EMAT0503 Discrete Probability Distribution PDF

Chapter 2 :Random Variable and its Probability Distribution DISCRETE PROBABILITY DISTRIBUTION EMAT0503  RANDOM VARIABLE Variable whose value is determines by a random experiment. DISCRETE PROBABILITY DISTRIBUTION Table of Formula that list...

Chapter 2 :Random Variable and its Probability Distribution DISCRETE PROBABILITY DISTRIBUTION EMAT0503  RANDOM VARIABLE Variable whose value is determines by a random experiment. DISCRETE PROBABILITY DISTRIBUTION Table of Formula that lists the probabilities for each outcome of the random variable, X. Example 1: Flip 3 coins at same time. Let random variable X be number of Heads showing. Solution: Possible Outcomes HHH → 3 Heads THH → 2 Heads HHT → 2 Heads THT → 1 Head HTH → 2 Heads TTH → 1 Head 1 3 3 1 HTT → 1 Head TTT → 0 Heads + + + =1 8 8 8 8 Discrete Probability Distribution 𝑛𝑜. 𝑜𝑓 𝐻𝑒𝑎𝑑𝑠(𝑥) 0 1 2 3 1 3 3 1 𝑃(𝑋 = 𝑥) 8 8 8 8  Example 2: Construct a probability distribution for sum of rolling two dice. Solution: Outcomes of First Die 1 2 3 4 5 6 1 1,1 2,1 3,1 4,1 5,1 6,1 Outcomes of 2 1,2 2,2 3,2 4,2 5,2 6,2 Second Die 3 1,3 2,3 3,3 4,3 5,3 6,3 4 1,4 2,4 3,4 4,4 5,4 6,4 5 1,5 2,5 3,5 4,5 5,5 6,5 6 1,6 2,6 3,6 4,6 5,6 6,6 ෍𝑃 𝑋 = 1 Discrete Probability Distribution Sum of 2 3 4 5 6 7 8 9 10 11 12 dice (x) P(X=x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36  TWO REQUIREMENTS FOR A PROBABILITY DISTRIBUTION 1. The sum of the probabilities of all the events in the sample space must equal to 1; that is, σ𝑃 𝑋 = 1 2. The probability of each event in the sample space must be between or equal to 0 and 1. That is, 0 ≤ 𝑃(𝑋) ≤ 1. Example 3: Determine whether each distribution is a Probability Distributions. Solution: b and c are probability distribution, a and d is not a probability distribution since P(X) cannot be negative. QUANTITATIVE MEASURES OF RANDOM VARIABLE Mean the mean for a sample or population was computed by adding the values and dividing by the total number of values, as shown in these formulas: Sample mean: Population mean: 𝚺𝑿 𝚺𝑿 𝑿= µ= 𝒏 𝑵  FORMULA FOR THE MEAN PROBABILITY DISTRIBUTION The mean of a random variable with a discrete probability distribution is 𝜇 = 𝑋1 ∙ 𝑃 𝑋1 + 𝑋2 ∙ 𝑃 𝑋2 + 𝑋3 ∙ 𝑃 𝑋3 … + 𝑋𝑛 ∙ 𝑃 𝑋𝑛 𝜇 = Σ𝑋 ∙ 𝑃 𝑋 Where 𝑋1 , 𝑋2 , 𝑋3 , … 𝑋𝑛 are the outcomes and 𝑃 𝑋1 , 𝑃 𝑋2 , 𝑃 𝑋3 … 𝑃 𝑋𝑛 are the corresponding probabilities. Note: Σ𝑋 ∙ 𝑃 𝑋 means to sum the product.  QUANTITATIVE MEASURES OF RANDOM VARIABLE Mean Example 4: Suppose from our first example, three coins are tossed at same time, and the number of heads that occurred is recorded. What will be the mean of the number of heads? Heads(x) 0 1 2 3 P(X) 1/8 3/8 3/8 1/8 0 Heads 1 Head 2 Heads 3 Heads 1 3 3 1 3 (0 𝑥 ) + (1 𝑥 ) + (2 𝑥 ) + (3 𝑥 ) = 𝑜𝑟 1.5 8 8 8 8 2 That is, if it were possible to toss the coins many times or an infinite number of times, the average of the number of