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===Answer 1=== | ===Answer 1=== | ||
| − | + | <math> | |
| + | Given: \mu = 1\sigma = 2 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | Prob (0 < x < 2) = \Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma}) | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | = \Phi(\frac{2-1}{2}) - \Phi(\frac{0-1}{2}) | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | =\Phi(\frac{1}{2}) - \Phi(\frac{1}{2}) = 0 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | Given: \mu = -1 \sigma = 3 | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | Prob (-2.5 < x < .5) = \Phi(\frac{b-\mu}{\sigma}) - \Phi(\frac{a-\mu}{\sigma}) | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | = \Phi(\frac{.5+1}{3}) - \Phi(\frac{-2.5+1}{3}) | ||
| + | </math> | ||
| + | |||
| + | <math> | ||
| + | =\Phi(\frac{1}{2}) - \Phi(\frac{1}{2}) = 0 | ||
| + | </math> | ||
| + | |||
| + | No, they are the same. | ||
===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Latest revision as of 04:48, 24 March 2013
[[Category:gaussian random variable
Contents
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.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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
Write it here.
Answer 3
Write it here.
