1. The Normal Probability Distribution
Chapter 7
The Normal Probability Distribution

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

3. Objectives Use the uniform probability distribution
Graph a normal curve
State the properties of the normal curve
Explain the role of area in the normal density function 3

4. Objective 1
Use the Uniform Probability Distribution 4

5. EXAMPLE Illustrating the Uniform Distribution
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. 5

6. A probability density function (pdf) is an equation used to compute probabilities of continuous random variables. It must satisfy the following two properties:
The total area under the graph of the equation over all possible values of the random variable must equal 1.
The height of the graph of the equation must be greater than or equal to 0 for all possible values of the random variable. 6

7. The graph below illustrates the properties for the “time” example
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

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. 8

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. 9

10. EXAMPLE Area as a Probability
The probability of choosing a time that is between 15 and 30 minutes after the hour is the area under the uniform density function.
Area
= P(15 < x < 30)
= 15/60
= 0.25 10

11. Objective 2
Graph a Normal Curve 11

12. Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve. 12

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). 13

14. 14

15. Objective 3
State the Properties of the Normal Curve 15

16. 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.
It has inflection points at μ – σ and μ – σ
The area under the curve is 1.
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. 16

17. 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.
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. 17

18. 18

19. Objective 4
Explain the Role of Area in the Normal Density Function 19

20. EXAMPLE A Normal Random Variable
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? 20

21. 21

22. 22

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? 23

24. 24

25. 25

26. 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. 26

27. Draw a normal curve with the parameters labeled.
EXAMPLE Interpreting 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.
Draw a normal curve with the parameters labeled.
Shade the area under the normal curve to the left of x = 2100 pounds.
Suppose that the area under the normal curve to the left of x = 2100 pounds is Provide two interpretations of this result. 27

28. μ = 2200 pounds and σ = 200 pounds (a), (b) (c)
EXAMPLE Interpreting 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
The probability that a randomly selected giraffe weighs less than 2100 pounds is 28

29. Applications of the Normal Distribution
Section
7.2
Applications of the Normal Distribution

30. Objectives Find and interpret the area under a normal curve
Find the value of a normal random variable 30

31. Objective 1
Find and Interpret the Area Under a Normal Curve 31

32. 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. 32

33. Standard Normal Curve 33

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. 34

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. 35

36. The area under the standard normal curve to the left of z = 1. 33 is 036

37. Use the Complement Rule to find the area to the right of z = 1.33.37

38. Areas Under the Standard Normal Curve38

39. Area to the left of z = –0.38 is 0.3520.
EXAMPLE Finding the Area Under the Standard Normal Curve
Find the area under the standard normal curve to the left of z = –0.38.
Area to the left of z = –0.38 is 39

40. Area under the normal curve to the right of zo = 1 – Area to the left of zo40

41. Area right of 1.25 = 1 – area left of 1.25 = 1 – 0.8944 = 0.1056
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 – = 41

42. = (Area left of z = 2.94) – (area left of z = –1.02) = 0.9984 – 0.1539
EXAMPLE Finding the Area Under the Standard Normal Curve
Find the area under the standard normal curve between z = –1.02 and z = 2.94.
Area between –1.02 and 2.94
= (Area left of z = 2.94) – (area left of z = –1.02)
= – = 42

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. 43

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. 44

45. Problem: Find the area between x1 and x2.
Approach: Shade the area between x1 and x2.
Solution:
• Convert the values of x to a z-scores. Use Table V to find the area to the left of z1 and to the left of z2. The area between z1 and z2 is (area to the left of z2) – (area to the left of z1).
• Use technology to find the area. 45

46. Objective 2
Find the Value of a Normal Random Variable 46

47. 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. 47

48. EXAMPLE Finding the Value of a Normal Random Variable
The combined (verbal + quantitative reasoning) score on the GRE is normally distributed with mean 1049 and standard deviation 189.
(Source:
What is the score of a student whose percentile rank is at the 85th percentile? 48

