2. Overview 6.1 Discrete Random Variables
6.2 Binomial Probability Distribution
6.3 Continuous Random Variables and the Normal Probability Distribution
6.4 Standard Normal Distribution
6.5 Applications of the Normal Distribution

3. 6.1 Discrete Random Variables
Objectives:
By the end of this section, I will be
able to…
Identify random variables.
Explain what a discrete probability distribution is and construct probability distribution tables and graphs.
Calculate the mean, variance, and standard deviation of a discrete random variable.

4. Random Variables A variable whose values are determined by chance
Chance in the definition of a random variable is crucial

5. Example 6.2 - Notation for random variables
Suppose our experiment is to toss a single fair die, and we are interested in the number rolled. We define our random variable X to be the outcome of a single die roll. a. Why is the variable X a random variable? b. What are the possible values that the random variable X can take? c. What is the notation used for rolling a 5? d. Use random variable notation to express the probability of rolling a 5.

6. Example 6.2 continued Solution
We don’t know the value of X before we toss the die, which introduces an element of chance into the experiment
Possible values for X: 1, 2, 3, 4, 5, and 6.
When a 5 is rolled, then X equals the outcome 5, or X = 5.
Probability of rolling a 5 for a fair die is 1/6, thus P(X = 5) = 1/6.

7. Main types of random variables
Discrete random variable - a finite or a countable number of values
Continuous random variable can take infinitely many values

8. Example 6.3 - Discrete and continuous random variables
For the following random variables, (i) determine whether they are discrete or continuous, and (ii) indicate the possible values they can take. a. The number of automobiles owned by a family b. The width of your desk in this classroom c. The number of games played in the next World Series d. The weight of model year 2007 SUVs

9. Example 6.3 continued Solution a.
Since the possible number of automobiles owned by a family is finite and may be written as a list of numbers, it represents a discrete random variable.
The possible values are
{0, 1, 2, 3, 4, . . .}.

10. Example 6.3 continued Solution b.
Width is something that must be measured,
not counted.
Width can take infinitely many different possible values, with these values forming an interval on the number line.
Thus, the width of your desk is a continuous
random variable.
The possible values might be 1 ft ≤ W ≤ 10 ft.

11. Example 6.3 continued Solution c.
The number of games played in the next World Series can be counted and thus represents a discrete random variable.
The possible values are {4, 5, 6, 7}.

12. Example 6.3 continued Solution d.
The weight of model year 2007 SUVs must be measured, not counted, and so represents a continuous random variable.
Weight can take infinitely many different possible values in an interval: possible values 2500 lb ≤ Y ≤ 7000 lb.

13. Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table, graph, or formula
Describe populations, not samples

14. Example 6.5 - Probability distribution table
Construct the probability distribution table of the number of heads observed when tossing a fair coin twice.

15. Example 6.5 continued Solution
The probability distribution table given in Table 6.2 uses probabilities we found in Example 6.4 page 261.
Table 6.2 Probability distribution of number
of heads on two fair coin tosses

16. Requirements Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1.
That is, ΣP(X) = 1.
The probability of each value of X must be between 0 and 1, inclusive.
That is, 0 ≤ P(X ) ≤ 1.

17. Discrete Probability Distribution as a Graph
Graphs show all the information contained in probability distribution tables
Identify patterns more quickly
FIGURE 6.1 Graph of probability distribution
for Kristin’s financial gain.

18. Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X.
Denoted as E(X )
Holds for discrete and continuous random variables

19. Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability.
Add the resulting products.

20. Example 6.8 - Calculating the mean of a discrete random variable
Find the mean age of the mother for the babies born to teenagers aged in 2004, from Example 6.7 page 265.

21. Example 6.8 continued Solution
Multiply each possible outcome (value of X) by its probability P(X).
Multiply the value X = 15 by its probability P(X) = 0.07, the value X = 16 by its probability P(X) = 0.17, and so on.
Add these four products to find the mean:

22. Example 6.8 continued Solution
μ=15(0.07) + 16(0.17) + 17(0.29) +18(0.47)
= 17.16
The mean age of the mother for the babies born to teenagers aged 15–18 is years old, which is near 17 years old

23. Variability of a Discrete Random Variable
Formulas for the Variance and Standard Deviation of a Discrete Random Variable

24. Summary
Section 6.1 introduces the idea of random variables, a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text.
Random variables are variables whose value is determined at least partly by chance.
Discrete random variables take values that are either finite or countable and may be put in a list.
Continuous random variables take an infinite number of possible values, represented by an interval on the number line.

25. Summary
Discrete random variables can be described using a probability distribution, which specifies the probability of observing each value of the random variable.
Such a distribution can take the form of a table, graph or formula.
Probability distributions describe populations, not samples.
We can find the mean μ, standard deviation σ, and variance σ2 of a discrete random variable using formulas.

