1. ## Normal Question

Hi, just doing a past paper and came across this question on the normal distributions that I'm not too sure about.

$Z - N(0,1)$, find $z$ such that $P(|Z| < z) = 0.790$

Normally I'd have no problem using the inverse normal function in my tables, not I'm not sure what that modulus sign does to the question?

Thanks

2. It means that you need to find the value of $z$ such that the Probability that $-z

{So the bounds ( endpoints of the given interval) are symmetric}

I think you can proceed from here.

3. Originally Posted by craig
Hi, just doing a past paper and came across this question on the normal distributions that I'm not too sure about.

$Z - N(0,1)$, find $z$ such that $P(|Z| < z) = 0.790$

Normally I'd have no problem using the inverse normal function in my tables, not I'm not sure what that modulus sign does to the question?

Thanks
$\Pr(-z < Z < z) = 0.790$. I suggest using symmetry to do the calculation.

If $P(-z < Z < z) = 0.790$, then it follows that $P(Z < -z) = \frac{1-0.790}{2} = 0.105$.