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Chapter 6 Discrete Probability Distributions

KVANLI PAVUR KEELING. Chapter 6 Discrete Probability Distributions. Chapter Objectives. At the completion of this chapter, you should be able to answer the following questions: ∙ What is meant by the term " probability distribution ?"

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Chapter 6 Discrete Probability Distributions

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  1. KVANLI PAVUR KEELING Chapter 6Discrete Probability Distributions

  2. Chapter Objectives • At the completion of this chapter, you should be able to answer the following questions: ∙ What is meant by the term "probability distribution?" ∙ What is the formula for the mean and variance of a discrete random variable? ∙ When, and how, is the binomial distribution used? ∙ When, and how, is the Poisson distribution used?

  3. Example • Flip a coin 2 times • Suppose we’re interested in the number of heads in the 2 flips • Chapter 5 way: Define A: H on 1st flip and H on 2nd flip B: H on 1st flip and T on 2nd flip C: T on 1st flip and H on 2nd flip D: T on 1st flip and T on 2nd flip

  4. Example • Chances of getting exactly one H in the 2 flips is P(B or C) = P(B) + P(C) since B and C are mutually exclusive • P(B) = P(H on 1st flip and T on 2nd flip) = P(H on 1st flip) · P(T on 2nd flip) = ½ · ½ = ¼ • Similarly, P(C) = ¼ • P(B or C) = ¼ + ¼ = ½

  5. An Easier Way • Define X to be the number of H’s in the 2 flips 0 with probability X = 1 with probability 2 with probability X=0 How can this occur? T on 1st flip and T on 2nd flip For a fair coin, this is ½ · ½ since coin flips are independent P(X = 0) is ½ · ½ = ¼ ¼

  6. Defining the Random Variable ¼ 0 with probability X = 1 with probability 2 with probability X=1 How can this occur? H on 1st flip and T on 2nd flip - OR - T on 1st flip and H on 2nd flip P(X = 1) is ¼ + ¼ = ½ ½ As before, this is ½ · ½ = ¼ And this is ½ · ½ = ¼

  7. Defining the Random Variable 0 with probability X = 1 with probability 2 with probability X=2 How can this occur? H on 1st flip and H on 2nd flip P(X = 2) is ½ · ½ = ¼ ¼ ½ ¼ Once again, this is ½ · ½ = ¼

  8. A Discrete Random Variable • X = number of H’s in 2 coin flips is called a discrete random variable • Why discrete? • Suppose you observe X • You would get something like {1, 1, 0, 2, 1, 2, 0, 1, …,1} • These are discrete data • Here, X = (counting something) This is a sample

  9. Continuous Random Variable • The other type of random variable is a continuous random variable • Here, X = (measuring something) • Examples: Height, weight, length, length of time • Consider a sample of heights: {5.82’, 6.45’,…., 5.41’} • These are continuous data • Continuous random variables will be discussed in Chapter 7

  10. Another Discrete Random Variable Container with 30 poker chips 10 of these 10 of these 10 of these 1 2 3 Select 2 chips with replacement Let X = total of the 2 chips

  11. X = Total of the 2 Chips 2 with probability 3 with probability X = 4 with probability 5 with probability 6 with probability X=2 How can this occur? on the 1st draw and on the 2nd draw The probability of selecting a on each draw is 1/3 since 10 of the 30 chips are numbered one • P(X = 2) is ⅓ · ⅓ = 1/9 Remember: We replace the 1st chip before drawing the 2nd chip

  12. X = Total of the 2 Chips 2 with probability 3 with probability X = 4 with probability 5 with probability 6 with probability X=3 How can this occur? on the 1st draw and on the 2nd draw -- OR -- on the 1st draw and on the 2nd draw • P(X = 3) is (1/3)(1/3) + (1/3)(1/3)= 1/9 + 1/9 = 2/9 1/9 2/9

  13. X = Total of the 2 Chips 2 with probability 3 with probability X = 4 with probability 5 with probability 6 with probability X=4 How can this occur? on the 1st draw and on the 2nd draw -- OR -- on the 1st draw and on the 2nd draw -- OR -- on the 1st draw and on the 2nd draw 1/9 It looks like P(X = x) is We’ll see if this holds up! 2/9 3/9 P(X = 4) is 1/9 + 1/9 + 1/9 = 3/9

  14. X = Total of the 2 Chips 2 with probability 3 with probability X = 4 with probability 5 with probability 6 with probability X=5 How can this occur? on the 1st draw and on the 2nd draw -- OR -- on the 1st draw and on the 2nd draw 1/9 According to the previous expression, P(X = 5) should be (5 – 1)/9 = 4/9 This expression will not work! 2/9 3/9 2/9 P(X = 5) is 1/9 + 1/9 = 2/9

  15. X = Total of the 2 Chips 2 with probability 3 with probability X = 4 with probability 5 with probability 6 with probability X=6 How can this occur? on the 1st draw and on the 2nd draw P(X = 6) is ⅓ · ⅓ = 1/9 Sum these values The total must equal one (and it does) 2/9 3/9 2/9 1/9

