1. © 2012 McGraw-Hill Ryerson Limited1 © 2009 McGraw-Hill Ryerson Limited

2. © 2012 McGraw-Hill Ryerson Limited2 Lind Marchal Wathen Waite

3. © 2012 McGraw-Hill Ryerson Limited3 Explain the difference between discrete and continuous probability distributions. Compute the mean and the standard deviation for a uniform probability distribution. Compute probabilities by using the uniform probability distribution. List the characteristics of the normal probability distribution. Define and calculate z-values. Learning Objectives LO 1 2 3 5 4

4. © 2012 McGraw-Hill Ryerson Limited4 Determine the probability an observation is between two points on a normal probability distribution. Determine the probability an observation is above or below a point on a normal probability distribution. Determine the value of a normally distributed random variable for a given probability. Use the normal probability distribution to approximate the binomial probability distribution. Learning Objectives LO 7 8 9 6

5. © 2012 McGraw-Hill Ryerson Limited5 Difference Between Discrete and Continuous Probability Distribution LO 1

6. © 2012 McGraw-Hill Ryerson Limited6 We consider two families of continuous probability distributions: 1.The uniform probability distribution 2.The normal probability distribution Types of Continuous Probability Distributions LO 1

7. © 2012 McGraw-Hill Ryerson Limited7 Usually result from measuring something. Usually interested in information such as the percentage of data above or below a certain point; or the percentage of data in a certain range. Important continuous probability distribution is the normal probability distribution. Describes the likelihood that a continuous random variable with an infinite number of possible values lies within a specified range. Continuous Probability Distributions LO 1

8. © 2012 McGraw-Hill Ryerson Limited8 The Mean and the Standard Deviation for a Uniform Probability Distribution LO 2

9. © 2012 McGraw-Hill Ryerson Limited9 The mean of a uniform distribution is located in the middle of the interval between the minimum and maximum values. It is computed as: MEAN OF THE UNIFORM DISTRIBUTION LO 2

10. © 2012 McGraw-Hill Ryerson Limited10 The standard deviation describes the dispersion of a distribution. In the uniform distribution, the standard deviation is also related to the interval between the maximum and minimum values. The equation for the uniform probability distribution is: Standard Deviation of The Uniform Distribution LO 2

11. © 2012 McGraw-Hill Ryerson Limited11 Compute Probabilities by Using the Uniform Probability Distribution LO 3

12. © 2012 McGraw-Hill Ryerson Limited12 The time Sue takes to travel from her home to evening class is uniformly distributed between 25 to 35 minutes. 1.Draw a graph of this distribution. 2.Show that the area of this uniform distribution is 1.00. 3.For how long will Sue “typically” have to travel? In other words what is the mean travelling time? What is the standard deviation of the travelling time? 4.What is the probability Sue will travel for more than 33 minutes? 5.What is the probability Sue will travel between 28 to 32 minutes? Example – Uniform Distribution LO 3

13. © 2012 McGraw-Hill Ryerson Limited13 In this case, the random variable is the length of time Sue travels. Time is measured on a continuous scale, and the travel times may range from 25 minutes up to 35 minutes. 1.The graph of the uniform distribution is shown below. The horizontal line is drawn at a height of.1, found by. The range of this distribution is 10 minutes. Solution – Uniform Distribution LO 3

14. © 2012 McGraw-Hill Ryerson Limited14 Solution – Uniform Distribution LO 3

15. © 2012 McGraw-Hill Ryerson Limited15 2.The times Sue must travel is uniform over the interval from 25 minutes to 35 minutes, so in this case a is 25 and b is 35. 3.To find the mean, we use formula (6–1). To find the standard deviation of the travel times, we use formula (6–2). Solution – Uniform Distribution LO 3 Continued

16. © 2012 McGraw-Hill Ryerson Limited16 4.From the area formula: This conclusion is illustrated by the following graph. Solution – Uniform Distribution LO 3 Continued

