1. The Normal Probability Distribution
Chapter 7
The Normal Probability Distribution
7.1
7.2
7.3
7.4

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
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. 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. 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. 6

7. 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. 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 8

9. Objective 2 Graph a Normal Curve
Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve. 9

10. 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). 10

11. Objective 3 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 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.
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. 11

12. 12

13. Objective 4
Explain the Role of Area in the Normal Density Function 13

14. 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? 14

15. 15

16. 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. 16

17. 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.
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 17

18. 7.1 Summary 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 18

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

20. Objectives Find and interpret the area under a normal curve
Find the value of a normal random variable 20

21. Objective 1 Standard Normal Curve
Find and Interpret the Area Under a Normal Curve
Standard Normal Curve 21

22. 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. 22

23. 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. 23

24. 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.
What does it mean to be 1.33 standard deviation above the mean? 24

25. The area under the standard normal curve to the left of z = 1. 33 is 025

26. Interpret the meaning:
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.
Interpret the meaning:
90.82% of the population has an IQ score below this individual.
9.18% of the population has an IQ score above this individual. 26

27. 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 27

28. 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 – = 28

29. = (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)
= – = 29

30. Objective 2 Find the Value of a Normal Random Variable
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 (or use InvNorm on calculator)
Step 3: Obtain the normal value from the formula x = μ + zσ. 30

31. 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 (Source:
What is the score of a student whose percentile rank is at the 85th percentile?
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. 31

32. 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.
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
The length of steel rods that make up the middle 90% of all steel rods manufactured would have lengths between cm and cm. 32

33. 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. 33

34. 7.2 Summary Find and interpret the area under a normal curve
Find the value of a normal random variable 34

35. Section
7.3
Assessing Normality

36. Objectives
Use Normal Probability Plots to Assess Normality 36

37. 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. 37

38. 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.
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. 38

39. 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? 39

40. The random variable “time between eruptions” is likely not normal.40

41. 41

42. 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 42

43. It is reasonable to believe that FEV/VC is normally distributed.43

44. 7.3 Summary
Use Normal Probability Plots to Assess Normality 44

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

46. Objectives
Approximate Binomial Probabilities Using the Normal Distribution 46

47. Criteria for a Binomial Probability Experiment
An experiment is said to be a binomial experiment if all of the following are true:
For each trial, there are two mutually exclusive outcomes - success or failure.
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.
The probability of success, p, is the same for each trial of the experiment. 47

48. 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. 48

49. 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
standard deviation 49

50. P(X = 18) ≈ P(17.5 < X < 18.5) 50

51. P(X < 18) ≈ P(X < 18.5) 51

52. Exact Probability Using Binomial: P(a)
Approximate
Probability Using Normal: P(a – 0.5 ≤ X ≤ a + 0.5)
Graphical Depiction 52

53. Exact Probability Using Binomial: P(X ≤ a)
Approximate Probability Using Normal: P(X ≤ a + 0.5)
Graphical Depiction 53

54. Exact Probability Using Binomial: P(X ≥ a)
Approximate Probability Using Normal: P(X ≥ a – 0.5)
Graphical Depiction 54

55. Exact Probability Using Binomial: P(a ≤ X ≤ b)
Approximate Probability Using Normal: P(a – 0.5 < X < b + 0.5)
Graphical Depiction 55

56. 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? 56

57. 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? 57

58. 7.4 Summary
Approximate Binomial Probabilities Using the Normal Distribution 58