  16. Representing a Discrete Random Variable • Make a list - - this was done in the previous slides • Make a histogram/bar chart 3/9 Prob. 2/9 1/9 2 3 4 5 6

  17. Finding the Probabilities • Earlier, we tried the expression P(X = x) is • This did not work for x = 5 and x = 6 • There is an expression that works here; namely P(X = x) is • You will never be asked to find such an expression! • Let’s just make sure it works

  18. The Probability Mass Function (PMF) • P(X = 3) = • P(X = 2) = • P(X = 4) = • P(X = 6) = • This expression is called a probability mass function • The PMF probabilities should be between 0 and 1 and sum to 1 • P(X = 5) = Everything checks out – this is the correct expression! PMF

  19. Another Look at Coin Flipping 0 with probability X = 1 with probability 2 with probability • To find the PMF for this situation, we need to define the combination function • This is written aCb • aCb counts the number of ways of getting “b” heads (H’s) in “a” flips of a coin ¼ ½ ¼

  20. The Combination Function a! is read “a factorial” • How to find: • For example, 5C2 counts the number of ways of getting 2 H’s (and 3 T’s) in 5 flips of a coin • This is • I would suggest you figure out how to find combinations on your calculator!!

  21. The PMF for this Example • X = number of H’s in 2 flips of a coin • The PMF here: P(X = x) is 2Cx(½)2 • Does this work? • P(X = 0) is • So, P(X = 0) is (1)(¼) = ¼ (this is correct) • P(X = 1) is = (2)(¼) = (½) (correct) • P(X = 2) is = (1)(¼) = ¼ (correct) 2C0 = 1 (Hint: 0! Is defined to be 1) It works!

  22. The PMF for any Number of Coin Flips • Does this pattern extend to any number of coin flips? • As we’ll see in the next section, it does • For example, the chances of getting 6 H’s in 10 flips of a coin is 10C6(½)10 = (210)(½)10 = .205 • When you flip a coin 10 times, 20% of the time you’ll get 6 H’s and 4 T’s There are 210 ways of getting 6 H’s in 10 flips

  23. Section 6.2 • Let X = total of the 2 chips • Suppose we observe X say, 10 times • Observations: {3, 5, 4, 4, 2, 5, 6, 3, 5,4} • Sample mean is x = 4.1 • Sample variance is s2 = 1.433 • Sample standard deviation is s = 1.197 • Question: What is the mean going to be if X is observed indefinitely? a sample

  24. Mean of a Random Variable • The answer to the question is referred to the mean of X • The symbol for the mean of X is µ • This is spelled mu and pronounced “myoo” • So, µ is the average value of X if it is observed indefinitely • How to find: µ is ∑x·P(x)

  25. X = Total of the 2 chips • µ is (2)(1/9) + (3)(2/9) + (4)(3/9) + (5)(2/9) + (6)(1/9) • This is µ = 36/9 = 4.0 • Question: The sample variance was s2 = 1.433 • What is the variance going to be if X is observed indefinitely? • This is called the variance of X • The symbol for the variance is σ2 Read as “sigma squared”

  26. X = Total of the 2 chips • How to find the variance: σ2 = ∑(x - µ)2 · P(x) • An easier way: σ2 = ∑x2P(x) - µ2 • Here, σ2 = [(2)2(1/9) + (3)2(2/9) + (4)2(3/9) + (5)2(2/9) + (6)2(1/9)] – (4)2 This is: σ2 = 156/9 – 16 = 1.333 • The standard deviation of X is • Here, This is the variance of X This is the standard deviation of X

  27. Summary • Every discrete random variable, X, has a ∙ mean (µ) ∙ variance (σ2) ∙ standard deviation (σ)

  28. Section 6.3 – Binomial Random Variables • Example X = number of heads (H’s) in 5 flips of a coin • As we will see, X is a special kind of discrete random variable, called a binomial random variable

  29. Conditions for a Binomial Random Variable 1. Have n = 5 coin flips 1. Have n trials 2. Each trial has two possible outcomes Called: success and failure 3. The chances of a success on each trial is p 2. Each flip has two possible outcomes success: flipping a H failure: flipping a T 3. p is ½ (assuming a fair coin)

  30. Conditions for a Binomial Random Variable 4. The trials are independent (don’t affect each other) 5. X = number of successes (out of the n trials) 4. Coin flips are always independent 5. X = number of H’s (out of the 5 flips) This is a binomial random variable

  31. X = Number of H’s in 5 Flips 0 with probability 1 with probability X = 2 with probability 3 with probability 4 with probability 5 with probability • It appears that P(X = x) is 5Cx(½)5 • Using this PMF, the values on the next slide are obtained Consider X = 2 (2 H’s and 3 T’s) How many ways can you get 2 H’s? There are 5C2 = 10 ways. Consider one of these: T H T T H Chances of this occurring is (½)(½)(½)(½)(½) = (½)5 = 1/32 Ditto for all 10 of these So, P(X = 2) is 5C2(½)5 = 10/32

