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

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

3. 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. 3

4. 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. 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. 5

6. 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. 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. Considering the example,
EXAMPLE Area as a Probability
Considering the example,
What is the probability that they will arrive between 10:22 am and 10:56 am?
It is 10:00 am, there is a 15% chance they will arrive in how many minutes? 9

10. Relative frequency histograms that are symmetric and bell-shaped are said to have the shape of a normal curve. 10

11. A continuous random variable is normally distributed, or has a normal probability distribution, if its relative frequency histogram of the random variable has the shape of a normal curve (bell-shaped and symmetric). 11

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13. 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. 13

14. 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. 14

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16. 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? 16

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19. 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? 19

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22. 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. 22

23. 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. 23

24. 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 24

25. Draw a normal curve with parameters labeled.
EXAMPLE Interpreting the Area Under a Normal Curve
The serum total cholesterol for males years old is approximately normally distributed with mean μ = 180 and standard deviation σ = 36.2, based on some data.
Draw a normal curve with parameters labeled.
An individual with cholesterol greater than 200 is considered to have high cholesterol. Shade the region under the normal curve to the right of x=200.
Suppose that the area under the normal curve to the right of x=200 is Provide two interpretations of this result. 25

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

27. 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. 27

28. Standard Normal Curve 28

29. 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. 29

30. 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. 30

31. The area under the standard normal curve to the left of z = 1. 33 is 031

32. Use the Complement Rule to find the area to the right of z = 1.33.32

33. Areas Under the Standard Normal Curve33

34. 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 34

35. 2. It can also be found using: normalcdf(lowerbound,upperbound) Or
1. Area under the normal curve to the right of zo = 1 – Area to the left of zo
2. It can also be found using:
normalcdf(lowerbound,upperbound) Or
Normalcdf(lowerbound,upperbound, mean, standard deviation) 35

36. 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 – = 36

37. = (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)
= – = 37

38. 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 table or technology to find the area. 38

39. 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. 39

40. 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 table or technology to find the area. 40

41. 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. 41

42. 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? 42

43. 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. 43

44. EXAMPLE Finding the Value of a Normal Random Variable
Suppose the height of three-year-old girls is normally distributed with mean inches and standard deviation 3.17 inches. Use the normal model to determine the proportion of the three-year-old females that have a height less than 35 inches. 44

45. EXAMPLE Finding the Value of a Normal Random Variable
Suppose the height of three-year-old girls is normally distributed with mean inches and standard deviation 3.17 inches. Find the probability that a randomly selected three-year-old girl is between 35 and 40 inches tall, inclusive. 45

46. EXAMPLE Finding the Value of a Normal Random Variable
Suppose the height of three-year-old girls is normally distributed with mean inches and standard deviation 3.17 inches. Find the height of a three-year-old female at the 20th percentile. 46

47. 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. 47

48. 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. 48

49. EXAMPLE Finding the Value of a Normal Random Variable
The scores earned on the mathematics portion of the SAT are approximately normally distributed with mean 516 and standard deviation What scores separate the middle 90% of test takers from the bottom and top 5%? 49

50. 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. 50

51. EXAMPLE
Find the value:
𝑧 0.10
𝑧 0.05
𝑧 0.01 51