Normal Distribution Calculator

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

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тИСMath Input
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 тАФ heights, test scores, measurement errors тАФ and is the foundation of most parametric inference.

Notation

XтИ╝N(╬╝,╧Г2)X \sim N(\mu, \sigma^2)

where ╬╝\mu is the mean (center of the bell) and ╧Г2\sigma^2 is the variance (╧Г\sigma is the standard deviation, the spread).

Probability Density Function (PDF)

f(x)=1╧Г2╧АexpтБб(тИТ(xтИТ╬╝)22╧Г2)f(x) = \frac{1}{\sigma\sqrt{2\pi}} \exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)

Key Properties

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

The Standard Normal Distribution and Z-Scores

The standard normal distribution is the special case ZтИ╝N(0,1)Z \sim N(0,1) тАФ mean 0, standard deviation 1. Its CDF is written ╬ж(z)\Phi(z).

Standardizing (Computing Z-Scores)

Any normal random variable XтИ╝N(╬╝,╧Г2)X \sim N(\mu, \sigma^2) can be converted to a standard normal variable:

Z=XтИТ╬╝╧ГZ = \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 ╬ж(z)\Phi(z) (the area to the left of zz):

P(X<x)=PтАЙтБг(Z<xтИТ╬╝╧Г)=╬жтАЙтБг(xтИТ╬╝╧Г)P(X < x) = P\!\left(Z < \frac{x-\mu}{\sigma}\right) = \Phi\!\left(\frac{x-\mu}{\sigma}\right)

P(X>x)=1тИТ╬жтАЙтБг(xтИТ╬╝╧Г)P(X > x) = 1 - \Phi\!\left(\frac{x-\mu}{\sigma}\right)

P(a<X<b)=╬жтАЙтБг(bтИТ╬╝╧Г)тИТ╬жтАЙтБг(aтИТ╬╝╧Г)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(XтЙдx)=pP(X \leq x) = p:

x=╬╝+╧ГтЛЕzpx = \mu + \sigma \cdot z_p

where zp=╬жтИТ1(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 X╦Й\bar{X} approaches N(╬╝,╧Г2/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, ╧З2\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, XтИ╝Binomial(n,p)X \sim \text{Binomial}(n, p) is approximately N(np,тАЙnp(1тИТp))N(np,\, np(1-p)):

P(XтЙдk)тЙИ╬жтАЙтБг(k+0.5тИТnpnp(1тИТp))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 npтЙе10np \geq 10 and n(1тИТp)тЙе10n(1-p) \geq 10.

Examples

Step 1: Standardize: z=(115тИТ100)/15=1.00z = (115 - 100)/15 = 1.00
Step 2: Look up ╬ж(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=(65тИТ70)/5=тИТ1.00z_1 = (65 - 70)/5 = -1.00
Step 2: Standardize upper bound: z2=(75тИТ70)/5=1.00z_2 = (75 - 70)/5 = 1.00
Step 3: P(тИТ1<Z<1)=╬ж(1)тИТ╬ж(тИТ1)=0.8413тИТ0.1587=0.6827P(-1 < Z < 1) = \Phi(1) - \Phi(-1) = 0.8413 - 0.1587 = 0.6827
Answer: P(65<X<75)тЙИ0.6827P(65 < X < 75) \approx 0.6827 тАФ the classic 68% rule for ┬▒1╧Г\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=╬╝+╧ГтЛЕz=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=(9.8тИТ10)/0.1=тИТ2z_1 = (9.8 - 10)/0.1 = -2, z2=(10.2тИТ10)/0.1=2z_2 = (10.2 - 10)/0.1 = 2
Step 2: P(9.8<X<10.2)=╬ж(2)тИТ╬ж(тИТ2)=0.9772тИТ0.0228=0.9545P(9.8 < X < 10.2) = \Phi(2) - \Phi(-2) = 0.9772 - 0.0228 = 0.9545
Step 3: P(out┬аof┬аtolerance)=1тИТ0.9545=0.0455P(\text{out of tolerance}) = 1 - 0.9545 = 0.0455
Answer: About 4.55% of bolts fall outside the tolerance тАФ consistent with the ┬▒2╧Г\pm 2\sigma rule.

Frequently Asked Questions

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

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

For any normal distribution N(╬╝, ╧Г┬▓): approximately 68% of values fall within ┬▒1╧Г of the mean, 95% within ┬▒2╧Г, and 99.7% within ┬▒3╧Г. 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 тЙе 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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