Normal distribution & z-score calculator

How to calculate a z-score and its percentile

A z-score counts how many standard deviations a value sits from the mean: z = (x − μ) ÷ σ. On an exam with mean 72 and standard deviation 8, a score of 84 has z = 12 ÷ 8 = 1.5. The normal area to its left, Φ(1.5) = 0.93319, puts it at the 93rd percentile.

1. Subtract the mean: 84 − 72 = 12 marks above average.
2. Divide by the standard deviation: z = 12 ÷ 8 = 1.5.
3. Look up the area to the left of z = 1.5 under the standard normal curve: Φ(1.5) = 0.93319.
4. Read it as a percentile: the score is at about the 93rd percentile, and the normal model puts 6.68% of students above it.

Scores below the mean give negative z-scores. A mark of 60 has z = (60 − 72) ÷ 8 = −1.5, and Φ(−1.5) = 0.06681 puts it near the 7th percentile.

Because z has no units, it compares results on different scales. A 610 on a test with mean 500 and SD 100 has z = 1.1, the 86th percentile, so the exam mark of 84, at z = 1.5, is the stronger result even though 610 is the bigger number.

To go from a z-score back to a raw value, reverse the formula: x = μ + zσ. A student aiming for the top 10% needs z = 1.2816, so a mark of 72 + 1.2816 × 8 = 82.25, or 82.2524 with z carried to more digits.

The 68–95–99.7 rule, extended

The rule gives the share of a normal distribution within 1, 2 and 3 standard deviations of the mean. For the exam, 68% of marks fall between 64 and 80 and 95% between 56 and 88. Further out, the tails thin quickly:

The "share outside" column is split equally between the two tails, so only half of it lies above the mean. About 1 value in 44 is more than 2σ above the mean, not 1 in 22.

Intervals that are not symmetric need two lookups. The share of exam marks between 60 and 80 runs from z = −1.5 to z = 1, so it is Φ(1) − Φ(−1.5) = 0.841345 − 0.066807 = 0.774538, or 77.45%.

Z-table: area to the left of z

A cumulative z-table gives Φ(z), the share of the distribution below z. The table below lists it for positive z with the two related areas beside it.

For a negative z, use the symmetry of the curve: Φ(−z) = 1 − Φ(z). So Φ(−1.5) = 1 − 0.93319 = 0.06681, the area above +1.5. Printed tables stop at two decimal places of z, so values in between need interpolation.

Percentiles and cut-offs

Working backwards from a percentile to a value is the inverse normal problem. Find the z for the percentile, then convert with x = μ + zσ.

For the exam, the lowest 5% of marks fall below 72 − 1.6449 × 8 = 58.84, and the middle half lies between the 25th and 75th percentiles, 72 ± 0.6745 × 8, or 66.60 to 77.40.

A percentile is a rank, not a mark: the 90th percentile beats 90% of the group, whatever share of the questions it got right. To rescale a whole class to a new mean and standard deviation instead, use the grade curve calculator.

When data are not normal

A z-score can be computed for any data set, but turning it into a percentile through Φ assumes the bell shape. Skewed data break that assumption in both tails at once.

Take bus waits that follow an exponential distribution, with mean and SD both 10 minutes. A normal model predicts 2.28% of waits above 30 minutes and 15.87% below 0; the exponential gives 4.98% above 30, more than twice as many, and none below 0.

Before trusting a percentile, check the data:

• Look at the histogram and skewness. The descriptive statistics calculator draws both; a skewness beyond about ±1 is a warning.
• Use a normal probability plot. Data from a normal distribution fall close to a straight line; curvature signals skew and bends at the ends signal heavy or light tails (NIST/SEMATECH e-Handbook, §1.3.3.21).
• Fall back on a guaranteed bound. Chebyshev's inequality holds for any distribution: at least 1 − 1/k² of values lie within k SDs of the mean, so at least 75% within 2σ and 88.9% within 3σ, against 95.45% and 99.73% for the normal.

With μ and σ estimated from a small sample, probabilities and intervals for the mean come from Student's t, in the t, chi-square and F distribution calculator. Counts, such as defective items in a batch, follow the binomial distribution.

Mistakes that change the answer

• Dividing by the variance. z uses the standard deviation. For the exam, 12 ÷ 64 gives z = 0.1875 and a 57th percentile instead of the 93rd.
• Reading the wrong tail. "Scored higher than 84" is P(X > 84) = 6.68%, the complement of P(X < 84). Sketch the curve and shade the area before looking anything up.
• Rounding z before the lookup. z = 1.96 and z = 2 differ by 0.04, but the areas above them are 2.50% and 2.28%, and the second is 9% smaller. Carry z to at least two decimal places.
• Worrying about < versus ≤. For a continuous distribution the chance of hitting one exact value is zero, so P(X < 84) and P(X ≤ 84) are equal.