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<math>
 
<math>
Given: Prob (0 < x < 2)
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Prob (0 < x < 2) = \Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma})
</math>
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+
<math>
+
= \Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma})
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</math>
 
</math>
  
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<math>
 
<math>
Given: Prob (-2.5 < x < .5)
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Prob (-2.5 < x < .5) = \Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma})
</math>
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+
<math>
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= \Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma})
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</math>
 
</math>
  

Latest revision as of 04:48, 24 March 2013

[[Category:gaussian random variable

Practice Problem: Compare Probabilities for different Gaussians


A (one-dimensional) random variable X is normally distributed with mean equal to one and standard deviation equal to two. Another (one-dimensional) random variable Y is normally distributed with mean equal to minus one and standard deviation equal to three.

Is $ \text{Prob } ( 0 < X < 2) $ greater than $ \text{Prob } ( -2.5 < Y < 0.5) \text{ ?} $ Explain.


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Answer 1

$ Given: \mu = 1\sigma = 2 $

$ Prob (0 < x < 2) = \Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma}) $

$ = \Phi(\frac{2-1}{2}) - \Phi(\frac{0-1}{2}) $

$ =\Phi(\frac{1}{2}) - \Phi(\frac{1}{2}) = 0 $

$ Given: \mu = -1 \sigma = 3 $

$ Prob (-2.5 < x < .5) = \Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma}) $

$ = \Phi(\frac{.5+1}{3}) - \Phi(\frac{-2.5+1}{3}) $

$ =\Phi(\frac{1}{2}) - \Phi(\frac{1}{2}) = 0 $

No, they are the same.

Answer 2

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Answer 3

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Back to ECE302 Spring 2013 Prof. Boutin

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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