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

- schrodinger

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- anonymous

what is the value of mu in normal distribution?

- anonymous

does this question use moments?

- anonymous

Um... I'm not sure, but I know that
N~ (0,1)
If that does any good?

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- anonymous

It just says that mu>0
So, you have to work with that.

- anonymous

okay it is positive.....

- anonymous

Yes

- y2o2

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

- anonymous

0.1587 is the answer... How do you get that?

- y2o2

i've just explained it, did you look only to the answer ?!

- anonymous

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?

- anonymous

And your answer doesn't seem to fit the answer in the text book... So I was wondering how you come to that answer...?

- y2o2

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:
\[if\ \ P(X > a)\ \ then\ \ P(Z > {a - mean \over standard deviation} )\]

- anonymous

Ok. That makes sense.

- y2o2

Have a look at this , it may help
http://en.wikipedia.org/wiki/Normal_distribution#Standardizing_normal_random_variables

- anonymous

Thanks, but I just don't know how to apply this to this question.?

- y2o2

You mean you don't understand the formula ?!

- anonymous

I understand the formula... But I don't know how to apply it with mu not having a value?

- y2o2

do you know how to get the value of P(z >a ) from the standard normal curve ?!

- anonymous

Yes. With (a) having a fixed value

- y2o2

Ok great

- y2o2

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
\[P(x > a) = P(z > {a - \mu \over \sigma}) = P(z > {1.5 \mu - \mu \over 0.5\mu}) = P(z > {\mu(1.5 - 1) \over 0.5\mu}) \]

- anonymous

Ok, but how does \[{1\over4}\mu^2 \] turn into 0.5mu?

- y2o2

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

- anonymous

Oh!

- anonymous

Ok.

- anonymous

So, how do you go on from there?

- y2o2

Sorry, I really don't have anymore to explain.
I already told you how to proceed....do a bit of thinking.

- anonymous

Do the \[\mu s\]cancel out?

- y2o2

yes

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