ABC Tools

Log-Normal Distribution, Explained

After reading this you can compute densities, cumulative probabilities and quantiles for a log-normal distribution, and you will know when your right-skewed data actually fits one.

What a log-normal distribution is

A random variable X is log-normal when its natural logarithm follows a normal distribution. If \ln(X) is normal with mean \mu and standard deviation \sigma, then X itself is log-normal. The variable X is always positive, and its shape is right-skewed: many small values, a long tail of large ones.

Here is the hook. Suppose household incomes have \mu = 11 and \sigma = 0.6 on the log scale (income measured in dollars). The typical income is e^{11} \approx 59{,}874, but the mean is e^{11 + 0.6^2/2} \approx 71{,}671. The mean sits above the median because the tail pulls it right. That gap is the signature of skew, and the log-normal captures it directly.

Note the two parameters describe the log of the data, not the data. This trips people up constantly, so keep it in front of you: \mu and \sigma live on the \ln(X) scale.

When to use it, and when not to

Reach for the log-normal when a quantity is positive, right-skewed, and arises from many small multiplicative effects. That last part matters. The normal distribution appears when you add many small independent effects; the log-normal appears when you multiply them, because the log of a product is a sum of logs.

Good candidates include incomes, file sizes, insurance claim amounts, biological measurements like the concentration of a substance, time-to-repair figures, and stock price ratios over a fixed horizon.

Do not use it for data that can be zero or negative. The log-normal has no mass at or below zero, since \ln(0) is undefined. Counts of events belong to the Poisson calculator, and waiting times with a constant hazard belong to the exponential calculator.

Also check the skew before committing. If a histogram of \ln(X) looks roughly symmetric and bell-shaped, the log-normal is a reasonable model. If \ln(X) is still skewed, try the gamma or Weibull distribution instead.

The formula and the intuition

The probability density function for x \gt 0 is:

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

Here \mu is the mean of \ln(X), \sigma is the standard deviation of \ln(X), and x is the value you evaluate. The 1/x factor is the density's stretch when you move from the log scale back to the original scale.

The cumulative distribution function uses the standard normal CDF \Phi:

F(x) = \Phi\left(\frac{\ln x - \mu}{\sigma}\right)

This is the key trick. To find P(X \le x), take \ln x, turn it into a z-score with (\ln x - \mu)/\sigma, and look it up in the normal table. Every log-normal probability question reduces to a normal question on the log scale.

The quantile (inverse) goes the other way. For a target probability p:

x_p = \exp\left(\mu + \sigma \, \Phi^{-1}(p)\right)

The mean, median and mode of X are all different, which is why "average" is ambiguous for skewed data:

\text{median} = e^{\mu}, \quad \text{mean} = e^{\mu + \sigma^2/2}, \quad \text{mode} = e^{\mu - \sigma^2}

With \mu = 0 and \sigma = 1, the median is 1, the mean is e^{0.5} \approx 1.649, and the mode is e^{-1} \approx 0.368. Three different centers, one distribution.

A worked example with the demo data

Probability from x with μ = 0, σ = 1, x = 1

The demo uses \mu = 0, \sigma = 1, and x = 1 in probability mode.

  1. Compute \ln x = \ln 1 = 0.
  2. Standardize: z = (0 - 0)/1 = 0.
  3. Look up the normal CDF: \Phi(0) = 0.5. So P(X \le 1) = 0.5.
  4. The density at x = 1 is \frac{1}{1 \cdot 1 \cdot \sqrt{2\pi}} \exp(0) = 0.3989.

The result P(X \le 1) = 0.5 confirms that 1 is the median here, matching e^{\mu} = e^0 = 1. Exactly half the mass sits below 1, even though the mean is 1.649. The half above 1 stretches far to the right.

x from probability with p = 0.95

Switch to critical mode with the same parameters and p = 0.95.

  1. Find the normal quantile: \Phi^{-1}(0.95) = 1.645.
  2. Apply the transform: x_{0.95} = \exp(0 + 1 \cdot 1.645) = e^{1.645}.
  3. Evaluate: x_{0.95} \approx 5.180.

