SLIDES PREPARED By Lloyd R. Jaisingh Ph.D. Morehead State University Morehead KY

1. 9-1 SLIDES PREPARED By Lloyd R. Jaisingh Ph.D. Morehead State University Morehead KY

2. 9-2 Chapter 9 The Normal Probability Distribution

3. 9-3 Outline Do I Need to Read This Chapter? You should read the Chapter if you would like to learn about: 9-1 The Normal Distribution 9-2 Properties of the Normal Distribution 9-3 The Standard Normal Distribution 9-4 Applications of the Normal Distribution

4. 9-4 Objectives Identify the properties of the normal distribution. Find the area under the standard normal distribution given various z values. Find probabilities for a normally distributed variable by transforming it into a standard normal variable.

5. 9-5 Getting Started Here we will focus on a special continuous variable that can assume any value in the interval (-?, +?). That is, the variable can assume any value on the real line. The distribution for this variable is called the normal distribution.

6. 9-6 Getting Started It was originally called the Gaussian distribution in honor of Carl Friedrich Gauss, who in 1833 published a work describing it. This distribution is considered the most important probability distribution in all of statistics.

7. 9-7 Getting Started - Carl Friedrich Gauss on German postage stamps

8. 9-8 9-1 The Normal Distribution The following data set summarizes the chest sizes of Scottish militiamen in the early 19th century. Chest sizes are measured in inches, and each observation reports the number of soldiers with that chest size. Source: http://lib.stat.cmu.edu/DASL/Datafiles/ MilitiamenChests.html

9. 9-9 9-1 The Normal Distribution

10. 9-10 9-1 The Normal Distribution

11. 9-11 9-1 The Normal Distribution The normal distribution can be viewed as the limiting distribution of a binomial random variable. That is, in a binomial experiment, if we use a fixed probability of success p, we can analyze what happens as the number of trials n increases.

12. 9-12 9-1 The Normal Distribution To visualize what happens, we can construct histograms for a fixed p and increasingly large n. The following displays show histograms for n = 5, 10, 25, and 50 with p = 0.1 and 0.5. Superimposed on the histograms are smooth curves which show the shapes of the distributions.

13. 9-13 9-1 The Normal Distribution

14. 9-14 9-1 The Normal Distribution

15. 9-15 9-1 The Normal Distribution

16. 9-16 9-1 The Normal Distribution

17. 9-17 9-1 The Normal Distribution

18. 9-18 As mentioned earlier, the normal distribution is considered the most important probability distribution in all of statistics. It is used to describe the distribution of many natural phenomena, such as the height of a person, IQ scores, weight, blood pressure etc. 9-1 The Normal Distribution

19. 9-19 The mathematical equation for the normal distribution is given below: 9-2 Properties of the Normal Distribution

20. 9-20 9-2 Properties of the Normal Distribution

21. 9-21 9-2 Properties of the Normal Distribution

22. 9-22 9-2 Properties of the Normal Distribution

23. 9-23 9-2 Properties of the Normal Distribution

24. 9-24 NOTE: These normal curves have similar shapes, but are located at different points along the x-axis. Also, the larger the standard deviation, the more spread out is the distribution, and the curves are symmetrical about the mean value. Explanation of the term – normal distribution: A normal distribution is a continuous, symmetrical, bell-shaped distribution of a normal random variable. 9-2 Properties of the Normal Distribution

25. 9-25 Summary of the Properties of the normal Distribution: The curve is continuous. The curve is bell-shaped. The curve is symmetrical about the mean. The mean, median, and mode are located at the center of the distribution and are equal to each other. 9-2 Properties of the Normal Distribution

26. 9-26 Summary of the Properties of the normal Distribution continued: The curve is unimodal (single mode) The curve never touches the x-axis. The total area under the normal curve is equal to 1. 9-2 Properties of the Normal Distribution

