2.  5-1 Introduction  5-2 Probability Distributions  5-3 Mean, Variance, and Expectation  5-4 The Binomial Distribution

3.  Construct a probability distribution for a random variable.  Find the mean, variance, and expected value for a discrete random variable.  Find the exact probability for X successes in n trials of a binomial experiment.

4.  Find the mean, variance, and standard deviation for the variable of a binomial distribution.

5.  A variable is defined as a characteristic or attribute that can assume different values.  A variable whose values are determined by chance is called a random variable. 5-2 Probability Distributions

6. Variable Discrete Assume only a specific number of values. Values can be counted.Continuous Assume all values in the interval between two given values. Values can be measured.

7. Number of calls Examples of Discrete Variables Outcomes for die rolling

8. TemperatureTime Examples of Continuous Variables

9.  A discrete probability distribution consists of the values a random variable can assume and the corresponding probabilities of the values.  The probabilities are determined theoretically or by observation.

10. H T H T H T First TossSecond Toss Tossing Two Coins

11.  From the three diagram, the sample space will be represented by HH, HT, TH, and TT.  If X is the random variable for the number of heads, then X assumes the value 0, 1, or 2.

12. Sample SpaceNumber of Heads TT TH HT HH 012012

13. OUTCOME X PROBABILITY P(X) 01/4 12/4 21/4

14. 210 1 0. 5 0.25 NUMBER OF HEADS P R O B A B I L I T Y Experiment: Toss Two Coins

15.  The mean of the random variable of a probability distribution is where X 1, X 2, …, X n are outcomes and P(X 1 ), P(X 2 ), …, P(X n ) are the corresponding probabilities. µ = X 1 P(X 1 ) + X 2 P(X 2 ) + … + X n P(X n ) = X P(X) 5-3 Mean, Variance, and Expectation for Discrete Variable

16. Example 1  Find the mean of the number of spots that appear when a die is tossed. The probability distribution is given below. X123456 P(X)1/6

17. µ = X P(X) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 21/6 = 3.5 The theoretical mean is 3.5.

18. Example 2  In a family with two children, find the mean number of children who will be girls. The probability distribution is given below. X012 P(X)1/41/21/4 5-3 Mean for Discrete Variable

19. µ = X P(X) = 0(1/4) + 1(1/2) + 2(1/4) = 1 The average number of girls in a two-child family is 1. 5-3 Mean for Discrete Variable

20.  The variance of a probability distribution is found by multiplying the square of each outcome by its corresponding probability, summing these products, and subtracting the square of the mean. 5-3 Variance for Discrete Variable

21. The variance of a probability distribution The standard deviation of a probability distribution 5-3 Variance for Discrete Variable

22. Example  Five balls numbered 0, 2, 4, 6, and 8 are placed in a bag. After the balls are mixed, one is selected, its number is noted, and then it is replaced. If the experiment is repeated many times, find the variance and standard deviation of the numbers on the balls. 5-3 Variance for Discrete Variable

23. Number on ball X 02468 Probability P(X) 1/5 5-3 Variance for Discrete Variable

24.  = (0)(1/5) + (2)(1/5) + (4)(1/5) + (6)(1/5) + (8)(1/5) = 4.0  X 2 P(X) = (0 2 )(1/5) + (2 2 )(1/5) + (4 2 )(1/5) + (6 2 )(1/5) + (8 2 )(1/5) = 0 + 4/5 + 16/5 + 36/15 + 64/5 = 120/5 = 24 5-3 Variance for Discrete Variable

25. XP(X)P(X)XP(X)X 2P(X) 00.2 0 0 2 0.4 0.8 40.2 0.8 3.2 60.2 1.2 7.2 80.2 1.6 12.8 4.0 24.0 5-3 Variance for Discrete Variable

27. The probability distribution for the number of customers one day at the Sunrise Coffee Shop is shown as below. Find the mean, variance, and standard deviation of the distribution. Number of customers X 4041424344 Probability P(X) 0.100.200.370.210.12

28.  The expected values of a discrete random variable of a probability distribution is the theoretical average of the variable. The symbol of E ( X ) is represented expected value. 5-3 Expectation µ = E(X) =  X P(X)