17. © 2012 McGraw-Hill Ryerson Limited17 5.The area within the distribution for the interval 28 to 32 represents the probability. We can illustrate this probability as follows. Solution – Uniform Distribution LO 3 Continued

18. © 2012 McGraw-Hill Ryerson Limited18 You Try It Out! The life of a light bulb follows a uniform distribution between 10 and 16 months. (a)Draw this uniform distribution. What are the height and base values? (b)Show the total area under the curve is 1.00. (c)Calculate the mean and the standard deviation of this distribution. (d)What is the probability that a light bulb will work for between 12 to 14 months? (e)What is the probability that it will work for less than 11 months? LO 3

19. © 2012 McGraw-Hill Ryerson Limited19 Characteristics of the Normal Probability Distribution LO 4

20. © 2012 McGraw-Hill Ryerson Limited20 1.It is bell-shaped and has a single peak at the centre of the distribution. a)The arithmetic mean, median, and mode are equal. b)The total area under the curve is 1.00; half the area under the normal curve is to the right of this centre point and the other half to the left of it 2.It is symmetrical about the mean. 3.It is asymptotic: The curve gets closer and closer to the X-axis but never actually touches it. a)To put it another way, the tails of the curve extend indefinitely in both directions. 4.The location of a normal distribution is determined by the mean (µ), and the dispersion or spread of the distribution is determined by the standard deviation (σ). Normal Probability Distribution Characteristics LO 4

21. © 2012 McGraw-Hill Ryerson Limited21 Characteristics of a normal distribution are shown graphically below: Characteristics of a Normal Distribution The Normal Distribution Graphically LO 4

22. © 2012 McGraw-Hill Ryerson Limited22 Normal Probability Distributions with Equal Means but Different Standard Deviations Equal Means but Different Standard Deviations LO 4

23. © 2012 McGraw-Hill Ryerson Limited23 Normal Probability Distributions with Different Means but Equal Standard Deviations Different Means but Equal Standard Deviations LO 4

24. © 2012 McGraw-Hill Ryerson Limited24 Normal Probability Distributions with Different Means and Standard Deviations Different Means and Standard Deviations LO 4

25. © 2012 McGraw-Hill Ryerson Limited25 The Standard Normal Probability Distribution LO 5

26. © 2012 McGraw-Hill Ryerson Limited26 The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. It is also called the z distribution. A z-value is the signed distance between a selected value, designated X, and the population mean µ, divided by the population standard deviation, σ. where: X is the value of any particular observation or measurement. is the mean of the distribution. is the standard deviation of the distribution. The Standard Normal Distribution LO 5

27. © 2012 McGraw-Hill Ryerson Limited27 Applications of the Standard Normal Distribution LO 6

28. © 2012 McGraw-Hill Ryerson Limited28 Suppose we wish to compute the probability that boxes of Sugar Yummies have a weight between 283 g and 285.4 g. From Chart 6–3, we know that the box weight of Sugar Yummies follows the normal distribution with a mean of 283 g and a standard deviation of 1.6 g. We want to know the probability or area under the curve between the mean, 283 g, and 285.4 g. The Standard Normal Value & Tables LO 6

29. © 2012 McGraw-Hill Ryerson Limited29 We can also express this problem using probability notation, similar to the style used in the previous chapter: P(283 < mass < 285.4). The Standard Normal Value & Tables LO 6

30. © 2012 McGraw-Hill Ryerson Limited30 Areas under the Normal Curve The Standard Normal Value & Tables Z0.000.010.020.030.040.05 1.30.40320.40490.40660.40820.40990.4115 1.40.41920.42070.42220.42360.42510.4265 1.50.43320.43450.43570.43700.43820.4394 1.60.44520.44630.44740.44840.44950.4505 1.70.45540.45640.45730.45820.45910.4599 1.80.46410.46490.46560.46640.46710.4678 1.90.47130.47190.47260.47320.47380.4744 LO 6

