Quick Answer
A value of 85 with a mean of 100 and standard deviation of 15 produces a z-score of -1.0, placing it at the 15.87th percentile with a cumulative probability of 0.1587.
Common Examples
| Input | Result |
|---|---|
| x = 85, mean = 100, SD = 15 | Z = -1.0, Percentile: 15.87% |
| x = 115, mean = 100, SD = 15 | Z = 1.0, Percentile: 84.13% |
| x = 130, mean = 100, SD = 15 | Z = 2.0, Percentile: 97.72% |
| x = 70, mean = 100, SD = 15 | Z = -2.0, Percentile: 2.28% |
| x = 100, mean = 100, SD = 15 | Z = 0.0, Percentile: 50.00% |
How It Works
The z-score (also called the standard score) measures how many standard deviations a particular value lies from the mean of a distribution. The formula is:
z = (x - μ) / σ
Where:
- x = the observed value
- μ = the population mean
- σ = the population standard deviation
A z-score of 0 means the value equals the mean. A z-score of +1 means the value is one standard deviation above the mean. A z-score of -2 means the value is two standard deviations below the mean.
Cumulative Probability
The cumulative probability (also called the p-value or CDF value) represents the proportion of values in a standard normal distribution that fall at or below a given z-score. This calculator uses the error function approximation (Abramowitz and Stegun method) to compute the cumulative distribution function:
P(Z ≤ z) = 0.5 × (1 + erf(z / √2))
Percentile
The percentile is the cumulative probability expressed as a percentage. A percentile of 84.13% means that approximately 84.13% of all values in the distribution fall at or below the observed value.
Common Z-Score Reference Points
In a standard normal distribution, approximately 68% of values fall within one standard deviation of the mean (z between -1 and +1), approximately 95% fall within two standard deviations (z between -2 and +2), and approximately 99.7% fall within three standard deviations (z between -3 and +3). This pattern is known as the 68-95-99.7 rule.
Worked Example
Suppose a student scores 85 on a test where the class mean is 100 and the standard deviation is 15. The z-score is (85 - 100) / 15 = -15 / 15 = -1.0. Using the cumulative distribution function, the probability P(Z ≤ -1.0) = 0.1587. This means the student scored at the 15.87th percentile, meaning approximately 15.87% of the class scored at or below 85. Conversely, a score of 115 produces z = (115 - 100) / 15 = +1.0 with a cumulative probability of 0.8413, placing it at the 84.13th percentile.
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