1. Statistics for Business and Economics
Random Variables & Probability Distributions

2. Learning Objectives
Distinguish Between the Two Types of Random Variables
Describe Discrete Probability Distributions
Describe the Binomial and Poisson Distributions
Describe the Uniform and Normal Distributions
As a result of this class, you will be able to...

3. Learning Objectives (continued)
Approximate the Binomial Distribution Using the Normal Distribution
Explain Sampling Distributions
Solve Probability Problems Involving Sampling Distributions

4. Thinking Challenge
You’re taking a 33 question multiple choice test. Each question has 4 choices. Clueless on 1 question, you decide to guess. What’s the chance you’ll get it right?
If you guessed on all 33 questions, what would be your grade? Would you pass?
The ‘pass’ question is meant to be a ‘teaser’ and not answered.

5. Types of Random Variables

6. Data Types
Data
Quantitative
Qualitative
Continuous
Discrete

7. Discrete Random Variables

8. Data Types
Data
Quantitative
Qualitative
Continuous
Discrete

9. Discrete Random Variable
A numerical outcome of an experiment
Example: Number of tails in 2 coin tosses
Discrete random variable
Whole number (0, 1, 2, 3, etc.)
Obtained by counting
Usually a finite number of values
Poisson random variable is exception ()

10. Discrete Random Variable Examples
Possible Values
Experiment
Make 100 Sales Calls
# Sales
0, 1, 2, ..., 100
Inspect 70 Radios
# Defective
0, 1, 2, ..., 70
Answer 33 Questions
# Correct
0, 1, 2, ..., 33
Count Cars at Toll
Between 11:00 & 1:00
# Cars
Arriving
0, 1, 2, ..., ∞

11. Continuous Random Variables

12. Data Types
Data
Quantitative
Qualitative
Continuous
Discrete

13. Continuous Random Variable
A numerical outcome of an experiment
Weight of a student (e.g., 115, 156.8, etc.)
Continuous Random Variable
Whole or fractional number
Obtained by measuring
Infinite number of values in interval
Too many to list like a discrete random variable

14. Continuous Random Variable Examples
Possible Values
Experiment
Weigh 100 People
Weight
45.1, 78, ...
Measure Part Life
Hours
900, 875.9, ...
Amount spent on food
$ amount
54.12, 42, ...
Measure Time
Between Arrivals
Inter-Arrival
Time
0, 1.3, 2.78, ...

15. Probability Distributions for Discrete Random Variables

16. Discrete Probability Distribution
List of all possible [x, p(x)] pairs
x = value of random variable (outcome)
p(x) = probability associated with value
Mutually exclusive (no overlap)
Collectively exhaustive (nothing left out)
0  p(x)  1 for all x
 p(x) = 1

17. Discrete Probability Distribution Example
Experiment: Toss 2 coins. Count number of tails.
Probability Distribution
Values, x Probabilities, p(x)
0 1/4 = .25
1 2/4 = .50
2 1/4 = .25
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18. Visualizing Discrete Probability Distributions
Listing
Table
{ (0, .25), (1, .50), (2, .25) }
f(x)
p(x)
# Tails
Count 1
.25 1 2
.50
Experiment is tossing 1 coin twice.
Graph 2 1
.25
p(x)
.50
Formula
.25 x n !
.00 p ( x ) =
px(1 – p)n - x 1 2
x!(n – x)!

19. Summary Measures Expected Value (Mean of probability distribution)
Weighted average of all possible values
 = E(x) = x p(x)
Variance
Weighted average of squared deviation about mean
2 = E[(x (x p(x)
population notation is used since all values are specified.
Standard Deviation ●

20. Summary Measures Calculation Tablex
p(x)
x p(x)
x – 
(x – )2
(x – )2p(x)
Total
xp(x)
(x p(x)

21. Thinking Challenge
You toss 2 coins. You’re interested in the number of tails. What are the expected value, variance, and standard deviation of this random variable, number of tails?
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22. Expected Value & Variance Solution*
p(x)
x p(x)
x – 
(x – ) 2
(x – ) 2p(x)
.25
.50 
= 1.0
-1.00
1.00
.25
2 = .50
 = .71 1
.50 2
.25
1.00
1.00

23. Discrete Probability Distributions
Binomial
Poisson

24. Binomial Distribution

25. Discrete Probability Distributions
Binomial
Poisson

26. Binomial Distribution
Number of ‘successes’ in a sample of n observations (trials)
Number of reds in 15 spins of roulette wheel
Number of defective items in a batch of 5 items
Number correct on a 33 question exam
Number of customers who purchase out of 100 customers who enter store

