1. Understanding Basic Statistics Fourth Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Chapter Seven Normal Curves and Sampling Distributions

2. Copyright © Houghton Mifflin Company. All rights reserved.7 | 2 Properties of The Normal Distribution The curve is bell-shaped with the highest point over the mean, μ.

3. Copyright © Houghton Mifflin Company. All rights reserved.7 | 3 The curve is symmetrical about a vertical line through μ. Properties of The Normal Distribution

4. Copyright © Houghton Mifflin Company. All rights reserved.7 | 4 The curve approaches the horizontal axis but never touches or crosses it. Properties of The Normal Distribution

5. Copyright © Houghton Mifflin Company. All rights reserved.7 | 5 The transition points between cupping upward and downward occur above μ + σ and μ – σ. Properties of The Normal Distribution

6. Copyright © Houghton Mifflin Company. All rights reserved.7 | 6 The Empirical Rule Applies to any symmetrical and bell- shaped distribution

7. Copyright © Houghton Mifflin Company. All rights reserved.7 | 7 The Empirical Rule Approximately 68% of the data values lie within one standard deviation of the mean.

8. Copyright © Houghton Mifflin Company. All rights reserved.7 | 8 Approximately 95% of the data values lie within two standard deviations of the mean. The Empirical Rule

9. Copyright © Houghton Mifflin Company. All rights reserved.7 | 9 Almost all (approximately 99.7%) of the data values will be within three standard deviations of the mean. The Empirical Rule

10. Copyright © Houghton Mifflin Company. All rights reserved.7 | 10 Application of the Empirical Rule The life of a particular type of light bulb is normally distributed with a mean of 1100 hours and a standard deviation of 100 hours. What is the probability that a light bulb of this type will last between 1000 and 1200 hours? Approximately 68%

11. Copyright © Houghton Mifflin Company. All rights reserved.7 | 11 Standard Score The z value or z score tells the number of standard deviations between the original measurement and the mean. The z value is in standard units.

12. Copyright © Houghton Mifflin Company. All rights reserved.7 | 12 Formula for z score

13. Copyright © Houghton Mifflin Company. All rights reserved.7 | 13 x Values and Corresponding z Values

14. Copyright © Houghton Mifflin Company. All rights reserved.7 | 14 Calculating z scores The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. Convert 21 minutes to a z score.

15. Copyright © Houghton Mifflin Company. All rights reserved.7 | 15 Mean delivery time = 25 minutes Standard deviation = 2 minutes Convert 29.7 minutes to a z score. Calculating z scores

16. Copyright © Houghton Mifflin Company. All rights reserved.7 | 16 Raw Score A raw score is the result of converting from standard units (z scores) back to original measurements, x values. Formula: x = z  + 

17. Copyright © Houghton Mifflin Company. All rights reserved.7 | 17 Mean delivery time = 25 minutes Standard deviation = 2 minutes Interpret a z score of 1.60. The delivery time is 28.2 minutes. Interpreting z-scores

18. Copyright © Houghton Mifflin Company. All rights reserved.7 | 18 Standard Normal Distribution:  = 0  = 1 Any x values are converted to z scores.

19. Copyright © Houghton Mifflin Company. All rights reserved.7 | 19 Importance of the Standard Normal Distribution: Areas will be equal. Any Normal Curve:

20. Copyright © Houghton Mifflin Company. All rights reserved.7 | 20 Areas of a Standard Normal Distribution Appendix Table 3 Pages A6 - A7

21. Copyright © Houghton Mifflin Company. All rights reserved.7 | 21 Use of the Normal Probability Table Appendix Table 3 is a left-tail style table. Entries give the cumulative areas to the left of a specified z.

22. Copyright © Houghton Mifflin Company. All rights reserved.7 | 22 To Find the area to the Left of a Given z score Find the row associated with the sign, units and tenths portion of z in the left column of Table 3. Move across the selected row to the column headed by the hundredths digit of the given z.

