Solving Probability Problem: P(Z^3 > 1)

[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.

 

[statdad]

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

You must have used the table incorrectly, since -3.99 is not a probability (or you have a typo here).

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.

If two inequalities are equivalent, they represent events that have the same probability. thus, for example,

<br /> P(3Z+5 &gt; 7) = P(Z &gt; 2/3)<br />

since 3Z+5&gt;7 and Z &gt; 2/3 are equivalent inequalities.

So, what inequality is equivalent to Z^3 &gt; 1? Perhaps you were on the correct track and it simply seemed too easy.