Skip to content

z-scores and percentiles

Statistics and data · level 7, lesson 3 of 7

A question to start

Maya scored 82 on her math test and 88 on her English test. Her English mark is higher, so it looks like her better result. But the math test was hard: the class mean was 70 with a standard deviation of 6. The English test was easier: the class mean was 80 with a standard deviation of 10.

Compared with her classmates, which result is more impressive? The raw marks cannot tell you, because the two tests are on different scales. This lesson gives you a way to put any two measurements on the same scale, even a test mark and a jump length.

The idea

How far from the mean, in standard deviations

A z-score (also called a standard score) says how many standard deviations a value is above or below the mean:

\[z = \frac{x - \text{mean}}{\text{standard deviation}}\]

For a population with mean \(\mu\) and standard deviation \(\sigma\) this is \(z = \dfrac{x - \mu}{\sigma}\); for a sample with mean \(\bar{x}\) and standard deviation \(s\) it is \(z = \dfrac{x - \bar{x}}{s}\).

  • \(z = 0\): the value equals the mean.
  • \(z = 2\): the value is 2 standard deviations above the mean.
  • \(z = -1.5\): the value is 1.5 standard deviations below the mean.

The units cancel: (centimetres minus centimetres) divided by centimetres is a plain number. That is why z-scores can compare things measured in different units.

For Maya: math \(z = \dfrac{82 - 70}{6} = 2\), English \(z = \dfrac{88 - 80}{10} = 0.8\). Her math mark is 2 standard deviations above her class; her English mark is less than one. The math result is the more impressive one.

Going the other way, a z-score turns back into a value with

\[x = \mu + z\sigma.\]

Why z-scores always have mean 0 and standard deviation 1

Turn every value of a data set into a z-score. The new list always has mean 0 and standard deviation 1. Here is the reason, in two steps.

Step 1: subtract the mean. Subtracting the same number \(c\) from every value slides the whole data set along the number line. The mean slides by \(c\) too (the total drops by \(nc\), so the mean drops by \(c\)). Distances between the values do not change, so the deviations from the mean and the standard deviation do not change. Subtracting the mean itself makes the new mean exactly 0.

Step 2: divide by the standard deviation. Dividing every value by a positive number \(k\) divides every deviation by \(k\), every squared deviation by \(k^2\), the mean of the squares by \(k^2\), and so (after the square root) the standard deviation by \(k\). Dividing by the standard deviation itself makes the new standard deviation exactly 1. The mean was 0 and stays 0.

Check with a small population: 2, 4, 4, 4, 5, 5, 7, 9. The mean is \(40 \div 8 = 5\). The squared deviations are 9, 1, 1, 1, 0, 0, 4, 16, total 32, so the (population) standard deviation is \(\sqrt{32 \div 8} = 2\). The z-scores are

\(-1.5,\ -0.5,\ -0.5,\ -0.5,\ 0,\ 0,\ 1,\ 2\)

Their total is 0, so their mean is 0. Their squares are 2.25, 0.25, 0.25, 0.25, 0, 0, 1, 4, total 8, so their standard deviation is \(\sqrt{8 \div 8} = 1\). ✓

This is why every normal curve becomes the same curve after standardising: the standard normal curve, with mean 0 and standard deviation 1. One table of areas serves every normal distribution.

Percentiles and percentile ranks

The percentile rank of a value is the percent of the data below it. If your score has percentile rank 85, then 85% of scores are lower than yours. The value with percentile rank \(p\) is called the \(p\)th percentile. The median is the 50th percentile, Q1 is the 25th and Q3 the 75th.

For a list of data, count: percentile rank \(= \dfrac{\text{number of values below}}{n} \times 100\%\). (Some books and programs count half of the values equal to it as well, so answers can differ by a little. Say which rule you use.)

For a normal distribution, use areas: the percentile rank of a value is the area to the left of its z-score.

The standard normal table

A z-table gives the percent of a standard normal distribution below each z. Calculators and spreadsheets give the same numbers (for example NORM.S.DIST(z, TRUE) in a spreadsheet). Some useful values, to two decimals:

\(z\) −2 −1.5 −1 −0.5 0 0.5 0.84 1 1.28 1.5 1.645 1.96 2 2.5 3
% below 2.28 6.68 15.87 30.85 50 69.15 79.95 84.13 89.97 93.32 95.00 97.50 97.72 99.38 99.87

Three uses:

  • Below a value: read the table at its z.
  • Above a value: \(100\%\) minus the table value. (Symmetry also helps: the percent above \(z\) equals the percent below \(-z\).)
  • Between two values: subtract the two table values.

Three z-scores worth remembering: 1.28 cuts off the top 10%, 1.645 the top 5%, and 1.96 the top 2.5% (so \(\pm 1.96\) holds the middle 95%). You will use 1.96 again in level 8.

Worked examples

Women's heights in a town: normal, \(\mu = 163\) cm, \(\sigma = 7\) cm.

1. A percentile from a value. What is the percentile rank of a height of 174 cm?

\(z = \dfrac{174 - 163}{7} = \dfrac{11}{7} \approx 1.57\). Technology gives about 94.2% below. So 174 cm is about the 94th percentile: taller than about 94% of the women.