27. 9-27 A very important property of any normal distribution is that within a fixed number of standard deviations from the mean, all normal distributions have the same fraction of their probabilities. We will illustrate for for ?1?, ?2?, and ?3? from the mean ?. 9-2 Properties of the Normal Distribution

28. 9-28 One-sigma rule: Approximately 68% of the data values should lie within one standard deviation of the mean. That is, regardless of the shape of the normal distribution, the probability that a normal random variable will be within one standard deviation of the mean is approximately equal to 0.68. The next slide illustrates this. 9-2 The Empirical Rule Revisited

29. 9-29 9-2 The Empirical Rule Revisited

30. 9-30 Two-sigma rule: Approximately 95% of the data values should lie within two standard deviations of the mean. That is, regardless of the shape of the normal distribution, the probability that a normal random variable will be within two standard deviations of the mean is approximately equal to 0.95. The next slide illustrates this. 9-2 The Empirical Rule Revisited

31. 9-31 9-2 The Empirical Rule Revisited

32. 9-32 Three-sigma rule: Approximately 99.7% of the data values should lie within three standard deviations of the mean. That is, regardless of the shape of the normal distribution, the probability that a normal random variable will be within three standard deviations of the mean is approximately equal to 0.997. The next slide illustrates this. 9-2 The Empirical Rule Revisited

33. 9-33 9-2 The Empirical Rule Revisited

34. 9-34 Quick Tips: The total area under the normal curve is equal to 1. The probability that a normal random variable is equal to a given discrete value is always zero, since the normal random variable is continuous. The probability that a normal random variable is between two values is given by the area under the normal curve between the two given values and the horizontal axis.

35. 9-35 Illustration for P(x1 ? X ? x2)

36. 9-36 Since each normally distributed random variable has its own mean and standard deviation, the shape and location of normal curves will vary. Thus, one would have to have information on the areas for all normal distributions. Impractical. Therefore, we use the information for a special normal distribution called the standard normal distribution to simplify this situation. 9-3 The Standard Normal Distribution

37. 9-37 Explanation of the term – standard normal distribution: The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. Any normal random variable can be converted to a standard normal random variable by computing the corresponding z-score. 9-3 The Standard Normal Distribution

38. 9-38 The z-score is computed from the following formula: In the equation, x is the value of a normal random variable X with mean ? and standard deviation ?. 9-3 The Standard Normal Distribution

39. 9-39 Quick Tips: The z-score is normally distributed with a mean of 0 and a standard deviation of 1. Recall that the z-score gives the number of standard deviations a specific value is above or below the mean.

40. 9-40 Extensive tables can be constructed for the standard normal random variable to aid in finding areas under the standard normal curve. 9-3 The Standard Normal Distribution

41. 9-41 Usually, standard normal tables give the area between the mean of 0 and a z-score value as shown below. 9-3 The Standard Normal Distribution

42. 9-42 A sample portion of a table using four decimal places is shown below. 9-3 The Standard Normal Distribution

43. 9-43 For example,if we want to find the area between z = 0 and z = 0.92, we find z = 0.9 in the first column, then look for z = 0.02 in the first row. Where the corresponding column and row intersect gives the value 0.3212. This is equivalent to finding P(0 ? Z ? 0.92). The area is depicted on the next slide. 9-3 The Standard Normal Distribution

44. 9-44 9-3 The Standard Normal Distribution

45. 9-45 Quick Tips: In solving problems relating to the standard normal distribution, it may be helpful if you use the following procedure: Write out the equivalent probability statement. Draw a normal curve. Shade in the desired area. Use the tables to find the shaded area.

