1. © 2010 Pearson Prentice Hall. All rights reserved Chapter The Normal Probability Distribution © 2010 Pearson Prentice Hall. All rights reserved 3 7

2. 7-2 Section 7.1 Properties of the Normal Distribution

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4. © 2010 Pearson Prentice Hall. All rights reserved Suppose that United Parcel Service is supposed to deliver a package to your front door and the arrival time is somewhere between 10 am and 11 am. Let the random variable X represent the time from 10 am when the delivery is supposed to take place. The delivery could be at 10 am (x = 0) or at 11 am (x = 60) with all 1-minute interval of times between x = 0 and x = 60 equally likely. That is to say your package is just as likely to arrive between 10:15 and 10:16 as it is to arrive between 10:40 and 10:41. The random variable X can be any value in the interval from 0 to 60, that is, 0 < X < 60. Because any two intervals of equal length between 0 and 60, inclusive, are equally likely, the random variable X is said to follow a uniform probability distribution. EXAMPLE Illustrating the Uniform Distribution 7-4

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6. © 2010 Pearson Prentice Hall. All rights reserved The graph below illustrates the properties for the “time” example. Notice the area of the rectangle is one and the graph is greater than or equal to zero for all x between 0 and 60, inclusive. Because the area of a rectangle is height times width, and the width of the rectangle is 60, the height must be 1/60. 7-6

7. © 2010 Pearson Prentice Hall. All rights reserved Values of the random variable X less than 0 or greater than 60 are impossible, thus the equation must be zero for X less than 0 or greater than 60. 7-7

8. © 2010 Pearson Prentice Hall. All rights reserved The area under the graph of the density function over an interval represents the probability of observing a value of the random variable in that interval. 7-8

9. © 2010 Pearson Prentice Hall. All rights reserved The probability of choosing a time that is between 15 and 30 seconds after the minute is the area under the uniform density function. 15 30 Area = P(15 < x < 30) = 15/60 = 0.25 EXAMPLEArea as a Probability 7-9

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11. © 2010 Pearson Prentice Hall. All rights reserved Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve. 7-11

12. © 2010 Pearson Prentice Hall. All rights reserved If a continuous random variable is normally distributed, or has a normal probability distribution, then a relative frequency histogram of the random variable has the shape of a normal curve (bell-shaped and symmetric). 7-12

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18. © 2010 Pearson Prentice Hall. All rights reserved The data on the next slide represent the heights (in inches) of a random sample of 50 two-year old males. (a) Draw a histogram of the data using a lower class limit of the first class equal to 31.5 and a class width of 1. (b) Do you think that the variable “height of 2- year old males” is normally distributed? EXAMPLE A Normal Random Variable 7-18

19. © 2010 Pearson Prentice Hall. All rights reserved 36.036.234.836.034.638.435.436.8 34.733.437.438.231.537.736.934.0 34.435.737.939.334.036.935.137.0 33.236.135.235.633.036.833.535.0 35.135.234.436.736.036.035.735.7 38.333.639.837.037.234.835.738.9 37.239.3 7-19

20. © 2010 Pearson Prentice Hall. All rights reserved 7-20

21. © 2010 Pearson Prentice Hall. All rights reserved In the next slide, we have a normal density curve drawn over the histogram. How does the area of the rectangle corresponding to a height between 34.5 and 35.5 inches relate to the area under the curve between these two heights? 7-21

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25. © 2010 Pearson Prentice Hall. All rights reserved EXAMPLEInterpreting the Area Under a Normal Curve The weights of giraffes are approximately normally distributed with mean μ = 2200 pounds and standard deviation σ = 200 pounds. (a)Draw a normal curve with the parameters labeled. (b)Shade the area under the normal curve to the left of x = 2100 pounds. (c)Suppose that the area under the normal curve to the left of x = 2100 pounds is 0.3085. Provide two interpretations of this result. (a), (b) (c) The proportion of giraffes whose weight is less than 2100 pounds is 0.3085 The probability that a randomly selected giraffe weighs less than 2100 pounds is 0.3085. 7-25

