# Probability Problem

1. Jan 20, 2010

### danniim

Can anyone help me with this?
If Z ~ N(0,1) determine P(Z^3 > 1)

I used the formula: x-miu/standard deviation where miu=0 and standard deviation is sq. root of 1.

I tired:

P(z^3>1)
P((x-0/1)^3>1)
P(x^3>1)
P(x>1)

I looked up the area of the standard normal distribution tables and got

=-3.99, which isn't right

I also tried

P(z^3>1)
P(z>1)

Looked it up using the tables and got

=0.15866

I don't think the question would be that easy though.

2. Jan 20, 2010

$$P(3Z+5 > 7) = P(Z > 2/3)$$
since $$3Z+5>7$$ and $$Z > 2/3$$ are equivalent inequalities.
So, what inequality is equivalent to $$Z^3 > 1$$? Perhaps you were on the correct track and it simply seemed too easy.