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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 \[P(X>1.5\mu).\]
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is the value of mu in normal distribution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does this question use moments?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Um... I'm not sure, but I know that N~ (0,1) If that does any good?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It just says that mu>0 So, you have to work with that.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay it is positive.....

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2Variance = (1/4) Mu² standard deviation = sqrt(Variance) = 0.5Mu \[P(X > 1.5 \mu) = p( Z > {1.5 \mu  \mu \over 0.5 \mu} ) = P(Z > 1) = 0.5  P(0<Z<1) = 0.5  0.3413 = 0.1587 \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.00.1587 is the answer... How do you get that?

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2i've just explained it, did you look only to the answer ?!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No. 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And your answer doesn't seem to fit the answer in the text book... So I was wondering how you come to that answer...?

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2Actually 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: \[if\ \ P(X > a)\ \ then\ \ P(Z > {a  mean \over standard deviation} )\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. That makes sense.

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2Have a look at this , it may help http://en.wikipedia.org/wiki/Normal_distribution#Standardizing_normal_random_variables

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks, but I just don't know how to apply this to this question.?

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2You mean you don't understand the formula ?!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand the formula... But I don't know how to apply it with mu not having a value?

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2do you know how to get the value of P(z >a ) from the standard normal curve ?!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. With (a) having a fixed value

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2now 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 \[P(x > a) = P(z > {a  \mu \over \sigma}) = P(z > {1.5 \mu  \mu \over 0.5\mu}) = P(z > {\mu(1.5  1) \over 0.5\mu}) \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, but how does \[{1\over4}\mu^2 \] turn into 0.5mu?

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2I 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, how do you go on from there?

y2o2
 4 years ago
Best ResponseYou've already chosen the best response.2Sorry, I really don't have anymore to explain. I already told you how to proceed....do a bit of thinking.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do the \[\mu s\]cancel out?
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