# Why does the average z-score for a standardized distribution always equal to zero?

My introductory statistics book mentioned this:

"When an entire distribution of scores is standardized, the average (i.e., mean) z score for the standardized distribution will always be 0, and the standard deviation of this distribution will always be 1.0."

Why does the average z-score always equal to zero?

• It is simply part of it being standardized. Subtract the mean value from a set of numbers and you get a set of numbers with mean 0 e.g. {1,2,3,4,5} has mean 3 and subtracting 3, {-2,-1,0,1,2} has mean 0..
– Paul
Commented May 7, 2019 at 13:29

Standardization is the process of applying a linear transformation of a random variable so that its mean is zero and its variance is one.

Taking a non-standardized variable, if you subtract the mean, the new mean is obviously zero.

The fact that the mean of the Z Distribution is zero, comes from the formula of calculating Z values. Check the following case with the set shown below, where the mean is $$58$$ and the standard deviation is $$5$$:

suppose we wanted to convert our mean (58) into a z score:

$$(58-58)/5 = 0/5 = 0$$

An algebraic proof can be shown if you want let me know.

To prove that the variance is equal to 1 is difficult for me!

Ref:

The Normal Distribution and Z Scores-University of YTAH

Z-Scores Paper

Similar Question

Suppose you have a shop where you sell, on average 100 pcs a day, with a standard deviation of 25. If your demand is normally distributed, you have a 50% probability that the demand per day will be less or more than 100. But suppose you want to know what the probability is that the demand will be less than 150 a day. You can calculate the z-score as $$(150-100)/25 = 2=> 97.72%$$ . This means that adding 2 times your standard deviation to your average, you will cover 97.72% of your normal distribution. So the general formula is $$average+z*standarddeviation$$. That's why the z-score always is zero on the average in a standard normal distribution table.