# Thread: Looking up Z-table "backwards"??

1. ## Looking up Z-table "backwards"??

Terribly stuck! How does one read a z-table 'backwards'?!

In one question, I'm required to find a value z*, such that $\displaystyle P(Z<z*)= 0.98$ and $\displaystyle P(0<Z<z*)= 0.48$ -- by reading the table 'backwards'.

And in the next one, in order to find $\displaystyle z_{a/2}$, I need to do the same.

Any help would be greatly appreciated!

2. Originally Posted by trackies
Terribly stuck! How does one read a z-table 'backwards'?!

In one question, I'm required to find a value z*, such that $\displaystyle P(Z<z*)= 0.98$ and $\displaystyle P(0<Z<z*)= 0.48$ -- by reading the table 'backwards'.

And in the next one, in order to find $\displaystyle z_{a/2}$, I need to do the same.

Any help would be greatly appreciated!
This is difficult to explain this way but look at the attachment, which is a portion of the z-table. The two highlighted figures bracket p=0.98, so the z-score corresponding to 0.98 lies between 2.05 and 2.06, you can refine this by linear interpolation to get z~=2.054

RonL

3. Hmm. Okay. But how do I find $\displaystyle P(0<Z<z*)=0.48$ using this method?

4. Originally Posted by trackies
Hmm. Okay. But how do I find $\displaystyle P(0<Z<z*)=0.48$ using this method?
$\displaystyle P(0<Z<z^*)=P(Z<z^*)-0.5=0.48$

so:

we seek $\displaystyle z^*$ such that: $\displaystyle p(Z<z^*)=0.98$

RonL

5. Originally Posted by CaptainBlack
$\displaystyle P(0<Z<z^*)=P(Z<z^*)-0.5=0.48$

so:

we seek $\displaystyle z^*$ such that: $\displaystyle p(Z<z^*)=0.98$

RonL
As always, you make the toughest questions look easy