1. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Chapter The Normal Probability Distribution 7

2. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Properties of the Normal Distribution 7.1

3. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Use the uniform probability distribution 2.Graph a normal curve 3.State the properties of the normal curve 4.Explain the role of area in the normal density function 5-3

4. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Use the Uniform Probability Distribution 5-4

5. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-5 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

6. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. A probability density function (pdf) is an equation used to compute probabilities of continuous random variables. It must satisfy the following two properties: 1.The total area under the graph of the equation over all possible values of the random variable must equal 1. 2.The height of the graph of the equation must be greater than or equal to 0 for all possible values of the random variable. A probability density function (pdf) is an equation used to compute probabilities of continuous random variables. It must satisfy the following two properties: 1.The total area under the graph of the equation over all possible values of the random variable must equal 1. 2.The height of the graph of the equation must be greater than or equal to 0 for all possible values of the random variable. 5-6

7. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-7 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.

8. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-8 Values of the random variable X less than 0 or greater than 60 are impossible, thus the function value must be zero for X less than 0 or greater than 60.

9. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-9 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.

10. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-10 The probability of choosing a time that is between 15 and 30 minutes after the hour is the area under the uniform density function. 15 30 Area = P(15 < x < 30) = 15/60 = 0.25 EXAMPLEArea as a Probability

11. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Graph a Normal Curve 5-11

12. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-12 Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve.

13. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-13 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).

14. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-14

15. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 State the Properties of the Normal Curve 5-15

16. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Properties of the Normal Density Curve 1.It is symmetric about its mean, μ. 2.Because mean = median = mode, the curve has a single peak and the highest point occurs at x = μ. 3.It has inflection points at μ – σ and μ – σ 4.The area under the curve is 1. 5.The area under the curve to the right of μ equals the area under the curve to the left of μ, which equals 1/2. Properties of the Normal Density Curve 1.It is symmetric about its mean, μ. 2.Because mean = median = mode, the curve has a single peak and the highest point occurs at x = μ. 3.It has inflection points at μ – σ and μ – σ 4.The area under the curve is 1. 5.The area under the curve to the right of μ equals the area under the curve to the left of μ, which equals 1/2. 5-16

17. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 6.As x increases without bound (gets larger and larger), the graph approaches, but never reaches, the horizontal axis. As x decreases without bound (gets more and more negative), the graph approaches, but never reaches, the horizontal axis. 7.The Empirical Rule: Approximately 68% of the area under the normal curve is between x = μ – σ and x = μ + σ; approximately 95% of the area is between x = μ – 2σ and x = μ + 2σ; approximately 99.7% of the area is between x = μ – 3σ and x = μ + 3σ. 6.As x increases without bound (gets larger and larger), the graph approaches, but never reaches, the horizontal axis. As x decreases without bound (gets more and more negative), the graph approaches, but never reaches, the horizontal axis. 7.The Empirical Rule: Approximately 68% of the area under the normal curve is between x = μ – σ and x = μ + σ; approximately 95% of the area is between x = μ – 2σ and x = μ + 2σ; approximately 99.7% of the area is between x = μ – 3σ and x = μ + 3σ. 5-17

18. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-18

19. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objective 4 Explain the Role of Area in the Normal Density Function 5-19

20. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-20 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

21. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-21 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

22. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-22

23. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-23 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?

24. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-24

25. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-25

26. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Area under a Normal Curve Suppose that a random variable X is normally distributed with mean μ and standard deviation σ. The area under the normal curve for any interval of values of the random variable X represents either the proportion of the population with the characteristic described by the interval of values or the probability that a randomly selected individual from the population will have the characteristic described by the interval of values. Area under a Normal Curve Suppose that a random variable X is normally distributed with mean μ and standard deviation σ. The area under the normal curve for any interval of values of the random variable X represents either the proportion of the population with the characteristic described by the interval of values or the probability that a randomly selected individual from the population will have the characteristic described by the interval of values. 5-26

27. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-27 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.

28. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-28 EXAMPLEInterpreting the Area Under a Normal Curve μ = 2200 pounds and σ = 200 pounds (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.

29. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Applications of the Normal Distribution 7.2

30. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Find and interpret the area under a normal curve 2.Find the value of a normal random variable 5-30

31. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Find and Interpret the Area Under a Normal Curve 5-31

32. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Standardizing a Normal Random Variable Suppose that the random variable X is normally distributed with mean μ and standard deviation σ. Then the random variable is normally distributed with mean μ = 0 and standard deviation σ = 1.The random variable Z is said to have the standard normal distribution. Standardizing a Normal Random Variable Suppose that the random variable X is normally distributed with mean μ and standard deviation σ. Then the random variable is normally distributed with mean μ = 0 and standard deviation σ = 1.The random variable Z is said to have the standard normal distribution. 5-32

33. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-33 Standard Normal Curve

34. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-34 The table gives the area under the standard normal curve for values to the left of a specified Z-score, z, as shown in the figure.

35. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-35 IQ scores can be modeled by a normal distribution with μ = 100 and σ = 15. An individual whose IQ score is 120, is 1.33 standard deviations above the mean.

36. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-36 The area under the standard normal curve to the left of z = 1.33 is 0.9082.

37. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-37 Use the Complement Rule to find the area to the right of z = 1.33.

38. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-38 Areas Under the Standard Normal Curve

39. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-39 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 to the left of z = –0.38 is 0.3520.

40. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-40 Area under the normal curve to the right of z o = 1 – Area to the left of z o

41. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-41 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

42. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-42 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

43. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-43 Problem: Find the area to the left of x. Approach: Shade the area to the left of x. Solution: Convert the value of x to a z-score. Use Table V to find the row and column that correspond to z. The area to the left of x is the value where the row and column intersect. Use technology to find the area.

44. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-44 Problem: Find the area to the right of x. Approach: Shade the area to the right of x. Solution: Convert the value of x to a z-score. Use Table V to find the area to the left of z (also is the area to the left of x). The area to the right of z (also x) is 1 minus the area to the left of z. Use technology to find the area.

45. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-45 Problem: Find the area between x 1 and x 2. Approach: Shade the area between x 1 and x 2. Solution: Convert the values of x to a z-scores. Use Table V to find the area to the left of z 1 and to the left of z 2. The area between z 1 and z 2 is (area to the left of z 2 ) – (area to the left of z 1 ). Use technology to find the area.

46. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Find the Value of a Normal Random Variable 5-46

47. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Procedure for Finding the Value of a Normal Random Variable Step 1:Draw a normal curve and shade the area corresponding to the proportion, probability, or percentile. Step 2:Use Table V to find the z-score that corresponds to the shaded area. Step 3:Obtain the normal value from the formula x = μ + zσ. Procedure for Finding the Value of a Normal Random Variable Step 1:Draw a normal curve and shade the area corresponding to the proportion, probability, or percentile. Step 2:Use Table V to find the z-score that corresponds to the shaded area. Step 3:Obtain the normal value from the formula x = μ + zσ. 5-47

48. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-48 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?

49. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-49 EXAMPLEFinding the Value of a Normal Random Variable 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: A person who scores 1246 on the GRE would rank in the 85 th percentile.

50. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-50 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

51. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-51 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.

52. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. The notation z α (pronounced “z sub alpha”) is the z-score such that the area under the standard normal curve to the right of z α is α. 5-52

53. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section The Normal Approximation to the Binomial Probability Distribution 7.4

54. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Approximate binomial probabilities using the normal distribution 5-54

55. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Approximate Binomial Probabilities Using the Normal Distribution 5-55

56. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Criteria for a Binomial Probability Experiment An experiment is said to be a binomial experiment if all of the following are true: 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. Criteria for a Binomial Probability Experiment An experiment is said to be a binomial experiment if all of the following are true: 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. 5-56

57. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-57 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 rule of thumb, if np(1 – p) > 10, the probability distribution will be approximately symmetric and bell-shaped.

58. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. The Normal Approximation to the Binomial Probability Distribution If np(1 – p) ≥ 10, the binomial random variable X is approximately normally distributed, with mean μ X = np and standard deviation The Normal Approximation to the Binomial Probability Distribution If np(1 – p) ≥ 10, the binomial random variable X is approximately normally distributed, with mean μ X = np and standard deviation 5-58

59. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-59 P(X = 18) ≈ P(17.5 < X < 18.5)

60. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-60 P(X < 18) ≈ P(X < 18.5)

61. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-61 Exact Probability Using Binomial: P(a) Approximate Probability Using Normal:P(a – 0.5 ≤ X ≤ a + 0.5) Graphical Depiction

62. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-62 Exact Probability Using Binomial: P(X ≤ a) Approximate Probability Using Normal: P(X ≤ a + 0.5) Graphical Depiction

63. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-63 Exact Probability Using Binomial: P(X ≥ a) Approximate Probability Using Normal: P(X ≥ a – 0.5) Graphical Depiction

64. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-64 Exact Probability Using Binomial: P(a ≤ X ≤ b) Approximate Probability Using Normal: P(X ≥ a – 0.5) Graphical Depiction

65. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-65 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?

66. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. 7-66 EXAMPLEUsing the Binomial Probability Distribution Function According to the Experian Automotive, 35% of all car-owning households 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?