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28. © 2010 Pearson Prentice Hall. All rights reserved EXAMPLERelation Between a Normal Random Variable and a Standard Normal Random Variable The weights of giraffes are approximately normally distributed with mean μ = 2200 pounds and standard deviation σ = 200 pounds. Draw a graph that demonstrates the area under the normal curve between 2000 and 2300 pounds is equal to the area under the standard normal curve between the Z- scores of 2000 and 2300 pounds. 7-28

29. © 2010 Pearson Prentice Hall. All rights reserved Section 7.2 The Standard Normal Distribution 7-29

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34. © 2010 Pearson Prentice Hall. All rights reserved The table gives the area under the standard normal curve for values to the left of a specified Z-score, z o, as shown in the figure. 7-34

35. © 2010 Pearson Prentice Hall. All rights reserved Find the area under the standard normal curve to the left of z = -0.38. EXAMPLE Finding the Area Under the Standard Normal Curve Area left of z = -0.38 is 0.3520. 7-35

36. © 2010 Pearson Prentice Hall. All rights reserved Area under the normal curve to the right of z o = 1 – Area to the left of z o 7-36

37. © 2010 Pearson Prentice Hall. All rights reserved EXAMPLE Finding the Area Under the Standard Normal Curve Find the area under the standard normal curve to the right of Z = 1.25. Area right of 1.25 = 1 – area left of 1.25 = 1 – 0.8944 = 0.1056 7-37

38. © 2010 Pearson Prentice Hall. All rights reserved Find the area under the standard normal curve between z = -1.02 and z = 2.94. EXAMPLE Finding the Area Under the Standard Normal Curve Area between -1.02 and 2.94 = (Area left of z = 2.94) – (area left of z = -1.02) = 0.9984 – 0.1539 = 0.8445 7-38

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41. © 2010 Pearson Prentice Hall. All rights reserved Find the z-score such that the area to the left of the z-score is 0.7157. EXAMPLE Finding a z-score from a Specified Area to the Left The z-score such that the area to the left of the z-score is 0.7157 is z = 0.57. 7-41

42. © 2010 Pearson Prentice Hall. All rights reserved EXAMPLE Finding a z-score from a Specified Area to the Right Find the z-score such that the area to the right of the z-score is 0.3021. The area left of the z-score is 1 – 0.3021 = 0.6979. The approximate z-score that corresponds to an area of 0.6979 to the left (0.3021 to the right) is 0.52. Therefore, z = 0.52. 7-42

43. © 2010 Pearson Prentice Hall. All rights reserved Find the z-scores that separate the middle 80% of the area under the normal curve from the 20% in the tails. EXAMPLE Finding a z-score from a Specified Area Area = 0.8 Area = 0.1 z 1 is the z-score such that the area left is 0.1, so z 1 = -1.28. z 2 is the z-score such that the area left is 0.9, so z 2 = 1.28. 7-43

44. © 2010 Pearson Prentice Hall. All rights reserved The notation z α (prounounced “z sub alpha”) is the z-score such that the area under the standard normal curve to the right of z α is α. 7-44

45. © 2010 Pearson Prentice Hall. All rights reserved Find the value of z 0.25 EXAMPLE Finding the Value of z  We are looking for the z-value such that the area to the right of the z-value is 0.25. This means that the area left of the z-value is 0.75. z 0.25 = 0.67 7-45

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47. © 2010 Pearson Prentice Hall. All rights reserved Notation for the Probability of a Standard Normal Random Variable P(a < Z < b) represents the probability a standard normal random variable is between a and b P(Z > a)represents the probability a standard normal random variable is greater than a. P(Z < a) represents the probability a standard normal random variable is less than a. 7-47

48. © 2010 Pearson Prentice Hall. All rights reserved Find each of the following probabilities: (a) P(Z < -0.23) (b) P(Z > 1.93) (c) P(0.65 < Z < 2.10) EXAMPLE Finding Probabilities of Standard Normal Random Variables (a) P(Z < -0.23) = 0.4090 (b) P(Z > 1.93) = 0.0268 (c) P(0.65 < Z < 2.10) = 0.2399 7-48

49. © 2010 Pearson Prentice Hall. All rights reserved For any continuous random variable, the probability of observing a specific value of the random variable is 0. For example, for a standard normal random variable, P(a) = 0 for any value of a. This is because there is no area under the standard normal curve associated with a single value, so the probability must be 0. Therefore, the following probabilities are equivalent: P(a < Z < b) = P(a < Z < b) = P(a < Z < b) = P(a < Z < b) 7-49

