In statistics, a z-score is a score's measure in standard deviations from the mean. So to find a the value of z for a particular score, take (score - mean) / SD. z-scores are sometimes referred to as standard units. The z-score tells how deviant a particular score is, and may be used to find outliers or significant data.

Given a normal distribution, by the 68-95-99 rule of thumb, about 68 percent of the data will have a Z-score on the range -1 to 1 inclusive (that is, will be between -1 and 1 standard deviations from the mean), about 95% will have a Z-score between -2 and 2, and about 99.7% will have a Z-score between -3 and 3. For more accurate values than the 68-95-99 rule, one should consult a table or calculator.

A Z-score is often looked up on a Z-table to determine what percent of the data should be within a certain range, or to find the probability that an arbitrary event occurred. On a TI-83 or TI-89, the normcdf and normpdf (for the former) or tistat.normcdf and tistat.normpdf (for the latter) can take the place of a Z-table.

When the standard deviation of the population is not known, a T-score can be used.
A followup to the node by Lagman, in which he states that "Given a normal distribution, by the 68-95-99 rule of thumb, about 68 percent of the data will have a Z-score on the range -1 to 1 inclusive (that is, will be between -1 and 1 standard deviations from the mean), about 95% will have a Z-score between -2 and 2, and about 99.7% will have a Z-score between -3 and 3."

I hate to be technical about this, but officially, the value for the 95% cutoff is 1.96 plus or minus standard deviations from the mean. I know this is touchy, but in the world or statistics, touchy makes a big difference. His phrasing is actually underestimating the true value of the percentage within the plus or minus 2 standard deviations, it would be higher than 95%.

A z score is used when the sample size is generally greater than 30, thus assuming normal distribution. Z scores are very useful in determining the relative ability of an individual when comparing two differently formatted tests. For example: A person in high school who scores 13 out of 20 on a AP Physics test, which has a mean of 10 and standard deviation of 3 test would thus have a z score of 1. Another student, who scores 90 out of 100 on a basic arithmetic exam with a mean of 80 and standard deviation of 20, would only be 1/2 standard deviation above the mean. Thus the student with the higher Z score (greater amount of standard deviations above the mean) would have performed relatively better.


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