49. EXAMPLE Finding the Value of a Normal Random Variable
The z-score that corresponds to the 85th percentile is the z-score such that the area under the standard normal curve to the left is This z-score is 1.04.
x = µ + zσ
= (189)
= 1246
Interpretation: A person who scores 1246 on the GRE would rank in the 85th percentile. 49

50. EXAMPLE Finding the Value of a Normal Random Variable
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. 50

51. z1 = –1.645 and z2 = 1.645 x1 = µ + z1σ = 100 + (–1.645)(0.45)
EXAMPLE Finding the Value of a Normal Random Variable
z1 = –1.645 and z2 = 1.645
Area = 0.05
Area = 0.05
x1 = µ + z1σ
= (–1.645)(0.45)
= cm
x2 = µ + z2σ
= (1.645)(0.45)
= cm
Interpretation: The length of steel rods that make up the middle 90% of all steel rods manufactured would have lengths between cm and cm. 51

52. The notation zα (pronounced “z sub alpha”) is the z-score such that the area under the standard normal curve to the right of zα is α. 52

53. Section
7.3
Assessing Normality

54. Objectives
Use normal probability to assess normality 54

55. 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. 55

56. Objective 1
Use Normal Probability Plots to Assess Normality 56

57. 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. 57

58. Drawing a Normal Probability Plot
Step 1 Arrange the data in ascending order.
Step 2 Compute where i is the index (the position of the data value in the ordered list) and n is the number of observations. The expected proportion of observations less than or equal to the ith data value is f. 58

59. Drawing a Normal Probability Plot
Step 3 Find the z-score corresponding to fi from Table V.
Step 4 Plot the observed values on the horizontal axis and the corresponding expected z-scores on the vertical axis. 59

60. The idea behind finding the expected z-score is that, if the data comes from normally distributed population, we could predict the area to the left of each of the data value. The value of fi represents the expected area left of the ith observation when the data come from a population that is normally distributed. For example, f1 is the expected area to the left of the smallest data value, f2 is the expected area to the left of the second smallest data value, and so on. 60

61. 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. 61

62. 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. 62

63. EXAMPLE Interpreting a Normal Probability Plot
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? 63

64. The random variable “time between eruptions” is likely not normal.64

65. 65

66. EXAMPLE Interpreting a Normal Probability Plot
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 66

67. EXAMPLE Interpreting a Normal Probability Plot67

68. It is reasonable to believe that FEV/VC is normally distributed.68

69. The Normal Approximation to the Binomial Probability Distribution
Section
7.4
The Normal Approximation to the Binomial Probability Distribution

70. Objectives
Approximate binomial probabilities using the normal distribution 70

71. Objective 1
Approximate Binomial Probabilities Using the Normal Distribution 71

72. 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. 72

73. 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. 73

74. The Normal Approximation to the Binomial Probability Distribution
If np(1 – p) ≥ 10, the binomial random variable X is approximately normally distributed, with mean μX = np and standard deviation 74

75. P(X = 18) ≈ P(17.5 < X < 18.5) 75

76. P(X < 18) ≈ P(X < 18.5) 76

77. Exact Probability Using Binomial: P(a)
Approximate
Probability Using Normal: P(a – 0.5 ≤ X ≤ a + 0.5)
Graphical Depiction 77

78. Exact Probability Using Binomial: P(X ≤ a)
Approximate Probability Using Normal: P(X ≤ a + 0.5)
Graphical Depiction 78

79. Exact Probability Using Binomial: P(X ≥ a)
Approximate Probability Using Normal: P(X ≥ a – 0.5)
Graphical Depiction 79

80. Using Binomial: P(a ≤ X ≤ b)
Exact Probability
Using Binomial: P(a ≤ X ≤ b)
Approximate Probability Using Normal: P(X ≥ a – 0.5)
Graphical Depiction 80

81. EXAMPLE Using the Binomial Probability Distribution Function
According to the Experian Automotive, 35% of all car-owning households have three or more cars.
In a random sample of 400 car-owning households, what is the probability that fewer than 150 have three or more cars? 81

82. EXAMPLE Using 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? 82