23. Copyright © Houghton Mifflin Company. All rights reserved.7 | 23 Find the area to the left of z = – 2.84

24. Copyright © Houghton Mifflin Company. All rights reserved.7 | 24 To find the area to the left of z = – 2.84

25. Copyright © Houghton Mifflin Company. All rights reserved.7 | 25 To find the area to the left of z = – 2.84

26. Copyright © Houghton Mifflin Company. All rights reserved.7 | 26 The area to the left of z = – 2.84 is.0023

27. Copyright © Houghton Mifflin Company. All rights reserved.7 | 27 Use Table 3 of the Appendix directly. To Find the Area to the Left of a Given Negative z Value:

28. Copyright © Houghton Mifflin Company. All rights reserved.7 | 28 Use Table 3 of the Appendix directly. To Find the Area to the Left of a Given Positive z Value:

29. Copyright © Houghton Mifflin Company. All rights reserved.7 | 29 Subtract the area to the left of z from 1.0000. To Find the Area to the Right of a Given z Value:

30. Copyright © Houghton Mifflin Company. All rights reserved.7 | 30 Use the symmetry of the normal distribution. Area to the right of z = area to left of –z. Alternate Way To Find the Area to the Right of a Given Positive z Value:

31. Copyright © Houghton Mifflin Company. All rights reserved.7 | 31 Subtract area to left of z 1 from area to left of z 2. (When z 2 > z 1.) To Find the Area Between Two z Values

32. Copyright © Houghton Mifflin Company. All rights reserved.7 | 32 Convention for Using Table 3 Treat any area to the left of a z value smaller than  3.49 as 0.000 Treat any area to the left of a z value greater than 3.49 as 1.000

33. Copyright © Houghton Mifflin Company. All rights reserved.7 | 33 a.P( z < 1.64 ) = __________ b. P( z < - 2.71 ) = __________.9495.0034 Use of the Normal Probability Table

34. Copyright © Houghton Mifflin Company. All rights reserved.7 | 34 Use of the Normal Probability Table c.P(0 < z < 1.24) = ______ d. P(0 < z < 1.60) = _______ e.P(  2.37 < z < 0) = ______.3925.4452.4911

35. Copyright © Houghton Mifflin Company. All rights reserved.7 | 35 Use of the Normal Probability Table f.P(  3 < z < 3 ) = ________ g. P(  2.34 < z < 1.57 ) = _____ h.P( 1.24 < z < 1.88 ) = _______.9974.9322.0774

36. Copyright © Houghton Mifflin Company. All rights reserved.7 | 36 Use of the Normal Probability Table i. P(  2.44 < z <  0.73 ) = _______ j.P( z > 2.39 ) = ________ k.P( z >  1.43 ) = __________.0084.2254.9236

37. Copyright © Houghton Mifflin Company. All rights reserved.7 | 37 To Work with Any Normal Distributions Convert x values to z values using the formula: Use Table 3 of the Appendix to find corresponding areas and probabilities.

38. Copyright © Houghton Mifflin Company. All rights reserved.7 | 38 Rounding Round z values to the hundredths positions before using Table 3. Leave area results with four digits to the right of the decimal point.

39. Copyright © Houghton Mifflin Company. All rights reserved.7 | 39 Application of the Normal Curve The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. If you order a pizza, find the probability that the delivery time will be: a.between 25 and 27 minutes. a. __________ b.less than 30 minutes.b. __________ c.less than 22.7 minutes.c. __________.3413.9938.1251

40. Copyright © Houghton Mifflin Company. All rights reserved.7 | 40 Inverse Normal Probability Distribution Finding z or x values that correspond to a given area under the normal curve

41. Copyright © Houghton Mifflin Company. All rights reserved.7 | 41 Look up area A in body of Table 3 and use corresponding z value. Inverse Normal Left Tail Case

42. Copyright © Houghton Mifflin Company. All rights reserved.7 | 42 Look up the number 1 – A in body of Table 3 and use corresponding z value. Inverse Normal Right Tail Case:

43. Copyright © Houghton Mifflin Company. All rights reserved.7 | 43 Look up the number (1 – A)/2 in body of Table 3 and use corresponding ± z value. Inverse Normal Center Case:

44. Copyright © Houghton Mifflin Company. All rights reserved.7 | 44 Using Table 3 for Inverse Normal Distribution Use the nearest area value rather than interpolating. When the area is exactly halfway between two area values, use the z value exactly halfway between the z values of the corresponding table areas.

45. Copyright © Houghton Mifflin Company. All rights reserved.7 | 45 When the area is exactly halfway between two area values When the z value corresponding to an area is smaller than  2, use the z value corresponding to the smaller area. When the z value corresponding to an area is larger than 2, use the z value corresponding to the larger area.

