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Engineering Statistics - IE 261
Chapter 3
Discrete Random Variables and
Probability Distributions
URL: http://home.npru.ac.th/piya/ClassesTU.html
http://home.npru.ac.th/piya/webscilab
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3-1 Discrete Random Variables
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3-1 Discrete Random Variables
Example 3-1
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3-2 Probability Distributions and
Probability Mass Functions
Figure 3-1 Probability distribution for bits in error.
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3-2 Probability Distributions and
Probability Mass Functions
Definition
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Example 3-5
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Example 3-5 (continued)
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3-3 Cumulative Distribution Functions
Definition
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Example 3-8
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Example 3-8
Figure 3-4 Cumulative distribution function for Example 3-12
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3-4 Mean and Variance of a Discrete
Random Variable
Definition
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3-4 Mean and Variance of a Discrete
Random Variable
Figure 3-5 A probability distribution can be viewed as a loading with
the mean equal to the balance point. Parts (a) and (b) illustrate equal
means, but Part (a) illustrates a larger variance.
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Proof of Variance:
 2 V X   EX   
2
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3-4 Mean and Variance of a Discrete
Random Variable
Figure 3-6 The probability distribution illustrated in Parts (a) and (b)
differ even though they have equal means and equal variances.
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Example 3-9
There is a chance that a bit transmitted through a digital transmission
channel is received in error. Let X equal the number of bits in error in
the next four bits transmitted. The possible values for X are {0, 1, 2, 3, 4}
Suppose:
P(X = 0) = 0.6561
P(X = 1) = 0.2916
P(X = 3) = 0.0036
P(X = 4) = 0.0001
P(X = 2) = 0.0486
Find the mean and the variance of X
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Example 3-9 (Solution)
  E( X ) 
SCILAB
-->x = [0 1 2 3 4];
-->fx = [0.6561 0.2916 0.0486 0.0036
-->MeanX = sum(x.*fx)
MeanX =
0.4
-->VarX = sum((x.^2).*fx) - MeanX^2
VarX =
0.36
0.0001];
   xk f  xk 
k
 2   xk2 f  xk  18 2
k
Example 3-9 (Solution)
 2  V (X ) 
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Example 3-11
-->x = [10:15];
fx = [0.08 0.15 0.3
-->MeanX = sum(x.*fx)
MeanX =
12.5
-->VarX = sum((x.^2).*fx) - MeanX^2
VarX =
1.85
0.2
0.2
0.07];
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