  32. P(X = x) is 5Cx(½)5 0 with probability 5C0(½)5 = (1)(1/32) = 1/32 1 with probability 5C1(½)5 = (5)(1/32) = 5/32 2 with probability 5C2(½)5 = (10)(1/32) = 10/32 3 with probability 5C3(½)5 = (10)(1/32) = 10/32 4 with probability 5C4(½)5 = (5)(1/32) = 5/32 5 with probability 5C5(½)5 = (1)(1/32) = 1/32 X = This is a binomial random variable with n = 5 and p = ½

  33. Any Number of Coin Flips • For any number of coin flips (say, n) the PMF is P(X = x) is nCx(½)n • Consider n = 20 coin flips • The chances of getting 12 H’s is 20C12(½)20 • 20C12 is very large – in fact, there are 125,970 ways of getting 12 H’s and 8 T’s • (½)20 is very small • It turns out that 20C12(½)20 = .120

  34. Using an Unfair Coin • Suppose you don’t have a fair coin • In fact, there is a 20% chance of flipping a head and an 80% chance of flipping a tail • These percentages are determined by flipping this coin many times and observing the results • Let X = number of H’s in 5 flips using this coin • X is still binomial, only now p = .2 rather than ½

  35. X = Number of H’s in 5 Flips of an Unfair Coin 0 with probability 1 with probability X = 2 with probability 3 with probability 4 with probability 5 with probability • It appears that P(X = x) is 5Cx(.2)x(.8)5-x • Using this PMF, the values on the next slide are obtained Consider X = 2 (2 H’s and 3 T’s) How many ways can you get 2 H’s? There are still 5C2 = 10 ways. Consider one of these: T H T T H Chances of this occurring is (.8)(.2)(.8)(.8)(.2) = (.2)2(.8)3 Ditto for all 10 of these So, P(X = 2) is 5C2(.2)2(.8)3

  36. P(X = x) is 5Cx(.2)x(.8)5-x This is p This is 1 -p 0 with probability 5C0(.2)0(.8)5 = .328 1 with probability 5C1(.2)1(.8)4 = .410 2 with probability 5C2(.2)2(.8)3 = .205 3 with probability 5C3(.2)3(.8)2 = .051 4 with probability 5C4(.2)4(.8)1 = .006 5 with probability 5C5(.2)5(.8)0 = .000 X = This is a binomial random variable with n = 5 and p = .2

  37. Any Binomial Random Variable • The PMF for any binomial random variable with n trials and probability of success on each trial equal to p is P(X = x) is nCxpx(1 – p)n-x • Table A-1 has this PMF worked out for n ≤ 20 and certain values of p • This table will be illustrated using the unfair coin example

  38. Table A-1

  39. Another Container Container with 20 poker chips 6 red ones 14 blue ones Select 4 chips with replacement Let X = number of red chips

  40. Is X Binomial? • There are n = 4 trials • There are 2 possible outcomes on each trial • drawing a red chip ← success • drawing a blue chip ← failure • What is p (chance of a success in each trial)? p is 6/20 = .3 • Are the trials independent? Yes, since the chips are selected with replacement • Yes, X is binomial with n = 4 and p = .3

  41. Table A-1

  42. X = Number of Red Chips This is 4C1(.3)1(.7)3 0 with probability .240 1 with probability .412 X = 2 with probability .265 3 with probability .076 4 with probability .008 • Buzz Words: “no more than” and “at least” • “no more than” means ≤ • “at least” means ≥

  43. X = Number of Red Chips • What are the chances of getting no more than 2 red chips? • P(X ≤ 2) = P(0) + P(1) + P(2) = .240 + .412 + .265 = .917 • An easier way: Use Table A-2 • It contains ≤ probabilities and will be slightly more accurate

  44. Table A-2

  45. X = Number of Red Chips • What are the chances of getting at least 2 red chips? • P(X ≥ 2) = P(2) + P(3) + P(4) = .265 + .076 + .008 = .349 • You can use Table A-2 • P(X ≥ 2) = 1 – P(X < 2) = 1 – P(X ≤ 1) • Using Table A-2 on the next slide, this is 1 - .652 = .348 (slightly more accurate) Using Table A-1 Don’t forget this step!

  46. Table A-2

  47. With or Without Replacement • When selecting items, your choice is with replacement or without replacement • To be a binomial random variable, items should be selected with replacement • But with this procedure, you could select one of the items more than once, generally not something you want to happen • For this reason, sampling without replacement is the usual procedure

  48. With or Without Replacement • It turns out that it doesn’t really matter whether you are sampling with replacement or without replacement provided your sample size (n) is less than 5% of the population size (N) - - that is, n/N < .05 • In the previous example, N = number of chips in the container (20)

  49. With or Without Replacement • In the chips example, 4/20 = .2 > .05 and this rule doesn’t apply • For this situation, the chips must be selected with replacement for X to be binomial • For many situations, you are selecting a sample from a huge population and the sample size is less than 5% of the population size • In this case, whether you select the items with or without replacement doesn’t really matter

  50. Shortcuts for the Mean and Variance • Some good news: There is no need to hammer out the mean and variance as discussed in section 2 for a binomial random variable • It turns out for a binomial random variable: mean = µ = np variance = σ2 = np(1 – p) standard deviation = σ =

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