50. © 2010 Pearson Prentice Hall. All rights reserved Section 7.3 Applications of the Normal Distribution 7-50

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53. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. * What is the probability that a randomly selected steel rod has a length less than 99.2 cm? * Based upon information obtained from Stefan Wilk. EXAMPLEFinding the Probability of a Normal Random Variable Interpretation: If we randomly selected 100 steel rods, we would expect about 4 of them to be less than 99.2 cm. 7-53

54. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. What is the probability that a randomly selected steel rod has a length between 99.8 and 100.3 cm? Interpretation: If we randomly selected 100 steel rods, we would expect about 42 of them to be between 99.8 cm and 100.3 cm. EXAMPLEFinding the Probability of a Normal Random Variable 7-54

55. © 2010 Pearson Prentice Hall. All rights reserved The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1049 and standard deviation 189. (Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.) The Department of Psychology at Columbia University in New York requires a minimum combined score of 1200 for admission to their doctoral program. (Source: www.columbia.edu/cu/gsas/departments/psychology/department.html.) EXAMPLEFinding the Percentile Rank of a Normal Random Variable What is the percentile rank of a student who earns a combined GRE score of 1300? The area under the normal curve is a probability, proportion, or percentile. Here, the area under the normal curve to the left of 1300 represents the percentile rank of the student. Area left of 1300 = Area left of (z = 1.33) = 0.91 (rounded to two decimal places) Interpretation: The student scored at the 91 st percentile. This means the student scored better than 91% of the students who took the GRE. 7-55

56. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. Suppose the manufacturer must discard all rods less than 99.1 cm or longer than 100.9 cm. What proportion of rods must be discarded? EXAMPLEFinding the Proportion Corresponding to a Normal Random Variable The proportion is the area under the normal curve to the left of 99.1 cm plus the area under the normal curve to the right of 100.9 cm. Area left of 99.1 + area right of 100.9 = (Area left of z = -2) + (Area right of z = 2) = 0.0228 + 0.0228 = 0.0456 Interpretation: The proportion of rods that must be discarded is 0.0456. If the company manufactured 1000 rods, they would expect to discard about 46 of them. 7-56

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59. © 2010 Pearson Prentice Hall. All rights reserved The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1049 and standard deviation 189. (Source: http://www.ets.org/Media/Tests/GRE/pdf/994994.pdf.) EXAMPLEFinding the Value of a Normal Random Variable What is the score of a student whose percentile rank is at the 85 th percentile? The z-score that corresponds to the 85 th percentile is the z-score such that the area under the standard normal curve to the left is 0.85. This z-score is 1.04. x = µ + zσ = 1049 + 1.04(189) = 1246 Interpretation: The proportion of rods that must be discarded is 0.0456. If the company manufactured 1000 rods, they would expect to discard about 46 of them. 7-59

60. © 2010 Pearson Prentice Hall. All rights reserved It is known that the length of a certain steel rod is normally distributed with a mean of 100 cm and a standard deviation of 0.45 cm. Suppose the manufacturer wants to accept 90% of all rods manufactured. Determine the length of rods that make up the middle 90% of all steel rods manufactured. EXAMPLEFinding the Value of a Normal Random Variable Area = 0.05 z 1 = -1.645 and z 2 = 1.645 x 1 = µ + z 1 σ = 100 + (-1.645)(0.45) = 99.26 cm x 2 = µ + z 2 σ = 100 + (1.645)(0.45) = 100.74 cm Interpretation: The length of steel rods that make up the middle 90% of all steel rods manufactured would have lengths between 99.26 cm and 100.74 cm. 7-60

61. © 2010 Pearson Prentice Hall. All rights reserved Section 7.4 Assessing Normality 7-61

62. © 2010 Pearson Prentice Hall. All rights reserved Suppose that we obtain a simple random sample from a population whose distribution is unknown. Many of the statistical tests that we perform on small data sets (sample size less than 30) require that the population from which the sample is drawn be normally distributed. Up to this point, we have said that a random variable X is normally distributed, or at least approximately normal, provided the histogram of the data is symmetric and bell-shaped. This method works well for large data sets, but the shape of a histogram drawn from a small sample of observations does not always accurately represent the shape of the population. For this reason, we need additional methods for assessing the normality of a random variable X when we are looking at sample data. 7-62

