Normal Distribution Calculator

Find probabilities, percentiles, and z-scores for any normal distribution with step-by-step solutions

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Find P(X < 72) where X ~ N(70, 5)
Find P(65 < X < 75) where X ~ N(70, 5)
Find the 90th percentile of N(100, 15)
Standardize X=85 for N(80, 10) and find P(X > 85)

What is the Normal Distribution?

The normal distribution (also called the Gaussian distribution or bell curve) is the most important probability distribution in statistics. It describes the distribution of many natural phenomena — heights, test scores, measurement errors — and is the foundation of most parametric inference.

Notation

XN(μ,σ2)X \sim N(\mu, \sigma^2)

where μ\mu is the mean (center of the bell) and σ2\sigma^2 is the variance (σ\sigma is the standard deviation, the spread).

Probability Density Function (PDF)

f(x)=1σ2πexp((xμ)22σ2)f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)

Key Properties

  • Symmetric about μ\mu: mean = median = mode
  • Bell-shaped: highest at μ\mu, tails extend infinitely
  • Total area under the curve = 1 (it's a probability distribution)
  • Empirical rule: ~68% of data within ±1σ\pm 1\sigma, ~95% within ±2σ\pm 2\sigma, ~99.7% within ±3σ\pm 3\sigma
IntervalApprox. probability
μ±σ\mu \pm \sigma68.27%
μ±2σ\mu \pm 2\sigma95.45%
μ±3σ\mu \pm 3\sigma99.73%

The Standard Normal Distribution and Z-Scores

The standard normal distribution is the special case ZN(0,1)Z \sim N(0,1) — mean 0, standard deviation 1. Its CDF is written Φ(z)\Phi(z).

Standardizing (Computing Z-Scores)

Any normal random variable XN(μ,σ2)X \sim N(\mu, \sigma^2) can be converted to a standard normal variable:

Z=XμσZ = \frac{X - \mu}{\sigma}

The z-score tells you how many standard deviations xx is above or below the mean.

Computing Probabilities

All normal probability problems reduce to looking up Φ(z)\Phi(z) (the area to the left of zz):

P(X<x)=P ⁣(Z<xμσ)=Φ ⁣(xμσ)P(X < x) = P\!\left(Z < \frac{x-\mu}{\sigma}\right) = \Phi\!\left(\frac{x-\mu}{\sigma}\right)

P(X>x)=1Φ ⁣(xμσ)P(X > x) = 1 - \Phi\!\left(\frac{x-\mu}{\sigma}\right)

P(a<X<b)=Φ ⁣(bμσ)Φ ⁣(aμσ)P(a < X < b) = \Phi\!\left(\frac{b-\mu}{\sigma}\right) - \Phi\!\left(\frac{a-\mu}{\sigma}\right)

Finding Percentiles (Inverse Normal)

To find the value xx such that P(Xx)=pP(X \leq x) = p:

x=μ+σzpx = \mu + \sigma \cdot z_p

where zp=Φ1(p)z_p = \Phi^{-1}(p) is the pp-th percentile of the standard normal.

Common percentiles:

Percentilezz
90th1.282
95th1.645
97.5th1.960
99th2.326

Why the Normal Distribution Matters

The normal distribution is central to statistics because of the Central Limit Theorem (CLT): the sampling distribution of the sample mean Xˉ\bar{X} approaches N(μ,σ2/n)N(\mu, \sigma^2/n) as nn grows, regardless of the original population's shape.

This means:

  • Confidence intervals and hypothesis tests for means use normal (or t) critical values
  • Many test statistics (t, z, F, χ2\chi^2) are derived from normal distributions
  • Even non-normal data can be analyzed using normal approximations when nn is large enough

Normal Approximation to the Binomial

When nn is large and pp is not too extreme, XBinomial(n,p)X \sim \text{Binomial}(n, p) is approximately N(np,np(1p))N(np,\, np(1-p)):

P(Xk)Φ ⁣(k+0.5npnp(1p))P(X \leq k) \approx \Phi\!\left(\frac{k + 0.5 - np}{\sqrt{np(1-p)}}\right)

(The +0.5+0.5 is the continuity correction for the discrete-to-continuous approximation.)

Validity: use this approximation when np10np \geq 10 and n(1p)10n(1-p) \geq 10.

示例题目

Step 1: Standardize: z=(115100)/15=1.00z = (115 - 100)/15 = 1.00
Step 2: Look up Φ(1.00)=0.8413\Phi(1.00) = 0.8413
Step 3: P(X<115)=0.8413P(X < 115) = 0.8413
Answer: About 84.1% of people have IQ scores below 115.

Step 1: Standardize lower bound: z1=(6570)/5=1.00z_1 = (65 - 70)/5 = -1.00
Step 2: Standardize upper bound: z2=(7570)/5=1.00z_2 = (75 - 70)/5 = 1.00
Step 3: P(1<Z<1)=Φ(1)Φ(1)=0.84130.1587=0.6827P(-1 < Z < 1) = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6827
Answer: P(65<X<75)0.6827P(65 < X < 75) \approx 0.6827 — the classic 68% rule for ±1σ\pm 1\sigma.

Step 1: The 90th percentile of the standard normal is z0.90=1.282z_{0.90} = 1.282
Step 2: Convert back: x=μ+σz=70+5(1.282)=70+6.41=76.41x = \mu + \sigma \cdot z = 70 + 5(1.282) = 70 + 6.41 = 76.41
Answer: The 90th percentile is approximately x=76.41x = 76.41.

Step 1: Standardize: z1=(9.810)/0.1=2z_1 = (9.8 - 10)/0.1 = -2, z2=(10.210)/0.1=2z_2 = (10.2 - 10)/0.1 = 2
Step 2: P(9.8<X<10.2)=Φ(2)Φ(2)=0.97720.0228=0.9545P(9.8 < X < 10.2) = \Phi(2) - \Phi(-2) = 0.9772 - 0.0228 = 0.9545
Step 3: P(out of tolerance)=10.9545=0.0455P(\text{out of tolerance}) = 1 - 0.9545 = 0.0455
Answer: About 4.55% of bolts fall outside the tolerance — consistent with the ±2σ\pm 2\sigma rule.

常见问题

The PDF (probability density function) f(x) gives the height of the bell curve at x — it does NOT directly give probabilities. The CDF Φ(z) gives the cumulative probability P(X ≤ x), i.e., the area under the PDF to the left of x. All probability computations use the CDF.

Standardize to get z = (x - μ)/σ, then P(X > x) = 1 - Φ(z). By symmetry of the standard normal, 1 - Φ(z) = Φ(-z), so you can also look up the left tail of the negative z-value.

For any normal distribution N(μ, σ²): approximately 68% of values fall within ±1σ of the mean, 95% within ±2σ, and 99.7% within ±3σ. This rule is useful for quick mental estimates without a calculator.

Many real-world measurements are approximately normal (heights, weights, test scores, measurement errors). For inference, the Central Limit Theorem guarantees that sample means are approximately normal for n ≥ 30 even if the population is not normal. Always check with a histogram or Q-Q plot before assuming normality for small samples.

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