46. 9-46 NOTE: More extensive tables are given in the Appendix of the text. Example 1: Find the area under the standard normal curve between z = 0 and z = 2.0. Solution: This is equivalent to finding P(0 ? z ? 2.0). From the standard normal tables, for z = 2.0, the corresponding value is 0.4772. Thus, P(0 ? z ? 2.0) = 0.4772. This area is shown in the next slide. 9-3 The Standard Normal Distribution

47. 9-47 Example 1 Continued: 9-3 The Standard Normal Distribution

48. 9-48 Example 2: Find the area under the standard normal curve between z = 0 and z = -1.8. Solution: This is equivalent to finding P(-1.8 ? z ? 0). From the symmetry of the distribution, P(-1.8 ? z ? 0) = P(0 ? z ? 1.8). From the standard normal tables, for z = 1.8, the corresponding value is 0.4641. Thus, P(-1.8 ? z ? 0) = 0.4641. This area is shown in the next slide. 9-3 The Standard Normal Distribution

49. 9-49 Example 2 Continued: 9-3 The Standard Normal Distribution

50. 9-50 Example 3: Find the area under the standard normal curve to the right of z = 1.5. Solution: This is equivalent to finding P(z ? 1.5). From the standard normal tables, for z = 1.5, the corresponding value is 0.4332. But this is the area between 0 and 1.5. Recall the total area to the right of 0 is 0.5. Thus, P(z ? 1.5) = 0.5 – 0.4332 = 0.0668 . This area is shown in the next slide. 9-3 The Standard Normal Distribution

51. 9-51 Example 3 Continued: 9-3 The Standard Normal Distribution

52. 9-52 Example 4: Find the area under the standard normal curve to the left of z = -1.75. Solution: This is equivalent to finding P(z ? -1.75). From the symmetry of the distribution, P(z ? -1.75) = P(z ? 1.75). This is a similar problem to Example 3. From the standard normal tables, for z = 1.75, the corresponding value is 0.4599. 9-3 The Standard Normal Distribution

53. 9-53 Example 4 Continued: But this is the area between 0 and 1.75. Recall the total area to the right of 0 is 0.5. Thus, P(z ? 1.75) = 0.5 – 0.4599 = 0.0401. This area is shown in the next slide. 9-3 The Standard Normal Distribution

54. 9-54 Example 4 Continued: 9-3 The Standard Normal Distribution

55. 9-55 Example 5: Find the area under the standard normal curve between z = 1.5 and z = 2.5. Solution: This is equivalent to finding P(1.5 ? z ? 2.5). This is equivalent to finding P(0 ? z ? 2.5) - P(0 ? z ? 1.5). From the standard normal tables, for z = 2.5, the corresponding value is 0.4938 and for z = 1.5, the corresponding value is 0.4332. 9-3 The Standard Normal Distribution

56. 9-56 Example 5 Continued: Thus, the required area for P(1.5 ? z ? 2.5) = P(0 ? z ? 2.5) - P(0 ? z ? 1.5) = 0.4938 – 0.4332 = 0.0606. This area is shown in the next slide. 9-3 The Standard Normal Distribution

57. 9-57 Example 5 Continued: 9-3 The Standard Normal Distribution

58. 9-58 Example 6: Find the area under the standard normal curve between z = -2.78 and z = -1.66. Solution: This is equivalent to finding P(-2.78 ? z ? -1.66). Because of the symmetry of the distribution, this is equivalent to finding P(1.66 ? z ? 2.78) = P(0 ? z ? 2.78) - P(0 ? z ? 1.66). 9-3 The Standard Normal Distribution

59. 9-59 Example 6 Continued: From the standard normal tables, for z = 2.78, the corresponding area is 0.4973 and for z = 1.66, the corresponding area is 0.4515. Thus, the required area for P(-2.78 ? z ? -1.66) = P(1.66 ? z ? 2.78) =P(0 ? z ? 2.78) - P(0 ? z ? 1.66) = 0.4973 – 0.4515 = 0.0458. This area is shown in the next slide. 9-3 The Standard Normal Distribution

60. 9-60 Example 6 Continued: 9-3 The Standard Normal Distribution

61. 9-61 Example 7: Find the area under the standard normal curve between z = -2.79 and z = 1.71. Solution: This is equivalent to finding P(-2.79 ? z ? 1.71). Now, P(-2.79 ? z ? 1.71) = P(-2.79 ? z ? 0) + P(0 ? z ? 1.71) = P(0 ? z ? 2.79) + P(0 ? z ? 1.71) = 0.4974 + 0.4564 = 0.9538. This area is shown in the next slide. 9-3 The Standard Normal Distribution

