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Week 5 Lecture, Statistics for Decision Making

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- 1. Week 5 Lecture Statistics for Decision Making<br />B. Heard<br />These charts are not to be posted or used without my written permission. Students can download a copy for their personal use.<br />
- 2. Preparing for the Week 5 Quiz<br />Factorials<br />Combinations/Permutations<br />Probability<br />Probability Distributions<br />Discrete/Continuous Distributions<br />Binomial Distribution<br />Poisson Distribution<br />Pivot Tables<br />Week 5 Lecture<br />
- 3. Factorials<br />n! simply means n x (n-1) x (n-2) x …. 3 x 2 x 1<br />For example 5! = 5x4x3x2x1 = 120, so 5!=120<br />Always remember that 0! = 1 (NOT ZERO)<br />Additional examples<br />3! + 4! = (3x2x1) + (4x3x2x1) = 6 + 24 = 30<br />5(5-3)! = 5 x 2! = 5 x (2x1) = 5x2 = 10<br />5(5! – 3!) = 5(120 – 6) = 5(114) = 570<br />4!(0!) = 24(1) = 24<br />3!/0! = 6/1 = 6<br />Week 5 Lecture<br />
- 4. Combinations/Permutations<br />Combinations – The number of ways of something happening when order DOES NOT matter.<br />Permutations – The number of ways of something happening when order DOES matter.<br />The number of ways to pick 5 out of 10 players to start on a basketball team would be a combination.<br />The number of ways to pick 5 out of 10 players to play 5 different positions would be a permutation.<br />Week 5 Lecture<br />
- 5. Combinations/Permutations (continued)<br />The number of ways to pick 3 committee members out of 11 people would be a combination because there are not distinct positions mentioned (order doesn’t matter).<br />The number of ways to pick 3 committee members out of 11 people where 1 would be the President, Vice-President and Treasurer would be a permutation because there are distinct positions mentioned (order does matter, you not only could be picked, but you could be one of three different positions).<br />Week 5 Lecture<br />
- 6. Combinations/Permutations (continued)<br />Let’s Do these in Minitab<br />The number of ways to pick 3 committee members out of 11 people would be a combination because there are not distinct positions mentioned (order doesn’t matter).<br />This is a combination where we are picking 3 from 11 (Sometimes noted 11C3).<br />The number of ways to pick 3 committee members out of 11 people where 1 would be the President, Vice-President and Treasurer would be a permutation because there are distinct positions mentioned (order does matter, you not only could be picked, but you could be one of three different positions).<br />This is a permutation where we are picking 3 from 11 (Sometimes noted 11P3).<br />Week 5 Lecture<br />
- 7. Combinations/Permutations (continued)<br />Minitab<br />Let’s look at the combination where we are picking 3 from 11 (Sometimes noted 11C3).<br />In Minitab, Choose “Calc”, then “Calculator”<br />Type C1, or any other blank column in the “Store result in variable” box<br />In the Expression box, type “Combinations(11,3)”<br />In the cell you chose, you will see your answer of 165<br />This means there are 165 ways to pick a committee of 3 from 11 people (remember order didn’t matter).<br />Week 5 Lecture<br />
- 8. Combinations/Permutations (continued)<br />Minitab<br />Week 5 Lecture<br />
- 9. Combinations/Permutations (continued)<br />Minitab<br />Let’s look at the permutation where we are picking 3 from 11 (Sometimes noted 11P3).<br />In Minitab, Choose “Calc”, then “Calculator”<br />Type C1, or any other blank column in the “Store result in variable” box<br />In the Expression box, type “Permutations(11,3)”<br />In the cell you chose, you will see your answer of 990<br />This means there are 990 ways to pick a committee of 3 from 11 people where there are THREE DISTINCT POSITIONS (order DID matter).<br />Week 5 Lecture<br />
- 10. Combinations/Permutations (continued)<br />Minitab<br />Week 5 Lecture<br />