31. © 2012 McGraw-Hill Ryerson Limited31 The values of 285.4 g and a z score of 1.50 are the same distance from µ = 283 and z = 0, respectively. The Standard Normal Value & Tables LO 6

32. © 2012 McGraw-Hill Ryerson Limited32 The monthly returns of certain mutual funds are normally distributed with a mean of $3000 and a standard deviation of $300. What is the z-value for return of $3200 on a mutual fund? For return of $2800 on a mutual fund? Example – Calculating a z-value LO 6

33. © 2012 McGraw-Hill Ryerson Limited33 The z-values for the two X values ($3200 and $2800) are: Solution – Calculating a z-value LO 6

34. © 2012 McGraw-Hill Ryerson Limited34 You Try It Out! From a certain source it has been found that mean for creativity is 12.85 and the standard deviation is 3.66. Convert: a) The raw creativity score of 7 to a z-value. b) The raw creativity score of 14 to a z-value. LO 6

35. © 2012 McGraw-Hill Ryerson Limited35 The Empirical Rule For a symmetrical, bell-shaped frequency distribution: Approximately 68 percent of the observations will lie within ±1 standard deviation of the mean. About 95 percent of the observations will lie within ±2 standard deviations of the mean. Practically all (99.7 percent) will lie within ±3 standard deviations of the mean. LO 6

36. © 2012 McGraw-Hill Ryerson Limited36 The Empirical Rule LO 6

37. © 2012 McGraw-Hill Ryerson Limited37 The Empirical Rule This information is summarized in the following graph. LO 6

38. © 2012 McGraw-Hill Ryerson Limited38 Example – The Empirical Rule A training department has conducted a refresher test for trainee employees. The mean of the score of employees is 34.5 and the standard deviation is 4.8. 1.µ ± 1σ of the employees’ score between what two values? 2. µ ± 2σ of the employees’ score between what two values? 3. µ ± 3σ of the employees’ score between what two values? LO 6

39. © 2012 McGraw-Hill Ryerson Limited39 Solution – The Empirical Rule We can use the results of the Empirical Rule to answer these questions. 1.µ ± 1σ of the employees’ score between 39.3 and 29.7 found by 34.5 ± 1(4.8). 2.µ ± 2σ of the employees’ score between 44.1 and 24.9, found by 34.5 ± 2(4.8). 3.µ ± 3σ of the employees’ score between 48.9 and 20.1, found by 34.5 ± 3(4.8). This information is summarized on the following chart. LO 6

40. © 2012 McGraw-Hill Ryerson Limited40 Solution – The Empirical Rule LO 6 µ +1 σ µ +2σµ +3σ µ - 3σ µ -2σ µ - 1σ µ 39.344.148.929.724.920.134.5

41. © 2012 McGraw-Hill Ryerson Limited41 You Try It Out! Adult women’s heights are normally distributed with μ = 65.5 inches and σ = 2.5 inches. a)μ ± 1σ of the women’s heights lie between what two values? b)μ ± 2σ of the women’s heights lie between what two values? c)μ ± 3σ of the women’s heights lie between what two values? d)What are the median and the modal heights? e)Is the distribution of height symmetrical? LO 6

42. © 2012 McGraw-Hill Ryerson Limited42 Finding Areas Under the Normal Curve LO 7

43. © 2012 McGraw-Hill Ryerson Limited43 The applications of the standard normal distribution involves finding the area in a normal distribution: 1.Between the mean and a selected value, which we identify as x. 2.Beyond x. 3.Between two points on different sides of the mean. 4.Between two points on the same side of the mean. 5.OR finding the value of the observation X when the percent above or below the observation is given. Finding Areas Under the Normal Curve LO 7

44. © 2012 McGraw-Hill Ryerson Limited44 In a gift store, there are different gift items with different prices. The mean value of these items is $1500 and the standard deviation is $155. What is the likelihood of selecting at random an item priced between $1500 and $1800? Example – Between the mean and X LO 7

