# Probability difficulties and Z scores

#### Deganveran

##### New Member
I am currently grappling with this question:

In a certain study the average age was 34.79 years. The SD of age was 14.78.

What probability do you have of observing someone 20 years old or younger?

I have determined that 20 years is a z score of 1.06.

However, here is what is throwing me off:

Is that -1.06 or positive?

And when looking at the unit normal table I am wondering if I am looking above Z or below z? I believe it is below z but the unit normal table is saying its 0.8554 or 85.54% chance of finding a 20 year old or younger probability at random. And that, I know is wrong as at least 75% should be within one standard deviation (or z = 1) of the mean.

#### Dragan

##### Super Moderator
I am currently grappling with this question:

In a certain study the average age was 34.79 years. The SD of age was 14.78.

What probability do you have of observing someone 20 years old or younger?

I have determined that 20 years is a z score of 1.06.

However, here is what is throwing me off:

Is that -1.06 or positive?
It is negative:

z=(20 - 34.79) / 14.78 = -1.000676...

#### Deganveran

##### New Member
It is negative:

z=(20 - 34.79) / 14.78 = -1.000676...
As a follow up question . . . twenty of below (with a mean 34.79) would be below the z score right? That would be .8413 (84.13). Above the z score is 0.1587 (15.87). Above the mean looks like the right answer numerically but below looks right contextually.

#### Dragan

##### Super Moderator
As a follow up question . . . twenty of below (with a mean 34.79) would be below the z score right? That would be .8413 (84.13). Above the z score is 0.1587 (15.87). Above the mean looks like the right answer numerically but below looks right contextually.
I'm having a difficult time understanding your follow-up question.

What probability do you have of observing someone 20 years old or younger?

As such, a Z-score of -1.0006 would correspond to a probability value of (approximately) p=0.1587. And, this prob. value would correctly answer the question.

#### Deganveran

##### New Member
I'm having a difficult time understanding your follow-up question.