Engineering Statistics - IE 261
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Transcript of Engineering Statistics - IE 261
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Engineering Statistics - IE 261
Chapter 3Discrete Random Variables andProbability Distributions
URL: http://home.npru.ac.th/piya/ClassesTU.htmlhttp://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 FunctionsDefinition
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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 3-8.
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3-4 Mean and Variance of a Discrete Random VariableDefinition
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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:
22 V X E X
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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 3-9There is a chance that a bit transmitted through a digital transmissionchannel is received in error. Let X equal the number of bits in error inthe 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 = 2) = 0.0486
P(X = 3) = 0.0036 P(X = 4) = 0.0001
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 0.0001];-->MeanX = sum(x.*fx) MeanX = 0.4 -->VarX = sum((x.^2).*fx) - MeanX^2 VarX = 0.36
k kk
x f x
2 2 2k k
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Example 3-9 (Solution)
2 ( )V X
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Example 3-11
-->x = [10:15]; fx = [0.08 0.15 0.3 0.2 0.2 0.07];-->MeanX = sum(x.*fx) MeanX = 12.5-->VarX = sum((x.^2).*fx) - MeanX^2 VarX = 1.85