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64. © 2010 Pearson Prentice Hall. All rights reserved A normal probability plot plots observed data versus normal scores. A normal score is the expected Z-score of the data value if the distribution of the random variable is normal. The expected Z-score of an observed value will depend upon the number of observations in the data set. 7-64

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66. © 2010 Pearson Prentice Hall. All rights reserved The idea behind finding the expected Z-score is that if the data comes from a population that is normally distributed, we should be able to predict the area left of each of the data values. The value of f i represents the expected area left of the i th data value assuming the data comes from a population that is normally distributed. For example, f 1 is the expected area left of the smallest data value, f 2 is the expected area left of the second smallest data value, and so on. 7-66

67. © 2010 Pearson Prentice Hall. All rights reserved If sample data is taken from a population that is normally distributed, a normal probability plot of the actual values versus the expected Z-scores will be approximately linear. 7-67

68. © 2010 Pearson Prentice Hall. All rights reserved We will be content in reading normal probability plots constructed using the statistical software package, MINITAB. In MINITAB, if the points plotted lie within the bounds provided in the graph, then we have reason to believe that the sample data comes from a population that is normally distributed. 7-68

69. © 2010 Pearson Prentice Hall. All rights reserved The following data represent the time between eruptions (in seconds) for a random sample of 15 eruptions at the Old Faithful Geyser in California. Is there reason to believe the time between eruptions is normally distributed? EXAMPLEInterpreting a Normal Probability Plot 7-69

70. © 2010 Pearson Prentice Hall. All rights reserved The random variable “time between eruptions” is likely not normal. 7-70

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72. © 2010 Pearson Prentice Hall. All rights reserved Suppose that seventeen randomly selected workers at a detergent factory were tested for exposure to a Bacillus subtillis enzyme by measuring the ratio of forced expiratory volume (FEV) to vital capacity (VC). NOTE: FEV is the maximum volume of air a person can exhale in one second; VC is the maximum volume of air that a person can exhale after taking a deep breath. Is it reasonable to conclude that the FEV to VC (FEV/VC) ratio is normally distributed? Source: Shore, N.S.; Greene R.; and Kazemi, H. “Lung Dysfunction in Workers Exposed to Bacillus subtillis Enzyme,” Environmental Research, 4 (1971), pp. 512 - 519. EXAMPLEInterpreting a Normal Probability Plot 7-72

73. © 2010 Pearson Prentice Hall. All rights reserved Reasonable to believe that FEV/VC is normally distributed. 7-73

74. © 2010 Pearson Prentice Hall. All rights reserved Section 7.5 The Normal Approximation to the Binomial Probability Distribution 7-74

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76. © 2010 Pearson Prentice Hall. All rights reserved Criteria for a Binomial Probability Experiment An experiment is said to be a binomial experiment provided: 1. The experiment is performed n independent times. Each repetition of the experiment is called a trial. Independence means that the outcome of one trial will not affect the outcome of the other trials. 2. For each trial, there are two mutually exclusive outcomes - success or failure. 3. The probability of success, p, is the same for each trial of the experiment. 7-76

77. © 2010 Pearson Prentice Hall. All rights reserved For a fixed p, as the number of trials n in a binomial experiment increases, the probability distribution of the random variable X becomes more nearly symmetric and bell-shaped. As a general rule of thumb, if np(1 – p) > 10, the probability distribution will be approximately symmetric and bell-shaped. 7-77

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79. © 2010 Pearson Prentice Hall. All rights reserved P(X = 18) ≈ P(17.5 < X < 18.5) 7-79

80. © 2010 Pearson Prentice Hall. All rights reserved P(X < 18) ≈ P(X < 18.5) 7-80

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82. © 2010 Pearson Prentice Hall. All rights reserved EXAMPLEUsing the Binomial Probability Distribution Function According to the Experian Automotive, 35% of all car-owning households have three or more cars. (a)In a random sample of 400 car-owning households, what is the probability that fewer than 150 have three or more cars? (b) In a random sample of 400 car-owning households, what is the probability that at least 160 have three or more cars? 7-82