29. Example 1 One thousand tickets are sold at RM 1 each for four prizes of RM100, RM50, RM25, and RM10. What is the expected value of the gain if a person purchases one ticket? 5-3 Expectation

30. Gain XRM 99RM 49RM 24RM 9- RM 1 Probability P(X) 1/1000 996/1000 E(X) =  X P(X) = RM99 1/1000 + RM49 1/1000 + RM24 1/1000 + RM9 1/1000 + (-RM1) 996/1000 = -RM0.815

31. Example 2 A lottery offers one RM1000 prize, one RM500 prize, and five RM100 prizes. One thousand tickets are sold at RM3 each. Find the expectation of the gain if a person purchases one ticket? 5-3 Expectation

32. Gain XRM 997RM 497RM 97- RM 3 Probability P(X) 1/1000 5/1000993/1000 E(X) =  X P(X) = RM997 1/1000 + RM497 1/1000 + RM97 5/1000 + (-RM3) 993/1000 = -RM1

33. Binomial experiment is a probability experiment that satisfies the following four requirements: 1. Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. These outcomes can be considered as either success or failure. 2. There must be fixed number of trials. 5-4 The Binomial Distribution

34. 3. The outcomes of each trial must be independent of each other. 4. The probability of a success must remain the same for each trial. 5-4 The Binomial Distribution

35.  The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called as binomial distribution. 5-4 The Binomial Distribution

36. Notation for the Binomial Distribution P(S) = p, probability of success P(F) = 1 - p = q, probability of failure n = number of trials X = number of successes

37.  In a binomial experiment, the probability of exactly X successes in n trials is 5-4 Binomial Probability Formula PX n nXX pq Xn X () ! ()!!   

38. Example 1  If a student randomly guesses at five multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices. 5-4 Binomial Probability

39. Solution n = 5, X = 3, and p = 1/5. Then, P (3) = [5!/(5-3)!3!] (1/5) 3 (4/5) 2 = 0.0512 ≈ 0.05 5-4 Binomial Probability

40. Example 2  A survey from Teenage Research Unlimited found that 30% of teenage consumers received their spending money from part-time jobs. If five teenagers are selected at random, find the probability that at least three of them will have part-time jobs. 5-4 Binomial Probability

41. Solution n = 5, X = 3, 4, 5 and p = 0.3. Then, P ( X ≥3) = P (3) + P (4) + P (5) = 0.132 + 0.028 + 0.002 = 0.162 5-4 Binomial Probability

42. P (4) = [5!/(5-4)!4!] (0.3) 4 (0.7) 1 = 0.028 P (5) = [5!/(5-5)!5!] (0.3) 5 (0.7) 0 = 0.002 5-4 Binomial Probability P(3) = [5!/(5-3)!3!] (0.3) 3 (0.7) 2 = 0.132

43. Example 3  Public Opinion reported that 5% of Malaysians are afraid of being alone in a house at night. If a random sample of 20 Malaysians is selected, find the probability that exactly 5 people in the sample who are afraid of being alone at night. 5-4 Binomial Probability

44. Solution n = 20, p = 0.05, X = 5 By using formula, P (5) = [20!/(20-5)!5!] (0.05) 5 (0.95) 15 = 0.002 5-4 Binomial Probability

45. From the table, P(5) = 0.002 nxp 0.050.10.20.30.40.5 200 1 2 3 4 50.002

46. If 90% of all people between the ages of 30 and 50 drive a car, find these probabilities for a sample of 20 people in that age group. a) Exactly 20 drive a car. b) At least 17 drive a car. c) At most 18 drive a car.

47. Mean µ = n p Variance = n p q Standard deviation = n p q 5-4 Mean, Variance, Standard Deviation for the Binomial Distribution

48. Example  A coin is tossed four times. Find the mean, variance, and standard deviation of the number of heads that will be obtained. 5-4 Mean, Variance, Standard Deviation for the Binomial Distribution

49. Solution: n = 4, p = 1/2, q = 1/2 µ = n p = (4)(1/2) = 2 Variance = n p q = (4)(1/2)(1/2) = 1 Standard deviation = n p q = 1 = 1 5-4 Mean, Variance, Standard Deviation for the Binomial Distribution

50. If 80% of the applicants are able to pass a driver’s proficiency road test, find the mean, variance, and standard deviation of the number of people who pass the test in a sample of 300 applicants.