27. Binomial Distribution Properties
Two different sampling methods
Infinite population without replacement
Finite population with replacement
Sequence of n identical trials
Each trial has 2 outcomes
‘Success’ (desired outcome) or ‘Failure’
Constant trial probability
Trials are independent

28. Binomial Probability Distribution Function
p(x) = Probability of x ‘Successes’
n = Sample Size
p = Probability of ‘Success’
x = Number of ‘Successes’ in Sample (x = 0, 1, 2, ..., n)

29. Binomial Probability Distribution Example
Experiment: Toss 1 coin 5 times in a row. Note number of tails. What’s the probability of 3 tails?
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30. Binomial Distribution Characteristics
n = 5 p = 0.1
Mean
Distribution has different shapes.
1st Graph:
If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 0 defective item is about 0.6 (60%).
If inspecting 5 items & the Probability of a defect is 0.1 (10%), the Probability of finding 1 defective items is about .35 (35%).
2nd Graph:
If inspecting 5 items & the Probability of a defect is 0.5 (50%), the Probability of finding 1 defective items is about .18 (18%).
Note:
Could use formula or tables at end of text to get Probabilities.
Standard Deviation
n = 5 p = 0.5

31. Binomial Distribution Thinking Challenge
You’re a telemarketer selling service contracts for Macy’s. You’ve sold 20 in your last 100 calls (p = .20). If you call 12 people tonight, what’s the probability of
A. No sales?
B. Exactly 2 sales?
C. At most 2 sales?
D. At least 2 sales?
Let’s conclude this section on the binomial with the following Thinking Challenge.

32. Binomial Distribution Solution*
n = 12, p = .20
A. p(0) = .0687
B. p(2) = .2835
C. p(at most 2) = p(0) + p(1) + p(2) = = .5584
D. p(at least 2) = p(2) + p(3)...+ p(12) = 1 – [p(0) + p(1)] = 1 – – = .7251
From the Binomial Tables:
A. p(0) = .0687
B. p(2) = .2835
C. p(at most 2) = p(0) + p(1) + p(2) = = .5584
D. p(at least 2) = p(2) + p(3)...+ p(12) = 1 - [p(0) + p(1)] = = .7251

33. Poisson Distribution

34. Discrete Probability Distributions
Binomial
Poisson

35. Poisson Distribution Number of events that occur in an interval
events per unit
Time, Length, Area, Space
Examples
Number of customers arriving in 20 minutes
Number of strikes per year in the U.S.
Number of defects per lot (group) of DVD’s
Other Examples:
Number of machines that break down in a day
Number of units sold in a week
Number of people arriving at a bank teller per hour
Number of telephone calls to customer support per hour

36. Poisson Process Constant event probability One event per interval
Average of 60/hr is 1/min for 60 1-minute intervals
One event per interval
Don’t arrive together
Independent events
Arrival of 1 person does not affect another’s arrival
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37. Poisson Probability Distribution Functionx ( ) !   e -
p(x) = Probability of x given 
 = Expected (mean) number of ‘successes’
e = (base of natural logarithm)
x = Number of ‘successes’ per unit

38. Poisson Distribution Characteristics
= 0.5
Mean
= 6
Standard Deviation

39. Poisson Distribution Example
Customers arrive at a rate of 72 per hour. What is the probability of 4 customers arriving in 3 minutes?
© 1995 Corel Corp.

40. Poisson Distribution Solution
72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval

41. Thinking Challenge
You work in Quality Assurance for an investment firm. A clerk enters 75 words per minute with 6 errors per hour. What is the probability of 0 errors in a 255-word bond transaction?
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42. Poisson Distribution Solution: Finding *
75 words/min = (75 words/min)(60 min/hr)
= 4500 words/hr
6 errors/hr = 6 errors/4500 words
= errors/word
In a 255-word transaction (interval):
 = ( errors/word )(255 words)
= .34 errors/255-word transaction

43. Poisson Distribution Solution: Finding p(0)*

44. Data Types
Data
Quantitative
Qualitative
Continuous
Discrete

45. Probability Distributions for Continuous Random Variables

46. Continuous Probability Density Function
Mathematical formula
Shows all values, x, and frequencies, f(x)
f(x) Is Not Probability
Value
(Value, Frequency)
Frequency
f(x) a b x
(Area Under Curve) f x dx ( )
All x a b     1 0,
Properties