46. Copyright © Houghton Mifflin Company. All rights reserved.7 | 46 Find the indicated z score:

47. Copyright © Houghton Mifflin Company. All rights reserved.7 | 47 Find the indicated z score: z = _______ – 2.57

48. Copyright © Houghton Mifflin Company. All rights reserved.7 | 48 z = _______ 2.33 Find the indicated z score:

49. Copyright © Houghton Mifflin Company. All rights reserved.7 | 49 Find the indicated z scores: z = ____–z = _____ –1.231.23

50. Copyright © Houghton Mifflin Company. All rights reserved.7 | 50 Find the indicated z scores: ± z =__________ ± 2.58

51. Copyright © Houghton Mifflin Company. All rights reserved.7 | 51 ± z = ________ ± 1.96 Find the indicated z scores:

52. Copyright © Houghton Mifflin Company. All rights reserved.7 | 52 Application of Determining z Scores The Verbal SAT test has a mean score of 500 and a standard deviation of 100. Scores are normally distributed. A major university determines that it will accept only students whose Verbal SAT scores are in the top 4%. What is the minimum score that a student must earn to be accepted?

53. Copyright © Houghton Mifflin Company. All rights reserved.7 | 53 The cut-off score is 1.75 standard deviations above the mean. Application of Determining z Scores

54. Copyright © Houghton Mifflin Company. All rights reserved.7 | 54 The cut-off score is 500 + 1.75(100) = 675. Mean = 500 standard deviation = 100 Application of Determining z Scores

55. Copyright © Houghton Mifflin Company. All rights reserved.7 | 55 Introduction to Sampling Distributions

56. Copyright © Houghton Mifflin Company. All rights reserved.7 | 56 Review of Statistical Terms Population Sample Parameter Statistic

57. Copyright © Houghton Mifflin Company. All rights reserved.7 | 57 Population the set of all measurements or counts (either existing or conceptual) under consideration

58. Copyright © Houghton Mifflin Company. All rights reserved.7 | 58 Sample a subset of measurements from a population

59. Copyright © Houghton Mifflin Company. All rights reserved.7 | 59 Parameter a numerical descriptive measure of a population

60. Copyright © Houghton Mifflin Company. All rights reserved.7 | 60 Statistic a numerical descriptive measure of a sample

61. Copyright © Houghton Mifflin Company. All rights reserved.7 | 61 We use a statistic to make inferences about a population parameter.

62. Copyright © Houghton Mifflin Company. All rights reserved.7 | 62 Some Common Statistics and Corresponding Parameters

63. Copyright © Houghton Mifflin Company. All rights reserved.7 | 63 Principal Types of Inferences Estimation: estimate the value of a population parameter Testing: formulate a decision about the value of a population parameter Regression: Make predictions or forecasts about the value of a statistical variable

64. Copyright © Houghton Mifflin Company. All rights reserved.7 | 64 Sampling Distribution a probability distribution for the sample statistic based on all possible random samples of the same size from the same population

65. Copyright © Houghton Mifflin Company. All rights reserved.7 | 65 Example of a Sampling Distribution Select samples with two elements each (in sequence with replacement) from the set {1, 2, 3, 4, 5, 6}.

66. Copyright © Houghton Mifflin Company. All rights reserved.7 | 66 Constructing a Sampling Distribution of the Mean for Samples of Size n = 2 List all samples (with 2 items in each) and compute the mean of each sample. sample:mean:sample:mean {1,1}1.0{1,6}3.5 {1,2}1.5{2,1}1.5 {1,3}2.0{2,2}4 {1,4}2.5…... {1,5}3.0 How many different samples are there?36

67. Copyright © Houghton Mifflin Company. All rights reserved.7 | 67 Sampling Distribution of the Mean p 1.01/36 1.52/36 2.03/36 2.54/36 3.05/36 3.56/36 4.05/36 4.54/36 5.03/36 5.52/36 6.01/36

68. Copyright © Houghton Mifflin Company. All rights reserved.7 | 68 Let x be a random variable with a normal distribution with mean  and standard deviation . Let x be the sample mean corresponding to random samples of size n taken from the distribution.

69. Copyright © Houghton Mifflin Company. All rights reserved.7 | 69 Theorem 7.1: The x distribution is a normal distribution. The mean of the x distribution is  (the same mean as the original distribution). The standard deviation of the x distribution is  (the standard deviation of the original distribution, divided by the square root of the sample size).

70. Copyright © Houghton Mifflin Company. All rights reserved.7 | 70 We can use this theorem to draw conclusions about means of samples taken from normal distributions. If the original distribution is normal, then the sampling distribution will be normal.

