## anonymous 4 years ago The random variable X has a normal distribution$Mean = \mu$$\mu > 0$$Variance= {1\over4}\mu^2$Find $P(X>1.5\mu).$

1. anonymous

what is the value of mu in normal distribution?

2. anonymous

does this question use moments?

3. anonymous

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

4. anonymous

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

5. anonymous

okay it is positive.....

6. anonymous

Yes

7. y2o2

Variance = (1/4) Mu² standard deviation = sqrt(Variance) = 0.5Mu $P(X > 1.5 \mu) = p( Z > {1.5 \mu - \mu \over 0.5 \mu} ) = P(Z > 1) = 0.5 - P(0<Z<1) = 0.5 - 0.3413 = 0.1587$

8. anonymous

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

9. y2o2

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

10. 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?

11. 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...?

12. 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: $if\ \ P(X > a)\ \ then\ \ P(Z > {a - mean \over standard deviation} )$

13. anonymous

Ok. That makes sense.

14. y2o2

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

15. anonymous

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

16. y2o2

You mean you don't understand the formula ?!

17. anonymous

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

18. y2o2

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

19. anonymous

Yes. With (a) having a fixed value

20. y2o2

Ok great

21. 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 $P(x > a) = P(z > {a - \mu \over \sigma}) = P(z > {1.5 \mu - \mu \over 0.5\mu}) = P(z > {\mu(1.5 - 1) \over 0.5\mu})$

22. anonymous

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

23. 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

24. anonymous

Oh!

25. anonymous

Ok.

26. anonymous

So, how do you go on from there?

27. y2o2

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

28. anonymous

Do the $\mu s$cancel out?

29. y2o2

yes