26. 6.2 Binomial Probability Distribution
Objectives:
By the end of this section, I will be
able to…
Explain what constitutes a binomial experiment.
Compute probabilities using the binomial probability formula.
Find probabilities using the binomial tables.
Calculate and interpret the mean, variance, and standard deviation of the binomial random variable.

27. Binomial Experiment Four Requirements:
Each trial of the experiment has only two possible outcomes (success or failure)
Fixed number of trials
Experimental outcomes are independent of each other
Probability of observing a success remains the same from trial to trial

28. Notation
Table 6.6 Notation for binomial experiments and the binomial distribution

29. Formula for the Number of Combinations
The number of combinations of X items chosen from n different items is given by
where n! represents n factorial, which equals n(n - 1)(n - 2) (2)(1); and 0! is defined to be 1.

30. Example 6.13 - How many team combinations in the intramural volleyball league?
Jeffrey is in charge of drawing up a schedule for his college’s intramural volleyball league. This year five teams have been fielded, and they must play each other once. How many games will be held?

31. Example 6.13 continued Solution Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held.

32. Binomial Probability Distribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is P(X) = (nCX) pX (1 - p)n-X

33. Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
See Example 6.16 on page 278
for more information
FIGURE 6.7 Excerpt from the binomial tables.

34. Binomial Mean, Variance, and Standard Deviation
Mean (or expected value): μ = n · p
Variance: σ2 = n · p · (1 - p)
Standard deviation:

35. Example 6.18 - Mean, variance, and standard deviation of left-handed students
Suppose we know that the population proportion p of left-handed students is a. In a sample of 200 students, how many would we expect to be left-handed?

36. Example 6.18 continued Solution The binomial random variable here is
X = the number of left-handed students.
a. Here, n = 200, and p = 0.10.
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n · p = (200)(0.10) = 20

37. Summary
The most important discrete distribution is the binomial distribution, where there are two possible outcomes, each with probability of success p, and n independent trials.
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula.

38. Summary
Binomial probabilities can also be found using the binomial tables or using technology.
There are formulas for finding the mean, variance, and standard deviation of a binomial random variable.

39. 6.3 Continuous Random Variables and the Normal Probability Distribution
Objectives:
By the end of this section, I will be
able to…
Identify a continuous probability distribution.
Explain the properties of the normal probability distribution.

40. FIGURE 6.15 Relatively small sample (n = 100) with large class
widths (0.5 lb).
(b) Large sample
(n = 200)
with smaller class widths (0.2 lb).

41. Figure 6.15 continued (c) Very large sample (n = 400)
with very small class widths
(0.1 lb).
(d) Eventually, theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small.

42. Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable

43. Requirements The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables).
The vertical height of the density curve can never be negative. That is, the density curve never goes below the horizontal axis.

44. Probability Probability for Continuous Distributions
is represented by area under the curve
above an interval.

45. The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed, the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution

46. FIGURE 6.19
The normal distribution is symmetric about its mean μ.

47. Properties of the Normal Density Curve (Normal Curve)
It is symmetric about the mean μ.
The highest point occurs at X = μ, because symmetry implies that the mean equals the median, which equals the mode of the distribution.
It has inflection points at μ-σ and μ+σ.
The total area under the curve equals 1.

48. Properties of the Normal Density Curve (Normal Curve) continued
Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 0.5 (Figure 6.19).
The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions. As X moves farther from the mean, the density curve approaches but never quite touches the horizontal axis.

49. Drawing a Graph to Solve Normal Probability Problems
Draw a “generic” bell-shaped curve, with a horizontal number line under it that is labeled as the random variable X.
Insert the mean μ in the center of the number line.

50. Steps in Drawing a Graph to Help You Solve Normal Probability Problems
Mark on the number line the value of X indicated in the problem.
Shade in the desired area under the
normal curve to the right or left of this
value.
Proceed to find the desired area or probability.

51. Summary
Continuous random variables assume infinitely many possible values, with no gap between the values.
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve.

52. Summary
The normal distribution is the most important continuous probability distribution.
It is symmetric about its mean μ and has standard deviation σ.
One should always sketch a picture of a normal probability problem to help solve it.

53. 6.4 Standard Normal Distribution
Objectives:
By the end of this section, I will be
able to…
Find areas under the standard normal curve, given a Z-value.
Find the standard normal Z-value, given an area.

