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anonymous

  • 4 years ago

The random variable X has a normal distribution\[Mean = \mu\]\[\mu > 0\]\[Variance= {1\over4}\mu^2\]Find \[P(X>1.5\mu).\]

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  1. anonymous
    • 4 years ago
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    what is the value of mu in normal distribution?

  2. anonymous
    • 4 years ago
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    does this question use moments?

  3. anonymous
    • 4 years ago
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    Um... I'm not sure, but I know that N~ (0,1) If that does any good?

  4. anonymous
    • 4 years ago
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    It just says that mu>0 So, you have to work with that.

  5. anonymous
    • 4 years ago
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    okay it is positive.....

  6. anonymous
    • 4 years ago
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    Yes

  7. y2o2
    • 4 years ago
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    Variance = (1/4) Mu² standard deviation = sqrt(Variance) = 0.5Mu \[P(X > 1.5 \mu) = p( Z > {1.5 \mu - \mu \over 0.5 \mu} ) = P(Z > 1) = 0.5 - P(0<Z<1) = 0.5 - 0.3413 = 0.1587 \]

  8. anonymous
    • 4 years ago
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    0.1587 is the answer... How do you get that?

  9. y2o2
    • 4 years ago
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    i've just explained it, did you look only to the answer ?!

  10. anonymous
    • 4 years ago
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    No. I see how you've explained it, and it makes sense. But how did you get the sqrt variance to equal 0.5mu instead of 1/4 mu^2?

  11. anonymous
    • 4 years ago
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    And your answer doesn't seem to fit the answer in the text book... So I was wondering how you come to that answer...?

  12. y2o2
    • 4 years ago
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    Actually the 0.5 Mu is the standard deviation not the variance. the standard deviation = sqrt (variance) and you can transform any normal random variable into a standard normal variable by the formula: \[if\ \ P(X > a)\ \ then\ \ P(Z > {a - mean \over standard deviation} )\]

  13. anonymous
    • 4 years ago
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    Ok. That makes sense.

  14. y2o2
    • 4 years ago
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    Have a look at this , it may help http://en.wikipedia.org/wiki/Normal_distribution#Standardizing_normal_random_variables

  15. anonymous
    • 4 years ago
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    Thanks, but I just don't know how to apply this to this question.?

  16. y2o2
    • 4 years ago
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    You mean you don't understand the formula ?!

  17. anonymous
    • 4 years ago
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    I understand the formula... But I don't know how to apply it with mu not having a value?

  18. y2o2
    • 4 years ago
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    do you know how to get the value of P(z >a ) from the standard normal curve ?!

  19. anonymous
    • 4 years ago
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    Yes. With (a) having a fixed value

  20. y2o2
    • 4 years ago
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    Ok great

  21. y2o2
    • 4 years ago
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    now you dont have the P(z>a) you have got p(x > a) so you've to convert the P(x) to P(x) so let a = 1.5 mu \[P(x > a) = P(z > {a - \mu \over \sigma}) = P(z > {1.5 \mu - \mu \over 0.5\mu}) = P(z > {\mu(1.5 - 1) \over 0.5\mu}) \]

  22. anonymous
    • 4 years ago
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    Ok, but how does \[{1\over4}\mu^2 \] turn into 0.5mu?

  23. y2o2
    • 4 years ago
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    I told you that (0.25μ²) is the variance(σ²) and the formula needs the standard deviation(σ) so you have to get σ = sqrt(σ²) = sqrt(0.25μ²) = 0.5μ

  24. anonymous
    • 4 years ago
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    Oh!

  25. anonymous
    • 4 years ago
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    Ok.

  26. anonymous
    • 4 years ago
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    So, how do you go on from there?

  27. y2o2
    • 4 years ago
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    Sorry, I really don't have anymore to explain. I already told you how to proceed....do a bit of thinking.

  28. anonymous
    • 4 years ago
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    Do the \[\mu s\]cancel out?

  29. y2o2
    • 4 years ago
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    yes

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