Discrete Distribution.pptx

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Discrete Distribution.pptx

• 1. The Discrete distribution MBMG-7104/ ITHS-2202/ IMAS-3101/ IMHS-3101 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 2. What is Probability Distribution? Probability Distribution is a statistical function which links or lists all the possible outcomes a random variable can take, in any random process, with its corresponding probability of occurrence. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 3. For example, if you roll a dice, the outcome is random (not fixed) and there are 6 possible outcomes, each of which occur with probability one-sixth. Or A random variable X takes on a defined set of values with different probabilities. For example, if you toss a fair coin, there are 2 possible outcomes head or tails, each of which occur with probability 1/2. The outcome of having head or tails is random (not fixed). @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 4. Mathematically :- For example, if you toss a fair coin, two times - Then sample Space for getting heads - 𝑆 = { 𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇} Let X = no. of heads., P(X) = probability of coming heads. T 4 possible outcomes T H Probability Distribution X Value Probability 0 1/4 = 0.25 1 2/4 = 0.50 2 1/4 = 0.25 T T H H H 0 1 2 X Probability 0.50 0.25 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 5. PROBABILITY DISTRIBUTION OF A DISCRETE RANDOM VARIABLE Let us consider a discrete r.v. X which can take the possible values x1, x2, x3,…, xn. With each value of the variable X, we associate a number, pi = P(X = Xi ) ; i = 1, 2,…, n which is known as the probability of Xi and satisfies the following conditions : (i) pi = P(X = Xi ) ≥ 0, (i = 1, 2,…, n) …(1) i.e., pi’s are all non-negative. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 6. PROBABILITY DISTRIBUTION OF A DISCRETE RANDOM VARIABLE and (ii) ∑ pi = p1 + p2 + … + pn = 1, …(2) i.e., the total probability is one. More specifically, let X be a discrete random variable and define : p (x) = P(X = xi) such that p(x) ≥ 0 and ∑ p(x) = 1, @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 7. Requirements for the Probability Distribution of a Discrete Random Variable x 1. p(x) ≥ 0 for all values of x 2.  p(x) = 1 where the summation of p(x) is over all possible values of x. A probability function maps the possible values of x against their respective probabilities of occurrence, p(x) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 8. Discrete example: roll of a die x p(x) 1/6 1 4 5 6 2 3   x all 1 P(x) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 9. Probability mass function (pmf) x p(x) 1 p(x=1)=1/6 2 p(x=2)=1/6 3 p(x=3)=1/6 4 p(x=4)=1/6 5 p(x=5)=1/6 6 p(x=6)=1/6 1.0 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 10. Cumulative distribution function (CDF) All random variables, discrete and continuous have a cumulative distribution function (CDF). Corresponding to any distribution function there is CDF denoted by F(x), which, for any value of xi, gives the probability of the event x≤ xi Therefore, if f(x) is the PMF of x , then CDF is given as CDF for Discrete random variable @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 11. Cumulative distribution function x P(x) 1/6 1 4 5 6 2 3 1/3 1/2 2/3 5/6 1.0 x P(x≤A) 1 P(x≤1)=1/6 2 P(x≤2)=2/6 3 P(x≤3)=3/6 4 P(x≤4)=4/6 5 P(x≤5)=5/6 6 P(x≤6)=6/6 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 12. Mean for discrete distribution @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 13. Mean for discrete distribution Example : - When throwing a normal die , let X be the random variable defined by X = the score shown on die Probability distribution X 1 2 3 4 5 6 P(x) 1/6 1/6 1/6 1/6 1/6 1/6 Or 𝐸(𝑋) = 1 × 1 6 + 2 × 1 6 + 3 × 1 6 + 4 × 1 6 + 5 × 1 6 + 6 × 1 6 = 1 6 1 + 2 + 3 + 4 + 5 + 6 = 21 6 = 7 2 = 3.5 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 14. Variance for discrete distribution Example : - for above example X = the score shown on die 𝐸(𝑋2) = 12 × 1 6 + 22 × 1 6 + 32 × 1 6 + 42 × 1 6 + 52 × 1 6 + 62 × 1 6 = 1 6 12 + 22 + 32 + 42 + 52 + 62 = 1 6 1 + 4 + 9 + 16 + 25 + 36 = 91 6 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 15. Variance for discrete distribution Hence Variance 𝑉 𝑋 = 𝐸 𝑋2 − 𝐸 𝑋 2 = 91 6 − 7 2 2 = 35 12 = 2.916 Thus , standard deviation 𝜎 = 𝑉(𝑋) = 2.916 = 1.7076 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 16. Mean & Variance for discrete distribution Example : - A company XYZ, estimates the net profit of a new product it is launching, to be Rs. 3,000,000 during the first year if it is successful, Rs. 1,000,000 if it is moderately successful and a loss of Rs. 1,000,000. The company assigns the following probabilities to first year prospectus for the product, successful: 0.15, moderately successful : 0.25, and unsuccessful : 0.60. What are the expected value and standard deviation of first year net profit for the product? Hence Variance 𝑉 𝑋 = 𝐸 𝑋2 − 𝐸 𝑋 2 standard deviation 𝜎 = 𝑉(𝑋) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 17. © 2011 Pearson Education, Inc @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 18. Common Probability Distribution @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 19. © 2011 Pearson Education, Inc Uniform Distribution @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 20. Discrete example: roll of a die x p(x) 1/6 1 4 5 6 2 3   x all 1 P(x) 𝑋 ~ 𝑈 𝑎, 𝑏 X~ U(1,6) All outcomes has equal probability @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 21. Bernoulli Distribution @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 22. Discrete example: True or Falls 𝑋 ~ 𝐵𝑒𝑟𝑛 𝑝 T H @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 23. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 24. Binomial Distribution @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 25. Binomial Distribution @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 26. Binomial Probability The Binomial distribution is also know as the outcome of Bernoulli process. A Bernoulli process is a random process in which : a. The process is performed under the same conditions for a fixed and finite number of trials, say, n. b. Each trial is independent of other trials, i.e the probability of an outcome for any particular trial is not influenced by the outcomes of the other trials. c. Each trial has two mutually exclusive possible outcomes, such as "success" or "failure", "good or "defective", "yes" or "no", "hit" or "miss", and so on. The outcomes are usually called success and failure for convenience. d. The probability of success, p, remains constant from trial to trial (so is the probability of failure q, where, q = 1-p). @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 27. Binomial Probability Characteristics of a Binomial Experiment 1. The experiment consists of n identical trials. 2. There are only two possible outcomes on each trial. We will denote one outcome by S (for success) and the other by F (for failure). 3. The probability of S remains the same from trial to trial. This probability is denoted by p, and the probability of F is denoted by q. Note that q = 1 – p. 4. The trials are independent. 5. The binomial random variable x is the number of S’s in n trials. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 28. Binomial Probability Distribution ! ( ) (1 ) ! ( )! x n x x n x n n p x p q p p x x n x             p(x) = Probability of x ‘Successes’ p = Probability of a ‘Success’ on a single trial q = 1 – p n = Number of trials x = Number of ‘Successes’ in n trials (x = 0, 1, 2, ..., n) n – x = Number of failures in n trials @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 29. Binomial Distribution .0 .5 1.0 0 1 2 3 4 5 X P(X) .0 .2 .4 .6 0 1 2 3 4 5 X P(X) n = 5 p = 0.1 n = 5 p = 0.5 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 30. Binomial Probability Distribution Example 3 5 3 ! ( ) (1 ) !( )! 5! (3) .5 (1 .5) 3!(5 3)! .3125 x n x n p x p p x n x p          Experiment: Toss 1 coin 5 times in a row. Note number of tails. What’s the probability of 3 tails? @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 31. Poisson Distribution @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 32. Poisson Distribution 1. Number of events that occur in an interval • events per unit — Time, Length, Area, Space 2. Examples  Number of customers arriving in 20 minutes  Number of strikes per year in the India.  Number of defects per lot (group) of DVD’s @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 33. Characteristics of a Poisson Random Variable 1. Consists of counting number of times an event occurs during a given unit of time or in a given area or volume (any unit of measurement). 2. The probability that an event occurs in a given unit of time, area, or volume is the same for all units. 3. The number of events that occur in one unit of time, area, or volume is independent of the number that occur in any other mutually exclusive unit. 4. The mean number of events in each unit is denoted by . @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 34. Poisson Probability Distribution Function  2  p(x) = Probability of x given   = Mean (expected) number of events in unit e = 2.71828 . . . (base of natural logarithm) x = Number of events per unit p x x ( ) !  x   e– (x = 0, 1, 2, 3, . . .) @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 35. Poisson Probability Distribution .0 .2 .4 .6 .8 0 1 2 3 4 5 X P(X) .0 .1 .2 .3 0 2 4 6 8 1 0 X P(X) = 0.5 = 6 Mean Standard Deviation    @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 36. Poisson Distribution Example At a departmental store, 72 customers arrive at store with a rate of 72 customers per hour. What is the probability for arriving 4 customers in 3 minutes? © 1995 Corel Corp. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 37. Poisson Distribution Solution 72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval   - 4 -3.6 ( ) ! 3.6 (4) .1912 4! x e p x x e p      @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 38. Binomial Distribution Thinking Challenge You’re a telemarketer selling service contracts for XYZ co., You’ve sold 20 in your last 100 calls (p = .20). If you call 12 people tonight, what’s the probability of A. No sales? B. Exactly 2 sales? C. At most 2 sales? D. At least 2 sales? @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 39. Binomial Distribution Solution* n = 12, p = .20 A. p(0) = .0687 B. p(2) = .2835 C. p(at most 2) = p(0) + p(1) + p(2) = .0687 + .2062 + .2835 = .5584 D. p(at least 2) = p(2) + p(3)...+ p(12) = 1 – [p(0) + p(1)] = 1 – .0687 – .2062 = .7251 @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 40. © 2011 Pearson Education, Inc Thinking Challenge – Poisson distribution You work in Quality Assurance for an investment firm. A clerk enters 75 words per minute with 6 errors per hour. What is the probability of 0 errors in a 255-word bond transaction? © 1984-1994 T/Maker Co. @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 41. Poisson Distribution Solution: Finding *  75 words/min = (75 words/min)(60 min/hr) = 4500 words/hr  6 errors/hr= 6 errors/4500 words = .00133 errors/word  In a 255-word transaction (interval):  = (.00133 errors/word )(255 words) = .34 errors/255-word transaction @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM
• 42. © 2011 Pearson Education, Inc Poisson Distribution Solution: Finding p(0)*   - 0 -.34 ( ) ! .34 (0) .7118 0! x e p x x e p      @Ravindra Nath Shukla (PhD Scholar) ABV-IIITM