54. The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1.

55. Table 6.7
Steps for finding areas under the
standard normal curve

56. Example 6.29 - Expressing areas under the standard normal curve as probabilities
Reexpress the following areas as probabilities. a. In Example 6.24 page 299, we found the area under the standard normal curve to the right of Z = to be b. In Example 6.25, we found the area under the standard normal curve between Z = -1 and Z = 1 to be

57. Example 6.29 continued Solution a.
The probability that Z is greater than
-1.25 is
That is, P(Z > 1.25) = b.
The probability that Z is between -1 and 1 is
That is, P(1 < Z < 1) =

58. Example 6.30
Find the value of Z with area 0.90 to the left Find the standard normal Z-value that has area 0.90 to the left of it.

59. Example 6.30 continued Solution Step 1:
Draw the standard normal curve.
Label the Z-value Z1.
Step 2:
Shade the area to the left of Z1.
Remember, that we are given an area and are looking for a value of Z.
Label the area to the left of Z1 with the given area (0.90), as shown in Figure 6.35.

60. Example 6.30 continued Step 3
Look up 0.97 on the inside of the Z table.
The closest area is
Move from left 1.8
Then move up from to 0.08 (see Figure 6.38 page 305).
Putting these values together,
Z = = 1.88.
Z-value with area 0.03 to its right is Z=1.88.

61. Example 6.30 continued
FIGURE 6.35

62. Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1.
This distribution is often called the Z distribution.
The Z table and technology can be used to find areas under the standard normal curve.

63. Summary
In the Z table, the numbers on the outside are values of Z, and the numbers inside are areas to the left of values of Z.
The Z table and technology can also be used to find a value of Z, given a probability or an area under the curve.

64. 6.5 Applications of the Normal Distribution
Objectives:
By the end of this section, I will be
able to…
Compute probabilities for a given value of any normal random variable.
Find the appropriate value of any normal random variable, given an area or probability.
Calculate binomial probabilities using the normal approximation to the binomial distribution.

65. Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula

66. Finding Probabilities for Any Normal Distribution
Step 1
Determine the random variable X, the mean μ, and the standard deviation σ.
Draw the normal curve for X, and shade the desired area.
Step 2
Standardize by using the formula
Z = (X - μ )/σ to find the values of Z
corresponding to the X-values.

67. Finding Probabilities for Any Normal Distribution
Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X.
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology.
This area is equal to the area under the normal curve for X drawn in Step 1.

68. Finding Normal Data Values for Specified Probabilities
Step 1
Determine X, μ, and σ, and draw the normal curve for X.
Shade the desired area.
Mark the position of X1, the unknown value of X.

69. Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area.
Look up the area you identified in Step 1 on the inside of the Z table.
If you do not find the exact value of your area, by convention choose the area that is closest.

70. Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X, which is the solution.
Use the formula X1 = Zσ + μ.

71. Example 6.38 - Finding the X-values that mark the Boundaries of the middle 95% of X-values
Edmunds.com reported that the average amount that people were paying for a 2007 Toyota Camry XLE was $23,400. Let X = price, and assume that price follows a normal distribution with μ = $23,400 and σ = $1000. Find the prices that separate the middle 95% of 2007 Toyota Camry XLE prices from the bottom 2.5% and the top 2.5%.

72. Example 6.38 continued Solution Step 1
Determine X, μ, and σ, and draw the normal curve for X.
Let X = price, μ = $23,400, and σ = $1000.
The middle 95% of prices are delimited by X1 and X2, as shown in Figure 6.49.
FIGURE 6.49

73. Example 6.38 continued Solution Step 2
Find the Z-values corresponding to the desired area.
The area to the left of X1 equals 0.025, and the area to the left of X2 equals
Looking up area on the inside of the Z table gives us Z1 = –1.96.
Looking up area on the inside of the Z table gives us Z2 = 1.96.

74. Example 6.38 continued Solution Step 3
Transform using the formula X1 = Zσ + μ.
X1 = Z1σ + μ
=(–1.96)(1000) + 23,400 = 21,440
X2 =Z2σ + μ
=(1.96)(1000) + 23,400 = 25,360
The prices that separate the middle 95% of Toyota Camry XLE prices from the bottom 2.5% of prices and the top 2.5% of prices are $21,440 and $25,360.

75. The Normal Approximation to the Binomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n:
if n · p ≥ 5 and n · (1 - p) ≥ 5, the binomial distribution may be approximated by a normal distribution with mean
μX = np
and
standard deviation σX =

76. Summary
Section 6.5 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 6.4.
Methods for finding probabilities for a given value of the normal random variable X were discussed.

77. Summary
For any normal probability distribution, values of X can be found for given probabilities using the “backward” methods of Section 6.4 and the formula X1 = Z1σ + μ.
The normal distribution can be used to approximate binomial probabilities when np ≥ 5 and np(1 - p) ≥ 5.