z=\frac{x-\bar{x}}{s}
Where x is the actual score, \bar{x} is the mean, and s is the standard deviation.
Example: using standard score to find percentages
Assume the height of a group of people was measured and found to be normally distributed with a mean of 1.5m and standard deviation of 0.1m. A standard normal distribution with z-score is shown below:
If we wanted to find the percentage of people below 1.7m, we could do so in the following way:
13.5+34+34+13.5+2.35+0.15=97.5 \%
Example: using z-scores to compare values
Using the same distribution (height with mean of 1.5m and standard deviation of 0.1m), we want to compare the heights of two people. Brent has a z-score of 0 and Jack has a z-score of -0.6.
Without converting these z-scores into the actual heights, we can say that Brent’s height is exactly average (1.5m), as this is what a z-score of 0 corresponds to, and that Jack’s height lies between 1.4m (z-score of -1) and 1.5m (z-score of 0), as -1<0.6<0. We can also say that Jack is shorter than Brent as 0.6<0.
We can also convert these z-scores into their actual heights. In order to do this, we will need to use the rearranged version of the z-score formula:
x=z s+\bar{x}
Applying this to Brent and Jack’s z-scores:
Brent’s actual height is: x=0 * 0.1+1.5=1.5 m
Jack’s actual height is: x=-0.6 * 0.1+1.5=1.44 m
Diagram of a Bell curve