How to use
- Enter the mean μ and standard deviation σ. The standard normal is 0 and 1.
- Choose what to find, such as below a, above a, between a and b, outside a and b, or x from a probability.
- Enter a (and b). The result and the shaded curve update as you type.
Options explained
- P(X < a)
- The probability of a value below a, the cumulative probability, which is the area under the curve to the left of a. Times 100 it is the percentile.
- From probability
- Left of x gives a quantile (such as the 95th percentile); centered gives the symmetric interval holding that probability, for example μ ± 1.96σ for 0.95.
- z-score
- z = (x − μ) / σ, how many standard deviations x lies from the mean.
Handy values (standard normal)
| Find | Result |
|---|---|
| P(Z < 1.5) | 0.9332 |
| P(Z > 1.5) | 0.0668 |
| P(−1 < Z < 1) | 0.6827 |
| P(−1.96 < Z < 1.96) | 0.9500 |
| P(−3 < Z < 3) | 0.9973 |
| 95th percentile (one-sided critical value) | 1.6449 |
| 97.5th percentile (two-sided 0.05 critical value) | 1.9600 |
This is the 68–95–99.7 rule: about 68% of values fall within one standard deviation of the mean, 95% within two and 99.7% within three.
Example: IQ scores
IQ tests are scaled to a mean of 100 and SD of 15 (pick the IQ preset).
- Share below 130: P(X < 130) = 0.9772, so about 2.3% score above 130.
- Between 85 and 115: P(85 < X < 115) = 0.6827, about two thirds of people.
- Top 10% cut-off: choose “From probability”, 0.10, “Right of x”, and get 119.2.
Reading a z-table
The standard normal table at the back of a textbook lists P(Z < z). The entry for z = 1.50 is 0.9332, the same as μ = 0, σ = 1, P(X < a), a = 1.5 here. For any other normal distribution, convert with z = (x − μ) / σ first; the calculator does that for you.
Frequently asked questions
Where does 1.96 come from?
In the standard normal distribution, 95% of the area lies between −1.96 and 1.96. Choose From probability, enter 0.95 and pick centered to see it.
Test scores average 75 with SD 10. What share scored below 90?
Choose P(X < a) with μ = 75, σ = 10, a = 90. The result is 0.9332, about 93%.
Can I use it for counts?
For large samples the binomial can be approximated by the normal. Use a continuity correction, so for "at most k" use k + 0.5.
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