Z-Score vs Percentile

A z-score rescales one value into a count of standard deviations. For a population it is $z = \frac{x - \mu}{\sigma}$, where $\mu$ (mu) is the mean and $\sigma$ (sigma) is the standard deviation; for a sample it is $z = \frac{x - \bar{x}}{s}$, with $\bar{x}$ (x-bar) and $s$ in those roles. The units cancel in the division, so a z-score has no units of its own and values from different scales become comparable. The sign carries the direction and the size carries the distance.

A student scores 88 on a test with mean 80 and standard deviation 4, then 92 on a test with mean 86 and standard deviation 8. The raw scores say the 92 was the better performance. The z-scores say otherwise: $z = (88 - 80)/4 = 2.00$ against $z = (92 - 86)/8 = 0.75$. Relative to the class each score came from, the 88 is much the stronger result, and that comparison is the whole reason to standardize.

The wrong sentence to watch for is "her z-score of 1.5 means she beat 93 percent of the class." That 93.3 percent is the area to the left of 1.5 under the standard normal curve, so it is a fact about the normal model rather than about z-scores. A z-score is defined for any distribution that has a mean and a standard deviation, including badly skewed ones, and only when the distribution is roughly normal does the z-table turn it into a percentile.

Standardizing cannot repair shape. Subtracting and dividing by constants shifts and stretches the values without reordering them, so a right-skewed set of z-scores is still right-skewed and its outliers are still outliers.

Small samples put a hard ceiling on the answer. Using the sample standard deviation, no value among $n$ observations can have $|z|$ larger than $\frac{n-1}{\sqrt{n}}$. For $n = 5$ that ceiling is 1.79, so a z-score of 2 computed from five data points is an arithmetic error and not a rare event. The formula also breaks down when $s = 0$, which happens only when every value is the same.

The $k$th percentile is the value at or below which about $k$ percent of the observations lie. The grammar of that sentence is worth slowing down for: $k$ is the percent, and the percentile is the value, carrying whatever units the data carry. If the 90th percentile of a test is 628 points, then 90 percent of scores are 628 or lower and about 10 percent are higher.

Run the definition backwards and you get the percentile rank, which starts from a value and returns a percent:

Take the ten scores 52, 58, 60, 64, 65, 69, 73, 79, 86, 91. Seven of them are 73 or lower, so the percentile rank of 73 is $\frac{7}{10} \times 100 = 70$, and 73 is the 70th percentile of this set. Move the same score into a different group and its percentile rank changes, because a percentile always describes a position within one particular set of data.

Percentiles work on a smooth distribution too, where they are read off the curve instead of counted. On a normal distribution with mean 500 and standard deviation 100, the 90th percentile sits about 1.28 standard deviations above the mean, at $500 + 1.28(100) = 628$ points. That is the same 628 from the opening paragraph, and it shows the two routes to a percentile: counting observations in a data set, or finding the area to the left under a density curve.

The misreading that costs the most marks is treating a percentile as a score. Standing at the 85th percentile is a claim about the other test-takers, not about how many questions you answered correctly. On a hard test you can get 40 percent of the items right and still sit at the 85th percentile of it.

Three percentiles have names of their own: the 25th is the first quartile $Q_1$, the 50th is the median, and the 75th is the third quartile $Q_3$. Percentile and z-score are the two standard ways of reporting relative standing, and they answer different questions. A percentile says how many values sit at or below yours; a z-score says how far above or below the mean yours is, measured in standard deviations.