- 11. Week 5 Lecture<br />Combinations/Permutations (continued)<br />For the same numbers, the number of permutations will always be larger! This is because there are distinct positions.<br />As an example<br />3C3 = 1 (there is only one way to pick three from three)<br />3P3 = 6 (there are 6 ways to pick 3 people from 3 to serve in 3 distinct positions, think about it 3 could be President, 2 are left to serve as Vice-President, one is left to serve as Treasurer, 3x2x1 = 6<br />In our previous example we had 11P3 = 990 which could be looked at as 11 x 10 x 9 = 990. Why? 11 could be Pres., 10 VP, and 9 Treasurer.<br />
- 12. Probability<br />Probability is simply the number of desired events over the total number of events that can happen.<br />Sound complicated? It’s not<br />What is the probability of rolling a 5 on six-sided die?<br />One side with a five/six sides = 1/6<br />Week 5 Lecture<br />
- 13. Probability (continued)<br />Sample Spaces (What can happen?)<br />For a regular light switch, the sample space would be {on, off}<br />For a six-sided die, the sample space would be {1,2,3,4,5,6}<br />For a new baby, the sample space would be {girl, boy}<br />For suits in a card deck, the sample space would be {hearts, diamonds, clubs, spades}<br />Week 5 Lecture<br />
- 14. Probability (continued)<br />Examples<br />What is the probability of drawing a Jack from a standard deck of cards? (There are 4 cards out of the 52 that are Jacks) The probability would be 4/52 or 1/13 simplified<br />What is the probability of drawing a red Jack from a standard deck of cards? (There are 2 cards out of the 52 that are red Jacks) The probability would be 2/52 or 1/26 simplified<br />What is the probability of drawing a Jack of Hearts from a standard deck of cards? (There is only 1 Jack of Hearts out of the 52) The probability would be 1/52<br />Week 5 Lecture<br />
- 15. Probability (continued)<br />Conditional Probability Examples<br />What is the probability of drawing a Jack from a standard deck of cards, if you first drew a 3 of clubs and didn’t replace it? (There are 4 cards out of the 51 that are Jacks) The probability would be 4/51 (Remember you didn’t replace the 3 of clubs)<br />What is the probability of drawing a Jack from a standard deck of cards, if you first drew a Jack of clubs and didn’t replace it? (There are 3 Jacks left out of the 51) The probability would be 3/51 or 1/17 simplified (Remember you didn’t replace the Jack of clubs)<br />Week 5 Lecture<br />
- 16. Probability Distributions<br />All Probabilities must be between 0 and 1and the sum of the probabilities must be equal to 1.<br />Examples<br />If X = {5, 10, 15, 20} and P(5) = 0.10, P(10) = 0.20, P(15) = 0.30, and P(20) = 0.40, can the distribution of the random variable X be considered a probability distribution?<br />YES, because all probabilities (0.10,0.20,0.30,0.40) are between 0 and 1 and they add up to 1 (0.10+0.20+0.30+0.40 = 1) <br />Week 5 Lecture<br />
- 17. Probability Distributions (continued)<br />Examples<br />If X = {5, 10, 15, 20} and P(5) = 0.20, P(10) = 0.20, P(15) = 0.20, and P(20) = 0.20, can the distribution of the random variable X be considered a probability distribution?<br />No, the probabilities (0.20,0.20,0.20,0.20) are between 0 and 1 BUT they DO NOT add up to 1 (0.20+0.20+0.20+0.20 = 0.80) <br />Week 5 Lecture<br />
- 18. Probability Distributions (continued)<br />Examples<br />If X = {-5, A, 7, 0} and P(-5) = 0.50, P(A) = 0.10, P(7) = 0.10, and P(0) = 0.30, can the distribution of the random variable X be considered a probability distribution?<br />YES, the probabilities (0.50,0.10,0.10,0.30) are between 0 and 1 and they add up to 1 (0.50+0.10+0.10+0.30 = 1) <br /> {-5, A, 7, 0} Does Not matter, you can have negative numbers, letters, colors, names, etc. in the sample space but you couldn’t have negative values as PROBABILITIES<br />Week 5 Lecture<br />
- 19. Probability Distributions (continued)<br />Examples<br />Given the random variable X = {100, 200} with P(100) = 0.7 and P(200) = 0.3. Find E(X).<br />Simple<br />E(X) = Sum of X(P(X) (add the values times their probabilities)<br />E(X) = 100(0.7) + 200(0.3) = 70 + 60 = 130<br />Week 5 Lecture<br />
- 20. Probability Distributions (continued)<br />Examples of Probability values<br />Which of these can be probability values?<br />3/5 YES<br />0.004 YES<br />1.32 NO<br />43% YES<br />-0.67 NO<br />1 YES<br />5 NO<br />0 YES<br />Week 5 Lecture<br />
