How do I find the mean from a normal probability?

Turn the probability into a z-score with the inverse normal, then rearrange x = μ + zσ to μ = x − zσ. If P(X < 60) = 0.9 and σ = 5, z = 1.2816 and μ = 60 − 1.2816 × 5 = 53.59. Solve for Mean μ shows each step.

How do I find P(a < X < b) for a normal distribution?

Turn each bound into a z-score with z = (x − μ)/σ, look up Φ(z) for each, and subtract: P(a < X < b) = Φ(z_b) − Φ(z_a). The simulation shows the same area shaded and the value to four decimal places.

Why is it 1.96 for 95%?

Φ(1.96) = 0.975, so 2.5% of the area lies above z = 1.96 and, by symmetry, 2.5% below z = −1.96. The remaining 95% sits in between.

When can I use a normal distribution instead of a binomial?

When n is large and p is not close to 0 or 1, so that np and n(1 − p) are both more than about 5. Use μ = np, σ² = np(1 − p), and widen each integer bound by 0.5 (the continuity correction). Turn on the approximation in Advanced to compare.

What does the inverse normal do?

It goes the other way: you give a probability p and it finds the value b with P(X < b) = p. For example the top 2.5% of a standard normal starts at z = 1.96.

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Normal and binomial distributions

The probability is the shaded area. Move the bounds, change μ and σ (or n and p), and watch random samples pile up into the same shape.

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Share of samples in the region

Readouts

What's happening

A continuous random variable has no probability at a single point, only over a range, and that probability is the area under the density curve. For a normal distribution the curve is set by its mean μ and standard deviation σ, and every normal curve becomes the standard one once you measure in z-scores, z = (x − μ)/σ. That is why 95% of values always lie within 1.96 standard deviations of the mean. The binomial counts successes in n independent trials with chance p: its bars add up the same way. Press Sample to draw random values: the gold histogram settles onto the curve and the share landing in your region settles on the probability, which is the law of large numbers at work.

z = (x − μ) / σP(a < X < b) = Φ(z_b) − Φ(z_a)P(X = k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ

A-Level Maths Statistics (AQA, Edexcel, OCR): binomial and normal distributions, inverse normal, normal approximation to the binomial. First-year probability and statistics.

Challenge

Predict first: what share of a normal distribution lies within 2 standard deviations of the mean? And within 1.96? Set a and b, read the shaded area, then sample 2000 values and see how close the share gets.

FAQ

How do I find the mean from a normal probability?: Turn the probability into a z-score with the inverse normal, then rearrange x = μ + zσ to μ = x − zσ. If P(X < 60) = 0.9 and σ = 5, z = 1.2816 and μ = 60 − 1.2816 × 5 = 53.59. Solve for Mean μ shows each step.
How do I find P(a < X < b) for a normal distribution?: Turn each bound into a z-score with z = (x − μ)/σ, look up Φ(z) for each, and subtract: P(a < X < b) = Φ(z_b) − Φ(z_a). The simulation shows the same area shaded and the value to four decimal places.
Why is it 1.96 for 95%?: Φ(1.96) = 0.975, so 2.5% of the area lies above z = 1.96 and, by symmetry, 2.5% below z = −1.96. The remaining 95% sits in between.
When can I use a normal distribution instead of a binomial?: When n is large and p is not close to 0 or 1, so that np and n(1 − p) are both more than about 5. Use μ = np, σ² = np(1 − p), and widen each integer bound by 0.5 (the continuity correction). Turn on the approximation in Advanced to compare.
What does the inverse normal do?: It goes the other way: you give a probability p and it finds the value b with P(X < b) = p. For example the top 2.5% of a standard normal starts at z = 1.96.