heads would be 1.5 QUANTITATIVE MEASURES OF RANDOM VARIABLE Rounding Rule for the Mean, Variance, and Standard Deviation for a Probability Distribution The rounding rule for the mean, variance, and standard deviation for variables of a probability distribution is this: The mean, variance, and standard deviation should be rounded to one more decimal place than the outcome X. When fractions are used, they should be reduced to lowest terms. Reference: “Elementary Statistics: A Step by Step Approach”, 9th Ed by Bluman QUANTITATIVE MEASURES OF RANDOM VARIABLE Mean Example 5: Find the mean of the number of spots that appear when die is tossed. Solution: Outcomes(x) 1 2 3 4 5 6 P(X) 1/6 1/6 1/6 1/6 1/6 1/6 µ = 𝜮𝑿 ∙ 𝑷(𝑿) 𝟏 𝟏 𝟏 𝟏 𝟏 𝟏 µ= 𝟏𝒙 + 𝟐𝒙 + 𝟑𝒙 + 𝟒𝒙 + 𝟓𝒙 + 𝟔𝒙 𝟔 𝟔 𝟔 𝟔 𝟔 𝟔 𝟐𝟏 𝟏 µ= = 𝟑 𝒐𝒓 𝟑. 𝟓 𝟔 𝟐  QUANTITATIVE MEASURES OF RANDOM VARIABLE M M Mean F F F M 1. 2. MFFMM MFFMF 3. MFFFM Example 6: F F 4. MFFFF M M F 5. MFMMM In families with five children, find the mean 6. MFMMF M M F F 7. MFMFM number of children who will be girls. 8. MFMFF M M M 9. MMMMM F Solution: M F M F 10. 11. MMMMF MMMFM 12. MMMFF Possible Outcomes = 2 x 2 x 2 x 2 x 2 = 32 M M M 13. MMFMM F 14. MMFMF F M 15. MMFFM F No. of girls F 16. MMFFF (x) 0 1 2 3 4 5 M M 17. FMMMM M F 18. FMMMF Probabilit F M 19. FMMFM y P(X) 1/32 5/32 10/32 10/32 5/32 1/32 M F M 20. 21. FMMFF FMFMM M F 22. FMFMF µ = 𝜮𝑿 ∙ 𝑷(𝑿) F F M F 23. 24. FMFFM FMFFF F 25. FFMMM 𝟏 𝟓 𝟏𝟎 𝟏𝟎 𝟓 𝟏 M M 26. FFMMF µ= 𝟎𝒙 + 𝟏𝒙 + 𝟐𝒙 + 𝟑𝒙 + 𝟒𝒙 + 𝟓𝒙 M F 27. FFMFM 𝟑𝟐 𝟑𝟐 𝟑𝟐 𝟑𝟐 𝟑𝟐 𝟑𝟐 F M F 28. FFMFF 𝟏 F M 29. 30. FFFMM FFFMF µ = 𝟐 𝒐𝒓 𝟐. 𝟓 M F F 31. FFFFM 𝟐 F M F 32. FFFFF  QUANTITATIVE MEASURES OF RANDOM VARIABLE Mean Example 7: The probability distribution shown represents the number of trips of five nights or more that Filipino adults take per year. (That is, 6% do not take any trips lasting five nights or more, 70% take one trip lasting five nights or more per year, etc.) Find the mean. Number of trips (x) 0 1 2 3 4 P(X) 0.06 0.70 0.20 0.03 0.01 Solution: µ = 𝜮𝑿 ∙ 𝑷(𝑿) µ = 𝟎 𝟎. 𝟎𝟔 + 𝟏 𝟎. 𝟕𝟎 + 𝟐 𝟎. 𝟐𝟎 + 𝟑 𝟎. 𝟎𝟑 + (𝟒)(𝟎. 𝟎𝟏) µ = 𝟏. 𝟐𝟑 QUANTITATIVE MEASURES OF RANDOM VARIABLE Variance and Standard Deviation the variance of a probability distribution can be computed by multiplying the square of each outcome by its corresponding probability, summing those products, and subtracting the square of the mean. The formula for the variance of a probability distribution is 2 2 2 𝜎 = Σ[ 𝑋 ∙ 𝑃 𝑋 −𝜇 The standard deviation of a probability distribution is 𝜎= 𝜎 2 𝑜𝑟 𝜎 = Σ[ 𝑋 2 ∙𝑃 𝑋 − 𝜇 2 Reference: “Elementary Statistics: A Step by Step Approach”, 9th Ed by Bluman  QUANTITATIVE MEASURES OF RANDOM VARIABLE Variance and Standard Deviation Example 7: Compute the variance and standard deviation for the probability distribution in example 5. Solution: Outcomes(x) 1 2 3 4 5 6 µ = 𝟑. 