Suppose a measurement is 72, with a reference mean of 65 and a standard deviation of 5. Its z-score is (72 − 65) / 5 = 1.4, so it is 1.4 standard deviations above the reference mean. A z-score standardizes distance; it does not, by itself, tell you whether a value is unusual. That depends on whether the reference group is appropriate and what its distribution looks like.
If the reference values are approximately normally distributed, z = 1.4 corresponds to a cumulative area of about 0.919, or roughly 91.9% below that value. That percentile interpretation may not hold for a strongly skewed or otherwise non-normal distribution.
For a quick arithmetic check, use a z score calculator.