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Chapter 8
Random Variables and
Probability Distributions
I Random Sampling
A. Population
1. Population element
2. Sampling with and without replacement
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B. Random Sampling Procedures
1. Table of random numbers (from Appendix D.1)
6 7 8 9 ...
71 50 54 36 . . .
64 85 27 20
59 19 18 97
82 82 11 54
16 86 20 26
21 22 23 24 25
26 78 25 47 47
94 76 62 11 89
84 97 50 87 46
42 34 43 39 28
52 01 63 01 59
6 91 76 21 64 64
7 00 97 79 08 06
44 91 13 32
37 30 28 59
56 08 25 70 29
30 19 99 85 48
50 87 41 60 76 83
44 88 96 07 . . .
30 56 10 48 59
1
2
3
4
5
1 2 3 4 5
10 27 53 96 23
28 41 50 61 88
34 21 42 57 02
61 81 77 23 23
61 15 18 13 54
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II Random Variables and Their Distributions
A. Random Variable: A Numerically Valued
Function Defined on a Sample Space
1. A function consists of two sets of elements
and a rule that assigns to each element in
the first set one and only one element in the
second set.
2. Examples: {(a, 1), (b, 5), (c, 6)}
{(Mike, tall), (Jim, short), (Joe, medium)}
3. If the second element is a number, the function is
numerically valued.
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4. A random variable associates one and only one
number with each point in a sample space; thus,
it is a numerically valued function defined on a
sample space.
Example: consider tossing a fair coin; points in the
sample space can be associated with numbers on
the real number line.
H
T
0
1
0 if coin is T
X  1 if coin is H

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5. The random variable X is the name for any one of
a set of permissible numerical values of a random
experiment.
6. Discrete random variable: range can assume only
a finite number of values or an infinite number of
values that is countable.
7. Continuous random variable: range is uncountably
infinite.
5
B. Probability Distribution
1. Probability distribution for tossing a fair coin
X
p(X = r)
0
1
1/2
1/2
Probability
2. Graph of the probability distribution
.5
0
1
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3. Three-section T maze
G
S
4. Correct series of turns: R L R
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5. Number of ways of traversing the T maze:
2  2  2 = 8 (fundamental counting rule)
_______________________
Turns
Number of errors, X
R, L, R
0
R, R, R
1
R, L, L
1
L, L, R
1
R, R, L
2
L, R, R
2
L, L, L
2
L, R, L
3
______________________________
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Probability Distribution for Number of
Errors in the Three-Choice T Maze
_________________________
Possible Values of the
Random Variable X
p(X = r)
0
.125
1
.375
2
.375
3
.125
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Probability
6. Graph of the probability distribution for the ThreeChoice T Maze
.4
.3
.2
.1
0 1 2 3
Number of errors
C. Expected Value of a Discrete Random
Variable
E(X) = p(X1)X1 + p(X2)X2 + . . . + p(Xn)Xn =  p X i X i
n
i 1
where p(X1) + p(X2) + . . . + p(Xn) = 1
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1. For the T maze example, the expected value is
E(X) = p(X1)X1 + p(X2)X2 + p(X3)X3 + p(X4)X4
Probability
= .125(0) + .375(1) + .375(2) + .125(3) = 1.5
.4
.3
.2
.1
0 1 2 3
Number of errors
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2. Expected value of a bet at the roulette table; you
pay $1.00 to win $35.00. The wheel has 38 slots.
Possible Winnings, Xi
p(Xi)
p(Xi)Xi
+ $35
1/38
1/38($35) =
– $1
37/38
37/38(–$1) = –37/38
35/38
 35 37   2
E ( X )   p( X i ) X i     
 .053
 38 38  38
i 1
n
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D. Standard Deviation of a Discrete Random
Variable
   p( X i )  X i  E( X i ) 
2
1. For the T maze example, the standard deviation,
 , is
 .125(0  1.5)2  .375(1 1.5)2  .375(2  1.5)2  .125(3  1.5)2
= 0.866
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E. Expected Value of a Continuous Random
Variable
1. A continuous random variable can assume an
infinite number of values. The probability that a
continuous random variable, X, has a particular
value is zero. Hence, we refer to the probability
that X lies in an interval between two values of the
random variable.
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2. Distribution for a continuous random variable
f (X)
a
b
X
The probability that X will assume a value between a and
b is equal to the area under the curve between those two
points.
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III Binomial Distribution
A. Three Characteristics of a Bernoulli Trial
B. Binomial Distribution
1. Binomial random variable: number of
successes observed on n ≥ 2 identical
Bernoulli trials
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2. Binomial function rule: probability of observing
exactly r heads (successes) in n trials is given by
p(X = r) = nCr prqn – r
where p(X = r) is the probability that the random
variable X equals r successes, nCr is the
combination of n objects taken r at a time, p is the
probability of a success, and q = 1 – p is the
probability of a failure.
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3. Consider tossing n = five fair coins: the
probability of observing r = 0, 1, . . . , 5 heads is
r
nr
 1  1
n!

r!(n  r)!  2   2 
0
50
 1  1
5!

0!(5  0)!  2   2 
 1  1
p( X  r)  nCr    
 2  2
 1  1
p( X  0)  5C0    
 2  2
1
 1  1
p( X  1)  5C1    
 2  2
51
r
nr
0
50
1
51
5!  1   1 

1!(5  1)!  2   2 
1

32
5

32
18
2
 1  1
p( X  2)  5C2    
 2  2
3
52
 1  1
p( X  3)  5C3    
 2  2
2

53
 1  1
5!
2!(5  2)!  2   2 
3
52
5!  1   1 

3!(5  3)!  2   2 

53
4
54
 1  1
5!

4!(5  4)!  2   2 
5
55
5!  1   1 

5!(5  5)!  2   2 
 1  1
p( X  4)  5C4    
 2  2
 1  1
p( X  5)  5C5    
 2  2
10
32
10

32
4
54
5
55
5

32
1

32
19
Probability
4. Binomial distribution for tossing five fair coins
10/32
8/32
6/32
4/42
2/32
0
1
2 3 4
5
Number of Heads, r
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C. Expected Value and Standard Deviation of a
Binomial Random Variable
1. Expected value
n
E( X )   p( X i ) X i  np
i 1
2. Standard deviation
2
   p( X i )  X i  E( X i )   npq
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3. For the coin tossing experiment where n = 5 and
p = q = 1/2
E( X )  np  (5)(.5)  2.5
Probability
  npq  5(1 / 2)(1 / 2)  1.118
10/32
8/32
6/32
4/42
2/32
0
1
2 3 4
5
Number of Heads, r
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D. Binomial Model Is Appropriate Under the
Following Conditions
1. There are n trials involving a population whose
elements belong to one of two classes
2. Probability of obtaining an element remains
constant from trial to trial, as when sampling with
replacement from a finite population
3. Outcomes of successive trials are independent
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E. Two Other Models
1. Multinomial distribution is an extension of the
binomial distribution for the case in which there
are more than two classes. It is identical to the
binomial distribution when there are only two
classes.
 Probability of obtaining an element remains
constant from trial to trial, as when sampling
with replacement
 Outcomes of successive trials are independent
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2. Hypergeometric distribution is appropriate
for the case in which there are more than two
classes but the probabilities associated with the
classes do not remain constant as when sampling
without replacement from a finite population.
25