47. Continuous Random Variable Probability( a  x  b )   f ( x ) dx
Probability Is Area Under Curve! a
f(x) x a b
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48. Continuous Probability Distributions
Uniform
Normal

49. Uniform Distribution

50. Continuous Probability Distributions
Uniform
Normal

51. Uniform Distribution x f(x) d c a b 1. Equally likely outcomes
2. Probability density function
3. Mean and Standard Deviation

52. Uniform Distribution Example
You’re production manager of a soft drink bottling company. You believe that when a machine is set to dispense 12 oz., it really dispenses 11.5 to 12.5 oz. inclusive. Suppose the amount dispensed has a uniform distribution. What is the probability that less than 11.8 oz. is dispensed?
SODA

53. Uniform Distribution Solution
f(x)
1.0 x
11.5
11.8
12.5
P(11.5  x  11.8) = (Base)(Height)
= ( )(1) = .30

54. Normal Distribution

55. Continuous Probability Distributions
Uniform
Normal

56. Importance of Normal Distribution
Describes many random processes or continuous phenomena
Can be used to approximate discrete probability distributions
Example: binomial
Basis for classical statistical inference

57. Normal Distribution f ( x ) x ‘Bell-shaped’ & symmetrical
Mean, median, mode are equal
‘Middle spread’ is 1.33
Random variable has infinite range f ( x ) x
Mean Median Mode

58. Probability Density Function
f(x) = Frequency of random variable x
 = Population standard deviation
 = ; e =
x = Value of random variable (– < x < )
 = Population mean

59. Effect of Varying Parameters ( & )
f(X) B A C X

60. Normal Distribution Probability
Probability is area under curve! f ( x ) x c d

61. The Standard Normal Table: P(0 < z < 1.96)
Standardized Normal Probability Table (Portion)
.06 Z
.04
.05 Z m
= 0 s
= 1
1.96
1.8
.4671
.4678
.4686
.4750
1.9
.4750
.4738
.4744
2.0
.4793
.4798
.4803
2.1
.4838
.4842
.4846
Shaded area exaggerated
Probabilities

62. The Standard Normal Table: P(–1.26  z  1.26)
Standardized Normal Distribution s
= 1
P(–1.26 ≤ z ≤ 1.26) =
= .7924
.3962
.3962
–1.26
1.26 Z m
= 0
Shaded area exaggerated

63. The Standard Normal Table: P(z > 1.26)
Standardized Normal Distribution s
= 1
P(z > 1.26)
= – .3962
= .1038
.5000
.3962
1.26 Z m
= 0

64. The Standard Normal Table: P(–2.78  z  –2.00)
Standardized Normal Distribution s
= 1
P(–2.78 ≤ z ≤ –2.00)
= – .4772
= .0201
.4973
.4772
–2.78
–2.00 Z m
= 0
Shaded area exaggerated

65. The Standard Normal Table: P(z > –2.13)
Standardized Normal Distribution s
= 1
P(z > –2.13) =
= .9834
.4834
.5000
–2.13 Z m
= 0
Shaded area exaggerated

66. Non-standard Normal Distribution
Normal distributions differ by mean & standard deviation.
Each distribution would require its own table.
That’s an infinite number of tables! X
f(X)

67. Standardize the Normal Distribution
One table! m
= 0 s
= 1 Z
Standardized Normal Distribution s m X

68. Non-standard Normal μ = 5, σ = 10: P(5 < X< 6.2)
Normal Distribution X m
= 5 s
= 10
6.2 Z m
= 0 s
= 1
.12
Standardized Normal Distribution
Shaded area exaggerated
.0478

69. Non-standard Normal μ = 5, σ = 10: P(3.8  X  5)
Shaded area exaggerated
Normal Distribution X m
= 5 s
= 10
3.8 Z m
= 0 s
= 1
-.12
Standardized Normal Distribution
.0478

70. Non-standard Normal μ = 5, σ = 10: P(2.9  X  7.1)
Shaded area exaggerated 5 s
= 10
2.9
7.1 X
Normal Distribution s
= 1
-.21 Z
.21
Standardized Normal Distribution
.1664
.0832

71. Non-standard Normal μ = 5, σ = 10: P(X  8)
Shaded area exaggerated X m
= 5 s
= 10 8
Normal Distribution Z
= 0
.30
Standardized Normal Distribution m s
= 1
.5000
.3821
.1179

72. Non-standard Normal μ = 5, σ = 10: P(7.1  X  8)
Shaded area exaggerated m
= 5 s
= 10 8
7.1 X
Normal Distribution m
= 0 s
= 1
.30 Z
.21
Standardized Normal Distribution
.1179
.0347
.0832

73. Normal Distribution Thinking Challenge
You work in Quality Control for GE. Light bulb life has a normal distribution with = 2000 hours and = 200 hours. What’s the probability that a bulb will last
A. between 2000 and hours?
B. less than 1470 hours?
Allow students about minutes to solve this.

