Question

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

Answer 1

what is the value of mu in normal distribution?

Answer 2

does this question use moments?

Answer 3

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

Answer 4

It just says that mu>0

So, you have to work with that.

Answer 5

okay it is positive.....

Answer 6

Yes

Answer 7

Variance = (1/4) Mu²

standard deviation = sqrt(Variance) = 0.5Mu

\[P(X > 1.5 \mu) = p( Z > {1.5 \mu - \mu \over 0.5 \mu} ) = P(Z > 1) = 0.5 - P(0<Z<1) = 0.5 - 0.3413 = 0.1587 \]

Answer 8

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

Answer 9

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

Answer 10

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?

Answer 11

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

Answer 12

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:

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

Answer 13

Ok. That makes sense.

Answer 14

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

Answer 15

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

Answer 16

You mean you don't understand the formula ?!

Answer 17

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

Answer 18

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

Answer 19

Yes. With (a) having a fixed value

Answer 20

Ok great

Answer 21

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

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

Answer 22

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

Answer 23

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

Answer 24

Oh!

Answer 25

Ok.

Answer 26

So, how do you go on from there?

Answer 27

Sorry, I really don't have anymore to explain.

I already told you how to proceed....do a bit of thinking.

Answer 28

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

Answer 29

yes