45. © 2012 McGraw-Hill Ryerson Limited45 Solution – Between the mean and X LO 7

46. © 2012 McGraw-Hill Ryerson Limited46 Solution – Between the mean and X The area under the normal curve between $1500 and $1800 is 0.4732. We estimate that 47.32 percent of the items range between $1500 and $1800. Or, the likelihood of selecting an item priced between $1500 and $1800 is 0.4732. LO 7 Continued

47. © 2012 McGraw-Hill Ryerson Limited47 Value of a Normally Distributed Random Variable for a Given Probability LO 8

48. © 2012 McGraw-Hill Ryerson Limited48 Normally distributed random variable for a given probability We find the probability of selecting an item costing between the mean of $1500 and $1800. This probability is 0.4732. Next, recall that half the area, or probability, is above the mean and half is below. So, the probability of selecting an item costing less than $1500 is 0.5000. Then add the two probabilities: 0.4732 + 0.5000 = 0.9732. So 97.32 percent of the items in the store cost less than $1500. LO 8

49. © 2012 McGraw-Hill Ryerson Limited49 Between the mean and X In Excel LO 8

50. © 2012 McGraw-Hill Ryerson Limited50 We reported that the mean price of the gift items in the store is normally distributed with a mean of $1500 and a standard deviation of $155. What is the probability of selecting an item that costs less than $1290? Example – Beyond X LO 8

51. © 2012 McGraw-Hill Ryerson Limited51 To find the area below –1.35, subtract from 0.50 the area from –1.35 to 0 = 0.50 – 0.4115 = 0.0885 Solution – Beyond X LO 8

52. © 2012 McGraw-Hill Ryerson Limited52 This means that 41.15 percent of the items cost between $1290 and $1500. Further, we can anticipate that 0.885 percent of the items are less than $1290. This information is summarized in the following diagram. Solution – Beyond X LO 8 Continued

53. © 2012 McGraw-Hill Ryerson Limited53 Beyond X In Excel LO 8

54. © 2012 McGraw-Hill Ryerson Limited54 You Try It Out! A telephone company has found that the lengths of its long distance telephone calls are normally distributed, with a mean of 230 seconds and a standard deviation of 60 seconds. a)What is the area under the normal curve between 230 and 255 seconds? Write this area in probability notation. b)What is the area under the normal curve for greater than 255 seconds? Write this area in probability notation. c)Show the details of this problem in a chart. LO 8

55. © 2012 McGraw-Hill Ryerson Limited55 We reported that the mean price of the items in the store is $1200 and the standard deviation is $155. What is the likelihood of selecting an item that costs between $1340 and $1700? Example – Between Two Points on Different Sides LO 8

56. © 2012 McGraw-Hill Ryerson Limited56 For the area between $1240 and the mean of $1500: For the area between the mean of $1500 and $1700: Solution – Between Two Points on Different Sides LO 8

57. © 2012 McGraw-Hill Ryerson Limited57 The area under the curve for a z of –1.03 is 0.0517. The area under the curve for a z of 1.29 is 0.4015. Adding the two areas: 0.3485 + 0.4015 = 0.7500. Thus, the probability of selecting an article whose price is between $1340 and $1700 is 0.7500. Solution – Between Two Points on Different Sides LO 8 Continued

58. © 2012 McGraw-Hill Ryerson Limited58 We reported that the mean price of the gift article in a store is normally distributed with a mean of $1500 and a standard deviation of $155. What is the likelihood of selecting a supervisor whose weekly income is between $1650 and $1750? Example – Between Two Points on The Same Side LO 8

59. © 2012 McGraw-Hill Ryerson Limited59 For x = 1650 For x = 1750 Solution – Between Two Points on The Same Side LO 8