74. Solution* P(2000  X  2400) X Z .4772 m = 2000 s = 200 2400 m = 0 s
Normal Distribution X m
= 2000 s
= 200
2400
Standardized Normal Distribution Z m
= 0 s
= 1
2.0
.4772

75. Solution* P(X  1470) X Z .0040 m = 2000 s = 200 1470 m = 0 s = 1
Normal Distribution Z m
= 0 s
= 1
-2.65
Standardized Normal Distribution
.5000
.0040
.4960

76. Finding Z Values for Known Probabilities
What is Z, given P(Z) = .1217?
Shaded area exaggerated Z m
= 0 s
= 1 ?
.1217
Standardized Normal Probability Table (Portion) Z
.00
0.2
0.0
.0000
.0040
.0080
0.1
.0398
.0438
.0478
.0793
.0832
.0871
.1179
.1255
.01
0.3
.1217
.31

77. Finding X Values for Known Probabilities
Normal Distribution
Standardized Normal Distribution
Shaded areas exaggerated Z m
= 0 s
= 1
.31
.1217 X m
= 5 s
= 10 ?
.1217
8.1

78. Assessing Normality

79. Assessing Normality
Draw a histogram or stem–and–leaf display and note the shape
Compute the intervals x + s, x + 2s, x + 3s and compare the percentage of data in these intervals to the Empirical Rule (68%, 95%, 99.7%)
Calculate
If ratio is close to 1.3, data is approximately normal

80. Assessing Normality Continued
Draw a Normal Probability Plot
Observed value
Expected Z–score

81. Normal Approximation of Binomial Distribution

82. Normal Approximation of Binomial Distribution
Not all binomial tables exist
Requires large sample size
Gives approximate probability only
Need correction for continuity
n = 10 p = 0.50 .0 .1 .2 .3 2 4 6 8 10 x
P(x)

83. Why Probability Is Approximate
Probability Added by Normal Curve
Probability Lost by Normal Curve
P(x) .3 .2
As the number of vertical bars (n) increases, the errors due to approximating with the normal decrease. .1 .0 x 2 4 6 8 10
Binomial Probability: Bar Height
Normal Probability: Area Under Curve from 3.5 to 4.5

84. Correction for Continuity
A 1/2 unit adjustment to discrete variable
Used when approximating a discrete distribution with a continuous distribution
Improves accuracy
4.5 (4 + .5)
3.5 (4 – .5) 4

85. Normal Approximation Procedure
1. Calculate the interval:
If interval lies in range 0 to n, normal approximation can be used
2. Express binomial probability in form
3. For each value of interest, a, use:

86. Normal Approximation Example
What is the normal approximation of p(x = 4) given n = 10, and p = 0.5?
P(x) .3 .2 .1 .0 x 2 4 6 8 10
3.5
4.5

87. Normal Approximation Solution
1. Calculate the interval:
Interval lies in range 0 to 10, so normal approximation can be used
2. Express binomial probability in form:

88. Normal Approximation Solution
3. Compute standard normal z values:
(a + .5)  n  p
(.5) Z     . 95 n  p ( 1  p )
10(.5)(1 - .5)
(b + .5)  n  p
(.5) Z     . 32 n  p ( 1  p )
10(.5)(1 - .5)

89. Normal Approximation Solution
4. Sketch the approximate normal distribution: 
= 0 
= 1
.1255
.3289
-.95
-.32 Z

90. Normal Approximation Solution
5. The exact probability from the binomial formula is (versus .2034)
P(x) .3 .2 .1 .0 x 2 4 6 8 10

92. Sampling Distributions

93. Parameter & Statistic Parameter Sample Statistic
Summary measure about population
Sample Statistic
Summary measure about sample
P in Population & Parameter
S in Sample & Statistic

94. Common Statistics & Parameters
Sample Statistic
Population Parameter
Mean  X 
Standard Deviation S 
Variance S2 2
Binomial Proportion p ^

95. Sampling Distribution
Theoretical probability distribution
Random variable is sample statistic
Sample mean, sample proportion, etc.
Results from drawing all possible samples of a fixed size
4. List of all possible [x, p(x)] pairs
Sampling distribution of the sample mean

