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,

$$P(3Z+5 > 7) = P(Z > 2/3)$$

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.

 

[Architeuthis Dux]

I can confirm that your approach is correct. However, the answer you obtained (0.15866) is the probability of a standard normal variable being greater than 1, which is not the same as P(Z^3 > 1). To solve this problem, we need to use the transformation property of normal distributions, which states that if X is a normal random variable with mean μ and standard deviation σ, then aX+b is also a normal random variable with mean aμ+b and standard deviation |a|σ.

In this case, we have Z^3, which can be rewritten as (Z^3)^(1/3), resulting in a new normal random variable with mean 0 and standard deviation |1/3|. Therefore, we can use the formula you mentioned, x-miu/standard deviation, to find the probability of Z^3 being greater than 1.

P(Z^3 > 1) = P((Z^3)^(1/3) > 1) = P(Z > 1^(1/3)) = P(Z > 1)

Using the standard normal distribution tables, we can find that the probability of a standard normal variable being greater than 1 is 0.15866, which is the same answer you obtained earlier. Therefore, the final answer to this probability problem is 0.15866.

In summary, your approach was correct, but you needed to use the transformation property of normal distributions to correctly solve for P(Z^3 > 1). I hope this helps!