- 21. Discrete/Continuous Distributions<br />Simple explanation<br />A discrete probability distribution has a finite number of possible outcomes. (People, items, distinct things that can not be measured infinitesimally)A continuous probability is based on a continuous random variable such as a persons height or weight. (GOOD EXAMPLES ARE UNITS OF MEASURE LIKE HEIGHTS, WEIGHTS, VOLUME, ETC.)<br />Week 5 Lecture<br />
- 22. Discrete/Continuous Distributions<br />Simple Example<br />The number of cans of soda in your refrigerator is discrete (0,1,2,3 etc.)<br />The amount of soda IN the can is continuous (ounces can be split and split and split, etc.)<br />Week 5 Lecture<br />
- 23. Binomial and Poisson Distributions<br />Know the difference between the two!<br />A good hint is that a Poisson usually give you an average number of something per time period and a Binomial gives you a probability and a number of times/trials/etc.<br />Week 5 Lecture<br />
- 24. Binomial Distribution using Minitab<br />The way is to show you an example.<br />Let’s say we have a binomial experiment with p = 0.2 and n = 6 and you are asked to set up the distribution and show all x values and the mean, variance and standard deviation.<br />Week 5 Lecture<br />
- 25. Binomial Distribution using Minitab <br />Open a new Minitab Project<br />Since n= 6, put 0,1,2,3,4,5,6 in column C1 (Don’t forget the zero)<br />Go to Calc>>Probability Distributions>>Binomial<br />Change Radial Button to “Probability”<br />Put 6 in for number of trials and 0.2 in for event probability<br />Put “C1” in for Input Column<br />Click “OK”<br />Week 5 Lecture<br />
- 26. Binomial Distribution using Minitab <br />Week 5 Lecture<br />You would write<br />X = {0,1,2,3,4,5,6}<br />P(x=0) = 0.2621<br />P(x=1) = 0.3932<br />P(x=2) = 0.2458<br />P(x=3) = 0.0819<br />P(x=4) = 0.0154<br />P(x=5) = 0.0015<br />P(x=6) = 0.0001<br />This is your distribution, we now have to calculate the mean, variance and standard deviation.<br />Etc.<br />
- 27. Binomial Distribution using Minitab <br />Remember we had p = 0.2 and n = 6<br />Mean? <br />E(X) = np so E(X) = 6(0.2) = 1.2 (the mean)<br />Why “E(X)”? Because we would “expect” the outcome 1.2 times out of the 6 times we did it.<br />Variance?<br />V(X) = npq (q is just “1-p”), so<br />V(X) = 6(0.2)(1-0.2) = 6(0.2)(0.8) = 0.96 (the variance)<br />Standard Deviation?<br />Std Dev. = Square Root of the Variance = √0.96 = 0.9216 (the standard deviation)<br />Week 5 Lecture<br />
- 28. Binomial Distribution using Minitab<br />Same Example, what if you were asked<br />P(X≥5)<br />P(X<3)<br />Etc.<br />Use your probabilities (next page) <br />Week 5 Lecture<br />
- 29. Binomial Distribution using Minitab<br />P(X≥5)<br />Week 5 Lecture<br />P(X≥5) = P(X=5) + P(X=6) = 0.001536 + 0.000064 = 0.0016<br />
- 30. Binomial Distribution using Minitab<br />P(X<3)<br />Week 5 Lecture<br />P(X<3) = P(X=0) + P(X=1) + P(X=2) = 0.262144 + 0.393216 = 0.245760 = 0.90112<br />
- 31. Poisson Distribution with Minitab<br />Week 5 Lecture<br />Find P(x=3)<br />For a Poisson Distribution with mean = 5<br />Probability Density Function <br />Poisson with mean = 5<br />x P( X = x )<br />3 0.140374<br />Poisson Distribution using Minitab <br />Go to Calc>>Probability Distributions>>Poisson<br />Change Radial Button to “Probability”<br />Put 5 in for number of trials and 3 in for “Input Constant”<br />Click “OK”<br />
- 32. Poisson Distribution with Minitab<br />Week 5 Lecture<br />What is the probability that X≤3? <br />Use Cumulative Distribution Function <br />Poisson with mean = 5<br />x P( X <= x )<br />3 0.265026<br />
- 33. Pivot Tables<br />Week 5 Lecture<br />
- 34. Pivot Tables<br />Week 5 Lecture<br />Find P(Girl)<br />P(Girl) = 19/41<br />Find P(Vanilla)<br />P(Vanilla) = 11/41<br />Find P(Girl who likes Chocolate)<br />13 out of the 41 so it is 13/41<br />
- 35. Pivot Tables<br />Week 5 Lecture<br />Find P(Girl given they like chocolate)<br />P(Girl|Choc) = 13/30<br />Find P(Like Vanilla given they are a boy)<br />P(Vanilla|Boy) = 5/22<br />
- 36. Come see me and download the charts atwww.facebook.com/statcave<br />You don’t have to be a Facebook person….<br />Week 5 Lecture<br />

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