71. Copyright © Houghton Mifflin Company. All rights reserved.7 | 71 The Mean of the Sampling Distribution

72. Copyright © Houghton Mifflin Company. All rights reserved.7 | 72 The mean of the sampling distribution is equal to the mean of the original distribution.

73. Copyright © Houghton Mifflin Company. All rights reserved.7 | 73 The Standard Deviation of the Sampling Distribution

74. Copyright © Houghton Mifflin Company. All rights reserved.7 | 74 The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

75. Copyright © Houghton Mifflin Company. All rights reserved.7 | 75 To Calculate z Scores

76. Copyright © Houghton Mifflin Company. All rights reserved.7 | 76 The time it takes to drive between cities A and B is normally distributed with a mean of 14 minutes and a standard deviation of 2.2 minutes. 1.Find the probability that a trip between the cities takes more than 15 minutes. 2.Find the probability that mean time of nine trips between the cities is more than 15 minutes.

77. Copyright © Houghton Mifflin Company. All rights reserved.7 | 77 Find the probability that a trip between the cities takes more than 15 minutes. Mean = 14 minutes, standard deviation = 2.2 minutes

78. Copyright © Houghton Mifflin Company. All rights reserved.7 | 78 Find the probability that the mean time of nine trips between the cities is more than 15 minutes. Mean = 14 minutes, standard deviation = 2.2 minutes

79. Copyright © Houghton Mifflin Company. All rights reserved.7 | 79 Mean = 14 minutes, standard deviation = 2.2 minutes Find the probability that mean time of nine trips between the cities is more than 15 minutes.

80. Copyright © Houghton Mifflin Company. All rights reserved.7 | 80 Standard Error of the Mean The standard error of the mean is the standard deviation of the sampling distribution

81. Copyright © Houghton Mifflin Company. All rights reserved.7 | 81 What if the Original Distribution Is Not Normal? Use the Central Limit Theorem.

82. Copyright © Houghton Mifflin Company. All rights reserved.7 | 82 Central Limit Theorem If x has any distribution with mean  and standard deviation , then the sample mean based on a random sample of size n will have a distribution that approaches the normal distribution (with mean  and standard deviation  divided by the square root of n) as n increases without limit.

83. Copyright © Houghton Mifflin Company. All rights reserved.7 | 83 How large should the sample size be to permit the application of the Central Limit Theorem? In most cases a sample size of n = 30 or more assures that the distribution will be approximately normal and the theorem will apply.

84. Copyright © Houghton Mifflin Company. All rights reserved.7 | 84 Central Limit Theorem For most x distributions, if we use a sample size of 30 or larger, the x distribution will be approximately normal.

85. Copyright © Houghton Mifflin Company. All rights reserved.7 | 85 Central Limit Theorem The mean of the sampling distribution is the same as the mean of the original distribution. The standard deviation of the sampling distribution is equal to the standard deviation of the original distribution divided by the square root of the sample size.

86. Copyright © Houghton Mifflin Company. All rights reserved.7 | 86 Central Limit Theorem Formula

87. Copyright © Houghton Mifflin Company. All rights reserved.7 | 87 Central Limit Theorem Formula

88. Copyright © Houghton Mifflin Company. All rights reserved.7 | 88 Central Limit Theorem Formula

89. Copyright © Houghton Mifflin Company. All rights reserved.7 | 89 Application of the Central Limit Theorem Records indicate that the packages shipped by a certain trucking company have a mean weight of 510 pounds and a standard deviation of 90 pounds. One hundred packages are being shipped today. What is the probability that their mean weight will be: a.more than 530 pounds? b.less than 500 pounds? c.between 495 and 515 pounds?

90. Copyright © Houghton Mifflin Company. All rights reserved.7 | 90 Are we authorized to use the Normal Distribution? Yes, we are attempting to draw conclusions about means of large samples.

91. Copyright © Houghton Mifflin Company. All rights reserved.7 | 91 Applying the Central Limit Theorem What is the probability that the mean weight will be more than 530 pounds? Consider the distribution of sample means: P( x > 530): z = 530 – 510 = 20 = 2.22 9 9 P(z > 2.22) = _______.0132

92. Copyright © Houghton Mifflin Company. All rights reserved.7 | 92 Applying the Central Limit Theorem What is the probability that their mean weight will be less than 500 pounds? P( x < 500): z = 500 – 510 = –10 = – 1.11 9 9 P(z < – 1.11) = _______.1335

93. Copyright © Houghton Mifflin Company. All rights reserved.7 | 93 Applying the Central Limit Theorem What is the probability that their mean weight will be between 495 and 515 pounds? P(495 < x < 515) : for 495: z = 495 – 510 = - 15 = - 1.67 9 9 for 515: z = 515 – 510 = 5 = 0.56 9 9 P(–1.67 < z < 0.56) = ______.6648

94. Copyright © Houghton Mifflin Company. All rights reserved.7 | 94 Normal Approximation of The Binomial Distribution: If n (the number of trials) is sufficiently large, a binomial random variable has a distribution that is approximately normal.