60. © 2012 McGraw-Hill Ryerson Limited60 1.To find the area between 0 and z (or –z), look up the probability directly in the table. 2.To find the area beyond z or (–z), locate the probability of z in the table and subtract that probability from 0.5000. 3.To find the area between two points on different sides of the mean, determine the z-values and add the corresponding probabilities. 4.To find the area between two points on the same side of the mean, determine the z-values and subtract the smaller probability from the larger. Summary of Normal Curve Applications LO 8

61. © 2012 McGraw-Hill Ryerson Limited61 You Try It Out! A psychologist finds that the intelligence quotients of a group of patients are normally distributed, with a mean of 104 and a standard deviation of 26. a)What percent of the patients have intelligence quotients between 89 and 120? Draw a normal curve and shade the desired area on your diagram. b)What percent of the patients have intelligence quotients between 110 and 120? Draw a normal curve and shade the desired area on your diagram. LO 8

62. © 2012 McGraw-Hill Ryerson Limited62 The Flash Bulbs Company wishes to set a minimum duration guarantee on its new range of bulbs. Tests reveal the mean number of minutes is 107 000 with a standard deviation of 3100 minutes and that the distribution of minutes follows the normal distribution. Flash Bulbs Company wants to set the minimum guaranteed number of minutes so that no more than 3 percent of the bulbs will have to be replaced. What minimum guaranteed minutes should Flash Bulbs announce? Example – Finding X When the Percentage is Given LO 8

63. © 2012 McGraw-Hill Ryerson Limited63 The details of the problem are shown in the below diagram, where X represents the minimum guaranteed number of minutes. Inserting the mean and the standard deviation for z: Solution – Finding X When the Percentage is Given LO 8

64. © 2012 McGraw-Hill Ryerson Limited64 Solution – Finding X When the Percentage is Given LO 8 Continued

65. © 2012 McGraw-Hill Ryerson Limited65 Knowing that the distance between µ and X is –1.88σ or z = –1.88, we can now solve for X: Solution – Finding X When the Percentage is Given LO 8 Continued

66. © 2012 McGraw-Hill Ryerson Limited66 Finding X When the Percentage is Given In Excel LO 8

67. © 2012 McGraw-Hill Ryerson Limited67 You Try It Out! An analysis of performance of the employees in a month follows normal distribution. The mean of the distribution is 95 and the standard deviation is 9. The team leader decided to award an A category to the employees whose performance are in the highest 20 percent. What is the dividing point for those employees who earn an A and those earning a B? LO 8

68. © 2012 McGraw-Hill Ryerson Limited68 The Normal Approximation to the Binomial Distribution LO 9

69. © 2012 McGraw-Hill Ryerson Limited69 As n gets large, the binomial distribution gets time- consuming to use. As n increases, a binomial distribution gets closer and closer to a normal distribution. Binomial Distributions for an n of 1, 3, and 20, where p =.50 The Normal Approximation to the Binomial LO 9 n = 1n = 3 n = 20

70. © 2012 McGraw-Hill Ryerson Limited70 When np and n(1 – p) are both at least 5. All binomial criteria are met: 1.There are only two mutually exclusive outcomes to an experiment: a “success” and a “failure” 2.The distribution results from counting the number of successes in a fixed number of trials. 3.Each trial is independent. 4.The probability, p, remains the same from trial to trial. The Normal Approximation to the Binomial – When To Use LO 9

71. © 2012 McGraw-Hill Ryerson Limited71 The value 0.5 subtracted or added, depending on the question, to a selected value when a discrete probability distribution is approximated by a continuous probability distribution. Only four cases may arise. These cases are: 1.For the probability that at least X occur, use the area above (X – 0.5). 2.For the probability that more than X occur, use the area above (X + 0.5). 3.For the probability that X or fewer occur, use the area below (X + 0.5). 4.For the probability that fewer than X occur, use the area below (X – 0.5). The Continuity Correction Factor LO 9

