anonymous
  • anonymous
The random variable X has a normal distribution\[Mean = \mu\]\[\mu > 0\]\[Variance= {1\over4}\mu^2\]Find \[P(X>1.5\mu).\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
what is the value of mu in normal distribution?
anonymous
  • anonymous
does this question use moments?
anonymous
  • anonymous
Um... I'm not sure, but I know that N~ (0,1) If that does any good?

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anonymous
  • anonymous
It just says that mu>0 So, you have to work with that.
anonymous
  • anonymous
okay it is positive.....
anonymous
  • anonymous
Yes
y2o2
  • 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
  • anonymous
0.1587 is the answer... How do you get that?
y2o2
  • y2o2
i've just explained it, did you look only to the answer ?!
anonymous
  • 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
  • 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
  • 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
  • anonymous
Ok. That makes sense.
y2o2
  • y2o2
Have a look at this , it may help http://en.wikipedia.org/wiki/Normal_distribution#Standardizing_normal_random_variables
anonymous
  • anonymous
Thanks, but I just don't know how to apply this to this question.?
y2o2
  • y2o2
You mean you don't understand the formula ?!
anonymous
  • anonymous
I understand the formula... But I don't know how to apply it with mu not having a value?
y2o2
  • y2o2
do you know how to get the value of P(z >a ) from the standard normal curve ?!
anonymous
  • anonymous
Yes. With (a) having a fixed value
y2o2
  • y2o2
Ok great
y2o2
  • 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
  • anonymous
Ok, but how does \[{1\over4}\mu^2 \] turn into 0.5mu?
y2o2
  • 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
  • anonymous
Oh!
anonymous
  • anonymous
Ok.
anonymous
  • anonymous
So, how do you go on from there?
y2o2
  • y2o2
Sorry, I really don't have anymore to explain. I already told you how to proceed....do a bit of thinking.
anonymous
  • anonymous
Do the \[\mu s\]cancel out?
y2o2
  • y2o2
yes

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