62. 9-62 Example 7 Continued: 9-3 The Standard Normal Distribution

63. 9-63 Example 8: Find the area under the standard normal curve to the right of z = -2.79. Solution: This is equivalent to finding P(z ? -2.79). Now, P(z ? -2.79) = 0.4974 + 0.5 = 0.9974. This area is shown in the next slide. 9-3 The Standard Normal Distribution

64. 9-64 Example 8 Continued: 9-3 The Standard Normal Distribution

65. 9-65 Quick Tips (again): Recall, in solving problems relating to the standard normal distribution, it may be helpful if you use the following procedure: Write out the equivalent probability statement. Draw a normal curve. Shade in the desired area. Use the tables to find the shaded area.

66. 9-66 To solve problems involving any normal random variable, we first will have to transform the original normal variable into a standard normal random variable by using the z-score transformation formula given below: 9-4 Applications of the Normal Distribution

67. 9-67 Example 9: If IQ scores are normally distributed with a mean of 100 and a standard deviation of 5, what is the probability that a person chosen at random will have an IQ score greater than 110? Solution: Let X = IQ score. Then we need to find P(X > 110). The equivalent z-score = (110 – 100)/5 =2. Thus, P(X > 110) = P(Z > 2) = 0.5 – 0.4772 = 0.0228. This area is shown in the next slide. 9-4 Applications of the Normal Distribution

68. 9-68 Example 9 Continued: 9-3 The Standard Normal Distribution

69. 9-69 Example 10: Suppose family incomes in a town are normally distributed with a mean of $1,200 and a standard deviation of $600 per month. What is the probability that a family has an income between $1,400 and $2,250? Solution: Let X = monthly income. Then we need to find P(1400 ? X ? 2250). The equivalent z-scores are (1400 – 1200)/600 = 0.33 and (2250 – 1200)/600 = 1.75. 9-4 Applications of the Normal Distribution

70. 9-70 Example 10 continued: Thus, P(1400 ? X ? 2250) = P(0.33 ? z ? 1.75) = P(0 ? z ? 1.75) - P(0 ? z ? 0.33) = 0.4599 – 0.1293 = 0.3306 This area is shown in the next slide. 9-4 Applications of the Normal Distribution

71. 9-71 Example 10 Continued: 9-3 The Standard Normal Distribution

72. 9-72 Example 11: A four-year college will accept any student ranked in the top 60% on a national examination. If the test scores are normally distributed with a mean of 500 and a standard deviation of 100, what is the cut off score for acceptance? Solution: Let X = test score and let x0 be the cut off score. Then we need to find x0 such that P(X > x0) = 0.6. 9-4 Applications of the Normal Distribution

73. 9-73 Example 11 continued: What we need to do is to find the corresponding z-score, say z0 such that an area of 0.6 is to the right of z0. The next slide shows the desired area. 9-4 Applications of the Normal Distribution

74. 9-74 Example 11 Continued: 9-3 The Standard Normal Distribution

75. 9-75 Example 11 continued: Using the area of 0.1, we have from the standard normal tables a corresponding z-score of -0.25 (negative since z0 will be to the left of 0). Thus z0 =(x0 -?)/? or -0.25 = (x0 –500)/100. Solving, gives x0 = 475. That is, the minimum score the college will accept is 475. 9-4 Applications of the Normal Distribution

76. 9-76 Quick Tips: Recall, in solving application problems relating to the normal distribution, it may be helpful if you use the following procedure: Define the appropriate random variable. Write out the equivalent probability statement in terms of this variable. Write out an equivalent transformed probability statement using the z-score. Draw a normal curve. Shade in the desired area. Use the tables to find the shaded area.