So 95% of the mass lies below about 5.18, and 5% lies above it. Compare that to the median of 1: the 95th percentile is more than five times the median. That long upper reach is what makes the log-normal useful for tail risk.

See how σ reshapes the curve

The density peaks at the mode (about 0.368) then trails off slowly. The vertical marker at 5.18 is the 95th percentile from the worked example.

With μ fixed at 0, raising σ from 0.25 to 1.5 stretches the right tail and lowers the peak. At σ = 0.25 the curve is nearly symmetric and bunched near x = 1; at σ = 1.5 the mode drops below 0.11 and the tail reaches far past x = 10.

Reading and interpreting the results

In probability mode the calculator returns two numbers: the density f(x) and the cumulative probability F(x) = P(X \le x). The density is not a probability. It is the height of the curve, useful for comparing likelihoods but not readable as a percentage. The cumulative value is the one you quote: F(5.18) = 0.95 means "95% of outcomes fall at or below 5.18."

To get an upper-tail probability, subtract from one. For the demo, P(X \gt 1) = 1 - 0.5 = 0.5. For an interval, subtract two CDF values: P(1 \lt X \le 5.18) = 0.95 - 0.5 = 0.45.

In critical mode you supply p and read back the value x_p such that P(X \le x_p) = p. Use it for percentiles: the 50th percentile is the median, the 90th tells you the threshold below which nine tenths of the data sit.

Common mistakes

Feeding in the mean of the raw data as μ
The parameter \mu is the mean of \ln(X), not of X. If your data average 60,000, then \mu \approx \ln(60000) = 11.0 only if the median is near that, and even then you must estimate \mu from \ln of the values, not from their arithmetic mean.
Confusing σ of the logs with the coefficient of variation
For a log-normal, \sigma is close to the coefficient of variation only when \sigma is small. At \sigma = 0.1 the coefficient of variation is about 0.100; at \sigma = 1 it is \sqrt{e^{1} - 1} \approx 1.311. They diverge fast.
Reading the density as a probability
The density at x = 1 is 0.3989. That is not a 40% chance of anything. Only areas under the curve are probabilities.
Using natural log versus log base 10 inconsistently
This calculator uses natural logs. If you estimate \mu and \sigma from base-10 logs, convert by multiplying both by \ln(10) \approx 2.303 before entering them.

Related calculators

The log-normal is the normal on a log scale, so the Normal distribution calculator and the Z-score calculator are the closest companions: every log-normal lookup is a normal lookup after taking \ln x. For other positive, skewed quantities compare the Gamma calculator and the Weibull calculator, which model failure and waiting times with different tail shapes. If your quantity is a count or a proportion instead, see the Binomial calculator or the Beta calculator.

Frequently asked questions

How do I estimate μ and σ from my data?

Take the natural log of every value, then compute the sample mean and sample standard deviation of those logs. Those two numbers are your \mu and \sigma. For example, if \ln x values average 0 with standard deviation 1, enter \mu = 0 and \sigma = 1.

Why is the mean larger than the median?

Because the distribution is right-skewed. The mean is e^{\mu + \sigma^2/2} and the median is e^{\mu}, so the mean exceeds the median by a factor e^{\sigma^2/2}. At \sigma = 1 that factor is e^{0.5} \approx 1.649.

Can x be exactly zero?

No. The support is x \gt 0, and f(0) = 0. As x approaches zero the density falls to zero, so the distribution never places mass at the origin.

What does σ = 0 mean?

It is a degenerate case: all mass collapses onto the single point e^{\mu}. The calculator needs \sigma \gt 0 to produce a spread.

How is this different from the exponential distribution?

Both are positive and right-skewed, but the exponential has a constant hazard rate and a single parameter, while the log-normal has a bell-shaped log scale and two parameters. The exponential's density is highest at zero; the log-normal's density is zero at the origin and peaks at the mode e^{\mu - \sigma^2}.

Try it yourself in the tool

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