 3 b. Remembers 1? i. Which one? 1. Must do rff, OR frf OR ffr a. r= remember b. f= forget c. Remembers 2? i. Which two? 1. Must do rrf, frr, rfr d. Remembers 3? i. Means rrr e. This is too much work 2. Shortcut a. 1- (don’t want) i. We don’t want her to remember 0 b. This is a synonym for forgets all 3 i. 1- (forgets all 3) 1. 1- (.1)(.1)(.1)= 1- .001= .999 iii. If Jeannie has sixchores, what’s the probabilitythat she will forget no more than five? 1. P(forgets no more than 5) 2. This is synonymous to forgetting zero, or forgetting 1, or forgetting 2, or forgetting 3, or forgetting 4, or forgetting 5 a. This would seem like too much work 3. What we don’t want: her to forget all six a. 1-(forgets 6) = 1- (.1)(.1)(.1)(.1)(.1)(.1) iv. If Jeannie still has ixchores, what is the probabilitythat she will remember less than 90% of her chores? 1. 90% of six is (.9)(6)=5.4 2. So, Jeannie remembers less than 5.4 chores a. So Jeannie remembers 0 or 1 chore or 2 chores or 3 chores or 4 chores or 5 chores b. Too much work 3. We don’t want Jeannie to remember 6 a. 1-rrrrrr b. Jeannie forgets 10% of the time, soshe remembers 90% of the time. c. This means it is i. 1- (.9)(.9)(.9)(.9)(.9)(.9)=.4686 v. What is the probabilitythat withthese six chores, Jeannie remembers more than 92%? 1. .92(6)=5.52 2. Jeannie remembers 5.52 chores 3. One higher than 5.52 is 6 4. So we can do the simple P( remembers all) a. (.9)(.9)(.9)(.9)(.9)(.9)=.5314 b. There are fifteen kids. The probabilitythat a kid has cat(s) is .2 i. Note: Always take care of all FIFTEEN kids ii. What is the probabilitythat at least 5% of kids have cats? 1. .05(15)= .75 kids 2. So at least.75 kids have cats a. This means at least 1 kid or 2 kids or 3 kids or 4 kids or 5 kids … or 15 kids have cats b. So it’s these probabilitiesall addedtogether 3. We do not want 0 kids to have cats a. In other words, 1- ALL don’t have cats i. 1- (.8)^15=.9648 iii. Probability of greater than 95% of the kids have cats 1. .95(15) = 14.25 a. = 15 have cats (all kids) 2. .2^15 = 0.000000000032768, or ZERO a. Know scientific notation b. 3.2768E -11 means to go 11 spaces to the left,to produce the answer that is seenabove iv. What is the probabilitythat only the first three kids have cats? 1. This means that three students have cats (.2) and the other twelve don’t have cats (.8) a. .2^3(.8^12) = .0005 Handout 8 Number eleven: For a sales promotion, Pepsi places winning symbols under the caps of 10% of all Pepsi bottles. You buy a 6-pack. What is the probabilitythat you LOSE no more than 5 times? I. Steps a. 10%= .10 (to win) b. This means to lose is .9 c. You lose 5 or less (lose 0, lose 1, lose 2, lose 3, lose 4, or lose 5) d. You don’t want to lose 6 i. 1-.9^6 = .4686

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