Week 5 Lecture Math 221 Mar 2012

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Week 5 Lecture Math 221 Mar 2012

  1. 1. B. HeardThese charts are not to be posted or used without my written permission. Students can download a copy for their personal use.
  2. 2.  Preparing for the Week 5 Quiz  Factorials  Combinations/Permutations  Probability  Probability Distributions  Discrete/Continuous Distributions  Binomial Distribution  Poisson Distribution  Pivot Tables
  3. 3.  Factorials  n! simply means n x (n-1) x (n-2) x …. 3 x 2 x 1  For example 5! = 5x4x3x2x1 = 120, so 5!=120  Always remember that 0! = 1 (NOT ZERO)  Additional examples  3! + 4! = (3x2x1) + (4x3x2x1) = 6 + 24 = 30  5(5-3)! = 5 x 2! = 5 x (2x1) = 5x2 = 10  5(5! – 3!) = 5(120 – 6) = 5(114) = 570  4!(0!) = 24(1) = 24  3!/0! = 6/1 = 6
  4. 4.  Combinations/Permutations  Combinations – The number of ways of something happening when order DOES NOT matter.  Permutations – The number of ways of something happening when order DOES matter.  The number of ways to pick 5 out of 10 players to start on a basketball team would be a combination.  The number of ways to pick 5 out of 10 players to play 5 different positions would be a permutation.
  5. 5.  Combinations/Permutations (continued)  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).  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).
  6. 6.  Combinations/Permutations (continued) Let’s Do these in Minitab  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).  This is a combination where we are picking 3 from 11 (Sometimes noted 11C3).  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).  This is a permutation where we are picking 3 from 11 (Sometimes noted 11P3).
  7. 7.  Combinations/Permutations (continued) Minitab  Let’s look at the combination where we are picking 3 from 11 (Sometimes noted 11C3).  In Minitab, Choose “Calc”, then “Calculator”  Type C1, or any other blank column in the “Store result in variable” box  In the Expression box, type “Combinations(11,3)” In the cell you chose, you will see your answer of 165 This means there are 165 ways to pick a committee of 3 from 11 people (remember order didn’t matter).
  8. 8.  Combinations/Permutations (continued) Minitab
  9. 9.  Combinations/Permutations (continued) Minitab  Let’s look at the permutation where we are picking 3 from 11 (Sometimes noted 11P3).  In Minitab, Choose “Calc”, then “Calculator”  Type C1, or any other blank column in the “Store result in variable” box  In the Expression box, type “Permutations(11,3)” In the cell you chose, you will see your answer of 990 This means there are 990 ways to pick a committee of 3 from 11 people where there are THREE DISTINCT POSITIONS (order DID matter).
  10. 10.  Combinations/Permutations (continued) Minitab
  11. 11.  Combinations/Permutations (continued) For the same numbers, the number of permutations will always be larger! This is because there are distinct positions. As an example 3C3 = 1 (there is only one way to pick three from three) 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 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.
  12. 12.  Probability  Probability is simply the number of desired events over the total number of events that can happen.  Sound complicated? It’s not  What is the probability of rolling a 5 on six-sided die?  One side with a five/six sides = 1/6
  13. 13.  Probability (continued)  Sample Spaces (What can happen?)  For a regular light switch, the sample space would be {on, off}  For a six-sided die, the sample space would be {1,2,3,4,5,6}  For a new baby, the sample space would be {girl, boy}  For suits in a card deck, the sample space would be {hearts, diamonds, clubs, spades}
  14. 14.  Probability (continued)  Examples  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  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  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
  15. 15.  Probability (continued)  Conditional Probability Examples  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)  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)
  16. 16.  Probability Distributions  All Probabilities must be between 0 and 1and the sum of the probabilities must be equal to 1.  Examples  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?  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)