72. © 2012 McGraw-Hill Ryerson Limited72 Suppose the management of the drama theatre found that 70 percent of its new audience return for another drama. For a week in which 80 new (first-time) audiences visited theatre, what is the probability that 60 or more will return for another drama? Example – Normal Approximation of the Binomial LO 9

73. © 2012 McGraw-Hill Ryerson Limited73 Step 1.Find the z corresponding to an X of 59.5 using formula (6–4), and formulas (5–4) and (5–5) for the mean and the variance of a binomial distribution: Solution – Normal Approximation of the Binomial LO 9

74. © 2012 McGraw-Hill Ryerson Limited74 Step 2. Determine the area under the normal curve between a µ of 56 and an X of 59.5. From step 1, we know that the z-value corresponding to 59.5 is 0.85. So we go to Appendix A.1 and read down the left margin to 0.8, and then we go horizontally to the area under the column headed by 0.05. That area is 0.3023. Solution – Normal Approximation of the Binomial LO 9 Continued

75. © 2012 McGraw-Hill Ryerson Limited75 Step 3. Calculate the area beyond 59.5 by subtracting 0.3023 from 0.5000, that is, 0.5000 – 0.3023 = 0.1977. Thus, 0.1977 is the probability that 60 or more first-time drama audience out of 80 will return for another drama. In probability notation: P(audience > 59.5) = 0.5000 – 0.3023 = 0.1977 The details of this problem are shown graphically: Solution – Normal Approximation of the Binomial LO 9 Continued

76. © 2012 McGraw-Hill Ryerson Limited76 You Try It Out! A survey revealed that 85 percent of customers in a certain store bargain while shopping. a)During a period in which 300 customers visited the store and 155 items were sold, what is the probability that the customers bargained? b)During a period in which 300 customers visited the store and 180 items were sold, what is the probability that the customers bargained? LO 9

77. © 2012 McGraw-Hill Ryerson Limited77 Chapter Summary I.The uniform distribution is a continuous probability distribution with the following characteristics. A.It is rectangular in shape. B.The mean and the median are equal. C.It is completely described by its minimum value a and its maximum value b. D.It is also described by the following equation for the region from a to b: [6–3]

78. © 2012 McGraw-Hill Ryerson Limited78 Chapter Summary E.The mean and standard deviation of a uniform distribution are computed as follows: [6–1] [6–2] II.The normal probability distribution is a continuous probability distribution with the following characteristics. A.It is bell-shaped and has a single peak at the centre of the distribution. B.The distribution is symmetrical.

79. © 2012 McGraw-Hill Ryerson Limited79 Chapter Summary C.It is asymptotic, meaning the curve approaches but never touches the X-axis. D.It is completely described by the mean and standard deviation. E.There is a family of normal distributions. 1.Another normal probability distribution is created when either the mean or the standard deviation changes. 2.The area under a normal curve expresses the probability of an outcome.

80. © 2012 McGraw-Hill Ryerson Limited80 Chapter Summary III.The standard normal probability distribution is a particular normal distribution. A.It has a mean of 0 and a standard deviation of 1. B.Any normal distribution can be converted to the standard normal distribution by the following formula. [6–4] C.By standardizing a normal distribution, we can report the distance from the mean in units of the standard deviation.

81. © 2012 McGraw-Hill Ryerson Limited81 Chapter Summary IV.The normal distribution can approximate a binomial distribution under certain conditions. A.np and n(1 – p) must both be at least 5. 1.n is the number of observations. 2.p is the probability of a success. B.The four conditions for a binomial distribution are: 1.There are only two possible outcomes. 2.p remains the same from trial to trial. 3.The trials are independent. 4.The distribution results from a count of the number of successes in a fixed number of trials.

82. © 2012 McGraw-Hill Ryerson Limited82 Chapter Summary C.The mean and variance of a binomial distribution are computed as follows: D.The continuity correction factor of.5 is used to extend the continuous value of X one-half unit in either direction. This correction compensates for estimating a discrete distribution by a continuous distribution.