𝟓 P(X) 1/6 1/6 1/6 1/6 1/6 1/6 2 2 2 𝜎 = 𝛴[(𝑋 ∙ 𝑃(𝑋)] − 𝜇 2 2 1 2 1 2 1 2 1 2 1 2 1 2 𝜎 = 1 𝑥 + 2 𝑥 + 3 𝑥 + 4 𝑥 + 5 𝑥 + 6 𝑥 − 3.5 6 6 6 6 6 6 𝝈𝟐 = 𝟐. 𝟗𝟏𝟕 𝜎 = 𝜎2 𝜎 = 2.917 = 𝟏. 𝟕𝟎𝟕𝟗 QUANTITATIVE MEASURES OF RANDOM VARIABLE Expectation Another concept related to the mean for a probability distribution is that of expected value or expectation. Expected value is used in various types of games of chance, in insurance, and in other areas, such as decision theory. The expected value of a discrete random variable of a probability distribution is the theoretical average of the variable. The formula is 𝝁 = 𝑬 𝑿 = 𝜮𝑿 ∙ 𝑷(𝑿) The symbol E(X) is used for the expected value. QUANTITATIVE MEASURES OF RANDOM VARIABLE Expectation Example 8: One thousand tickets are sold at $1 each for a color television valued at $350. What is the expected value of the gain if you purchase one ticket? Solution: Win Lose Gain X $349 -$1 Probability P(X) 1/1000 999/1000 𝐄(𝐗) = 𝜮𝑿 ∙ 𝑷(𝑿) 1 999 E X = $349 ∙ + (−$1) ∙ = −$𝟎. 𝟔𝟓 1000 1000  DISCRETE RANDOM VARIABLE BINOMIAL RANDOM VARIABLE A specific type of discrete random variable in binomial experiment is called a binomial random variable. A binomial experiment is comprised of n independent trials. Each trial has only two non-overlapping possible outcomes: “success” or “failure”. The probability of observing a “success”, denoted by p, is constant from one trial to another. A binomial random variable is the number “successes” out of the n independent trials. In functional form, the probability distribution of a binomial random variable, X can be expressed as: 𝒓 𝒏−𝒓 𝑷 𝑿 = 𝒏𝑪𝒓 ∙ 𝒑 ∙ 𝒒  DISCRETE RANDOM VARIABLE BINOMIAL RANDOM VARIABLE 𝒓 𝒏−𝒓 𝑷 𝑿 = 𝒏𝑪𝒓 ∙ 𝒑 ∙ 𝒒 Where: n - number of independent trials in a performance of the random experiment. r – number of success that occur in the n trials. p - the probability of “success”. q - the probability of “failure”. It is defined as q = 1 – p. Mean Variance 𝛍 = 𝒏𝒑 𝟐 𝝈 = 𝒏𝒑𝒒 = 𝒏𝒑(𝟏 − 𝒑) QUANTITATIVE MEASURES OF RANDOM VARIABLE Binomial Random Variable Example 9: Faith had a career battering average of 0.325. What was the probability she would get at least one hit in three official times at bat? Given: 𝑃𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1 = 𝑃1 + 𝑃2 + 𝑃3 n=3 1 3−1 p = 0.325 𝑃𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 1 = 3𝐶1 ∙ 0.325 ∙ 1 − 0.325 2 3−2 q = 1-0.325 + 3𝐶2 ∙ 0.325 ∙ 1 − 0.325 3 + 3𝐶3 ∙ 0.325 ∙ 1 − 0.325 3−3 𝑷 𝒂𝒕 𝒍𝒆𝒂𝒔𝒕 𝟏 = 𝟎. 𝟔𝟗𝟐  QUANTITATIVE MEASURES OF RANDOM VARIABLE Binomial Random Variable Example 9: Faith had a career battering average of 0.325. What was the probability she would get at least one hit in three official times at bat? However, we can get the same answer if use the complement of at least one”. Its complement is “zero hits”. The probability of no hit is: 0 3−0 = 0.308 𝑃0 = 3𝐶0 ∙ 0.325 ∙ 1 − 0.325 Thus, the probability of at least one hit in three times at bat is: 𝑷 𝒂𝒕 𝒍𝒆𝒂𝒔𝒕 𝟏 = 𝟏 − 𝑷𝟎 = 1 − 0.308 = 𝟎. 𝟔𝟗𝟐