95. Copyright © Houghton Mifflin Company. All rights reserved.7 | 95 Define “sufficiently large” The sample size, n, is considered to be "sufficiently large" if np and nq are both greater than 5.

96. Copyright © Houghton Mifflin Company. All rights reserved.7 | 96 Mean and Standard Deviation: Binomial Distribution

97. Copyright © Houghton Mifflin Company. All rights reserved.7 | 97 Experiment: tossing a coin 20 times Problem: Find the probability of getting exactly 10 heads. Distribution of the number of heads appearing should look like: 10 20 0

98. Copyright © Houghton Mifflin Company. All rights reserved.7 | 98 Using the Binomial Probability Formula n = x = p = q = 1 – p = P(10) = 0.176197052 20 10 0.5

99. Copyright © Houghton Mifflin Company. All rights reserved.7 | 99 Normal Approximation of the Binomial Distribution First calculate the mean and standard deviation:  = np = 20 (.5) = 10

100. Copyright © Houghton Mifflin Company. All rights reserved.7 | 100 The Continuity Correction Continuity Correction allows us to approximate a discrete probability distribution with a continuous distribution.

101. Copyright © Houghton Mifflin Company. All rights reserved.7 | 101 The Continuity Correction We are using the area under the curve to approximate the area of the rectangle.

102. Copyright © Houghton Mifflin Company. All rights reserved.7 | 102 The Continuity Correction If r is the left-point of an interval, subtract 0.5 to obtain the corresponding normal variable. x = r  0.5 If r is the right-point of an interval, add 0.5 to obtain the corresponding normal variable. x = r + 0.5

103. Copyright © Houghton Mifflin Company. All rights reserved.7 | 103 The Continuity Correction Continuity Correction: to compute the probability of getting exactly 10 heads, find the probability of getting between 9.5 and 10.5 heads.

104. Copyright © Houghton Mifflin Company. All rights reserved.7 | 104 Using the Normal Distribution P(9.5 < x < 10.5 ) = ? For x = 9.5: z = – 0.22 For x = 10.5: z = 0.22

105. Copyright © Houghton Mifflin Company. All rights reserved.7 | 105 P(9.5 < x < 10.5 ) = P( – 0.22 < z < 0.22 ) =.5871 –.4129 =.1742 Using the Normal Distribution

106. Copyright © Houghton Mifflin Company. All rights reserved.7 | 106 Application of Normal Distribution If 22% of all patients with high blood pressure have side effects from a certain medication, and 100 patients are treated, find the probability that at least 30 of them will have side effects. Using the Binomial Probability Formula we would need to compute: P(30) + P(31) +... + P(100) or 1 - P( x < 29)

107. Copyright © Houghton Mifflin Company. All rights reserved.7 | 107 Using the Normal Approximation to the Binomial Distribution Is n sufficiently large? Check: np = nq =

108. Copyright © Houghton Mifflin Company. All rights reserved.7 | 108 Is n sufficiently large? np = 22 nq = 78 Both are greater than five. Using the Normal Approximation to the Binomial Distribution

109. Copyright © Houghton Mifflin Company. All rights reserved.7 | 109 Find the mean and standard deviation  = 100(.22) = 22 and  =

110. Copyright © Houghton Mifflin Company. All rights reserved.7 | 110 Applying the Normal Distribution To find the probability that at least 30 of them will have side effects, find P(x  29.5)

111. Copyright © Houghton Mifflin Company. All rights reserved.7 | 111 z = 29.5 – 22 = 1.81 4.14 Find P(z  1.81) The probability that at least 30 of the patients will have side effects is 0.0351. Applying the Normal Distribution

112. Copyright © Houghton Mifflin Company. All rights reserved.7 | 112 Reminders: Use the normal distribution to approximate the binomial only if both np and nq are greater than 5. Always use the continuity correction when using the normal distribution to approximate the binomial distribution.