Enter the number of trials and the probability of success. Add a number of successes to get the exact probability of that outcome, or leave it blank for the summary alone.
Each bar is P(X = k). The darker bars lie within one standard deviation (σ = 2.049) of the mean.
Show the working, step by step
The formulas
mean: μ = np variance: σ² = npq std dev: σ = √(npq) where q = 1 − p P(X = k) = C(n,k) · p^k · q^(n−k)
The standard deviation here needs no data at all — it follows from n and p alone, because the shape of the distribution is completely determined by those two numbers.
A worked example
A test has 20 questions with four options each. A student guesses every one, so n = 20 and p = 0.25.
- Mean: μ = 20 × 0.25 = 5 correct answers.
- Variance: σ² = 20 × 0.25 × 0.75 = 3.75.
- Standard deviation: σ = √3.75 = 1.936.
So a guesser typically scores 5, and scoring 9 is about two standard deviations above that — unusual but not extraordinary. Scoring 13 is over four standard deviations out, which is strong evidence they were not guessing.
How p drives the spread
| p | μ = np | σ = √(npq) |
|---|---|---|
| 0.01 | 1.0 | 0.995 |
| 0.10 | 10 | 3.000 |
| 0.25 | 25 | 4.330 |
| 0.50 | 50 | 5.000 |
| 0.75 | 75 | 4.330 |
| 0.90 | 90 | 3.000 |
| 0.99 | 99 | 0.995 |
All rows use n = 100. The standard deviation peaks at p = 0.5 and is symmetric either side — p = 0.25 and p = 0.75 give identical spread, because uncertainty depends on how far p sits from certainty, not on which outcome is the likely one.
The four conditions
The binomial model requires all four, and the third and fourth are the ones that usually fail:
- Fixed n. The number of trials is decided in advance.
- Two outcomes. Each trial is a success or a failure.
- Constant p. The probability is the same every trial. Sampling without replacement from a small population breaks this.
- Independence. One trial's result tells you nothing about the next. Clustered data — pupils in classes, repeated measures on one person — breaks this, and the usual symptom is a real spread noticeably wider than √(npq) predicts.
Related calculators
-
Normal distribution
The continuous approximation, valid when np and nq are large.
-
Standard deviation
The general case, computed from raw data.
-
Variance calculator
The npq that sits under the square root.
-
Z-score calculator
How unusual a particular count is.
Common questions
What is the standard deviation of a binomial distribution?
σ = √(npq), where n is the number of trials, p the probability of success
and q = 1 − p. The mean is np and the variance npq.
When is the binomial the right model?
Four conditions: a fixed number of trials, two outcomes per trial, the same probability every trial, and trials independent of each other. Coin flips, pass/fail tests and conversion counts fit. Drawing cards without replacement does not, because the probability changes as you go.
Why is the standard deviation largest at p = 0.5?
Because that is where the outcome is least predictable. With p = 0.99 almost every trial succeeds, so there is little to vary; at p = 0.5 each trial is a genuine coin flip. At p = 0 or p = 1 the outcome is certain and σ is exactly zero.
When can I approximate the binomial with a normal distribution?
The usual rule of thumb is that both np ≥ 10 and nq ≥ 10. Then
the binomial is close enough to a normal with the same mean and standard deviation to use
the normal calculator instead.
The approximation is poor when p is near 0 or 1, because the true distribution is skewed and bounded while the normal is neither.