2. A value from a percentile. How tall must a woman be to be in the tallest 10%?

The tallest 10% are above the 90th percentile, which is \(z \approx 1.28\). Then \(x = \mu + z\sigma = 163 + 1.28 \times 7 = 163 + 8.96 \approx\) 172 cm.

3. Between two values. What percent are between 159.5 cm and 173.5 cm?

\(z_1 = \dfrac{159.5 - 163}{7} = -0.5\) and \(z_2 = \dfrac{173.5 - 163}{7} = 1.5\). From the table: \(93.32 - 30.85 =\) 62.47%.

4. Percentile rank from a list. Twenty test scores, in order:

52, 58, 61, 63, 65, 67, 68, 70, 71, 72, 74, 75, 77, 78, 80, 82, 85, 88, 91, 96

A score of 80 has 14 scores below it: percentile rank \(\dfrac{14}{20} = 70\%\), the 70th percentile. A score of 63 has 3 below it: the 15th percentile. Notice that the percentile rank is not the score: 80 marks out of 100 is the 70th percentile in this class.

5. Comparing across sports. In a school competition, the long jump has mean 4.2 m and standard deviation 0.5 m; the 100 m run has mean 14.0 s and standard deviation 0.8 s. Ari jumps 5.0 m and runs 12.8 s. Which result is better relative to the others?

  • Long jump: \(z = \dfrac{5.0 - 4.2}{0.5} = 1.6\).
  • 100 m: \(z = \dfrac{12.8 - 14.0}{0.8} = -1.5\). A negative z is good here, because a lower time is better: Ari is 1.5 standard deviations faster than the mean.

The jump (1.6) is a little more exceptional than the run (1.5). Decathlon scoring uses tables in a similar spirit, so that a great throw and a great sprint can be added fairly.

Try it

Use the table above or technology.

  1. A value has \(z = -2\) in a distribution with mean 50 and standard deviation 4. What is the value?
  2. Convert to z-scores: (a) 130 when \(\mu = 100, \sigma = 15\); (b) 9.1 when \(\mu = 10, \sigma = 0.6\).
  3. Scores are normal with mean 100 and standard deviation 15. What percent score below 92.5? Above 124? Between 85 and 122.5?
  4. Same scores. What score is needed to be in the top 5%?
  5. Lee scored 74 in a class with mean 65 and standard deviation 6. Sam scored 84 in a class with mean 72 and standard deviation 10. Who did better relative to their class?
  6. In the list of 20 scores above, find the percentile rank of 88 and of 52. Which score is the 50th percentile?
  7. A data set has mean 30 and standard deviation 5. Every value is converted to a z-score, and then each z-score is multiplied by 10 and has 50 added. What are the mean and standard deviation of the final list?
  8. A z-score of 0.67 is about the 75th percentile of a normal distribution. Use this to find the IQR of a normal distribution in terms of \(\sigma\). Then use the 1.5 × IQR rule to find how many standard deviations from the mean the outlier fences are.
Answers
  1. \(x = \mu + z\sigma = 50 + (-2)(4) = 42\).
  2. (a) \(z = \dfrac{130 - 100}{15} = 2\). (b) \(z = \dfrac{9.1 - 10}{0.6} = \dfrac{-0.9}{0.6} = -1.5\).
  3. 92.5 gives \(z = -0.5\): 30.85% below. 124 gives \(z = 1.6\): technology gives 94.52% below, so 5.48% above. 85 gives \(z = -1\) and 122.5 gives \(z = 1.5\): \(93.32 - 15.87 =\) 77.45%.
  4. The top 5% starts at \(z = 1.645\): \(100 + 1.645 \times 15 \approx 124.7\). A score of about 125.
  5. Lee: \(z = \dfrac{74 - 65}{6} = 1.5\). Sam: \(z = \dfrac{84 - 72}{10} = 1.2\). Lee did better relative to the class.
  6. 88 has 17 scores below: \(\dfrac{17}{20} = 85\%\), the 85th percentile. 52 has none below: the 0th percentile (by this rule). The median is the average of the 10th and 11th values, \((72 + 74) \div 2 = 73\): that is the 50th percentile (10 of 20 values are below it).
  7. The z-scores have mean 0 and standard deviation 1. Multiplying by 10 gives mean 0 and standard deviation 10. Adding 50 slides the list: mean 50, standard deviation still 10. (This is how some exams report "T-scores".)
  8. Q1 and Q3 are at \(\mu - 0.67\sigma\) and \(\mu + 0.67\sigma\), so IQR \(\approx 1.35\sigma\). The upper fence is \(Q3 + 1.5 \times \text{IQR} \approx 0.67\sigma + 2.02\sigma = 2.7\sigma\) above the mean (and the lower fence \(2.7\sigma\) below). In normal data only about 0.7% of values are beyond the fences.

Common mistake

Thinking a bigger z-score is always better. A z-score says "how unusual and in which direction". For running times, golf scores or prices, a negative z-score is the good one.

Mixing up percent and percentile. A score of 80% on a test and the 80th percentile are different things: one is how many marks you got, the other is how many people you beat.

Where it leads