96. Developing Sampling Distributions
Suppose There’s a Population ...
Population size, N = 4
Random variable, x
Values of x: 1, 2, 3, 4
Uniform distribution
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97. Population Characteristics
Summary Measures
Population Distribution
P(x) .3 .2
Have students verify these numbers. .1 .0 x 1 2 3 4

98. All Possible Samples of Size n = 2
2nd Observation 1 2 3 4
1st
Obs
16 Sample Means
2nd Observation 1 2 3 4
1st
Obs
1,1
1,2
1,3
1,4
1.0
1.5
2.0
2.5
2,1
2,2
2,3
2,4
1.5
2.0
2.5
3.0
3,1
3,2
3,3
3,4
2.0
2.5
3.0
3.5
4,1
4,2
4,3
4,4
2.5
3.0
3.5
4.0
Sample with replacement

99. Sampling Distribution of All Sample Means
2nd Observation 1 2 3 4
1st
Obs
16 Sample Means
Sampling Distribution of the Sample Mean .0 .1 .2 .3
1.0
1.5
2.0
2.5
3.0
3.5
4.0
P(x) x
1.0
1.5
2.0
2.5
1.5
2.0
2.5
3.0
2.0
2.5
3.0
3.5
2.5
3.0
3.5
4.0

100. Summary Measures of All Sample Means
Have students verify these numbers.

101. Sampling Distribution
Comparison
Population
Sampling Distribution
P(x) .0 .1 .2 .3 1 2 3 4 .0 .1 .2 .3
1.0
1.5
2.0
2.5
3.0
3.5
4.0
P(x) x x

102. Standard Error of the Mean
1. Standard deviation of all possible sample means, x ● Measures scatter in all sample means, x
Less than population standard deviation
3. Formula (sampling with replacement)

103. Properties of the Sampling Distribution of x

104. Properties of the Sampling Distribution of x
Regardless of the sample size,
The mean of the sampling distribution equals the population mean
An estimator is a random variable used to estimate a population parameter (characteristic).
Unbiasedness
An estimator is unbiased if the mean of its sampling distribution is equal to the population parameter.
Efficiency
The efficiency of an unbiased estimator is measured by the variance of its sampling distribution.
If two estimators, with the same sample size, are both unbiased, then the one with the smaller variance has greater relative efficiency.
Consistency
An estimator is a consistent estimator of a population parameter if the larger the sample size, the more likely it is that the estimate will come close to the parameter.
The standard deviation of the sampling distribution equals

105. Sampling from Normal Populations

106. Sampling from Normal Populations
Central Tendency
Dispersion
Sampling with replacement
Population Distribution m
= 50 s
= 10 X
Sampling Distribution
n =16 X = 2.5
n = 4 X = 5 m X
= 50 -

107. Standardizing the Sampling Distribution of x
Standardized Normal Distribution m
= 0 s
= 1 Z

108. Thinking Challenge
You’re an operations analyst for AT&T. Long-distance telephone calls are normally distribution with  = 8 min. and  = 2 min. If you select random samples of 25 calls, what percentage of the sample means would be between 7.8 & 8.2 minutes?
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109. Sampling Distribution Solution*8 s ` X
= .4
7.8
8.2 s
= 1
–.50 Z
.50
.3830
Standardized Normal Distribution
.1915

110. Sampling from Non-Normal Populations

111. Sampling from Non-Normal Populations
Central Tendency
Dispersion
Sampling with replacement
Population Distribution s
= 10 m
= 50 X
Sampling Distribution
n = 4 X = 5
n =30 X = 1.8 m -
= 50 X X

112. Central Limit Theorem X As sample size gets large enough (n  30) ...
sampling distribution becomes almost normal. X

113. Central Limit Theorem Example
The amount of soda in cans of a particular brand has a mean of 12 oz and a standard deviation of .2 oz. If you select random samples of 50 cans, what percentage of the sample means would be less than oz?
SODA

114. Central Limit Theorem Solution*
Shaded area exaggerated
Sampling Distribution 12 s ` X
= .03
11.95 s
= 1
–1.77 Z
.0384
Standardized Normal Distribution
.4616

115. Conclusion Distinguished Between the Two Types of Random Variables
Described Discrete Probability Distributions
Described the Binomial and Poisson Distributions
Described the Uniform and Normal Distributions
Approximated the Binomial Distribution Using the Normal Distribution
As a result of this class, you will be able to...

116. Conclusion (continued)
Explained Sampling Distributions
Solved Probability Problems Involving Sampling Distributions