  17. 17.  Probability Distributions (continued)  Examples  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?  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)
  18. 18.  Probability Distributions (continued)  Examples  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?  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)  {-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
  19. 19.  Probability Distributions (continued)  Examples  Given the random variable X = {100, 200} with P(100) = 0.7 and P(200) = 0.3. Find E(X).  Simple  E(X) = Sum of X(P(X) (add the values times their probabilities)  E(X) = 100(0.7) + 200(0.3) = 70 + 60 = 130
  20. 20.  Probability Distributions (continued)  Examples of Probability values  Which of these can be probability values?  3/5 YES  0.004 YES  1.32 NO  43% YES  -0.67 NO  1 YES  5 NO  0 YES
  21. 21.  Discrete/Continuous Distributions Simple explanation 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.)
  22. 22.  Discrete/Continuous Distributions Simple Example  The number of cans of soda in your refrigerator is discrete (0,1,2,3 etc.)  The amount of soda IN the can is continuous (ounces can be split and split and split, etc.)
  23. 23.  Binomial and Poisson Distributions Know the difference between the two!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.
  24. 24.  Binomial Distribution using Minitab The way is to show you an example. 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.
  25. 25.  Binomial Distribution using Minitab  Open a new Minitab Project  Since n= 6, put 0,1,2,3,4,5,6 in column C1  (Don’t forget the zero)  Go to Calc>>Probability Distributions>>Binomial  Change Radial Button to “Probability”  Put 6 in for number of trials and 0.2 in for event probability  Put “C1” in for Input Column  Click “OK”
  26. 26.  Binomial Distribution using Minitab You would write X = {0,1,2,3,4,5,6} P(x=0) = 0.2621 P(x=1) = 0.3932 P(x=2) = 0.2458 This is your Etc. distribution, we P(x=3) = 0.0819 now have to calculate the P(x=4) = 0.0154 mean, variance and standard P(x=5) = 0.0015 deviation. P(x=6) = 0.0001
  27. 27.  Binomial Distribution using Minitab  Remember we had p = 0.2 and n = 6  Mean?  E(X) = np so E(X) = 6(0.2) = 1.2 (the mean)  Why “E(X)”? Because we would “expect” the outcome 1.2 times out of the 6 times we did it.  Variance?  V(X) = npq (q is just “1-p”), so  V(X) = 6(0.2)(1-0.2) = 6(0.2)(0.8) = 0.96 (the variance)  Standard Deviation?  Std Dev. = Square Root of the Variance = √0.96 = 0.9216 (the standard deviation)
  28. 28.  Binomial Distribution using Minitab  Same Example, what if you were asked  P(X≥5)  P(X<3)  Etc.  Use your probabilities (next page)
  29. 29.  Binomial Distribution using Minitab  P(X≥5) P(X≥5) = P(X=5) + P(X=6) = 0.001536 + 0.000064 = 0.0016
  30. 30.  Binomial Distribution using Minitab  P(X<3) P(X<3) = P(X=0) + P(X=1) + P(X=2) = 0.262144 + 0.393216 + 0.245760 = 0.90112
  31. 31.  Poisson Distribution with Minitab Find P(x=3) For a Poisson Distribution with mean = 5 Probability Density Function Poisson with mean = 5 x P( X = x ) 3 0.140374 Poisson Distribution using Minitab Go to Calc>>Probability Distributions>>Poisson Change Radial Button to “Probability” Put 5 in for number of trials and 3 in for “Input Constant” Click “OK”
  32. 32.  Poisson Distribution with MinitabWhat is the probability that X≤3?Use Cumulative Distribution FunctionPoisson with mean = 5x P( X <= x )3 0.265026
  33. 33.  Pivot Tables Chocolate Vanilla Total Girls 13 6 19 Boys 17 5 22 Total 30 11 41
  34. 34.  Pivot TablesFind P(Girl) Chocolate Vanilla TotalP(Girl) = 19/41 Girls 13 6 19Find P(Vanilla)P(Vanilla) = 11/41 Boys 17 5 22Find P(Girl who likesChocolate) Total 30 11 4113 out of the 41 so it is13/41
  35. 35.  Pivot TablesFind P(Girl given they Chocolate Vanilla Totallike chocolate)P(Girl|Choc) = 13/30 Girls 13 6 19Find P(Like Vanilla Boys 17 5 22given they are a boy)P(Vanilla|Boy) = 5/22 Total 30 11 41
  36. 36.  Pleasenote that I posted a presentation in the Statcave last week that shows how to do factorials, combinations and permutations in Excel AlsoI posted a presentation on doing binomial, geometric and Poisson distribution calculations in Excel. Both of these may be helpful to you!
  37. 37.  Comesee me and download the charts at www.facebook.com/statcave You don’t have to be a Facebook person….

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