How is it related to the normal?

If X is lognormal, then ln X follows a normal distribution. Its parameters μ and σ are the mean and standard deviation of the logarithm, not of the original variable.

Why is it right-skewed?

Because it arises from multiplying many independent positive factors (percentage growths), which produces occasional very large values. As a result the mean exceeds the median, and the median e^μ often describes the typical case better.

What phenomena does it model well?

Positive quantities with a long right tail: incomes and salaries, prices, response times, particle or city sizes, and disability durations in insurance.

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Explanation

A random variable X follows a log-normal distribution if its natural logarithm ln(X) follows a normal distribution. This occurs naturally in processes where the outcome is the product of many independent random factors, just as the sum of factors leads to the normal distribution via the central limit theorem. The distribution is always positive and has a long right tail, which makes it suitable for modeling incomes, financial asset prices, repair times, particle sizes and task durations. Note: μ and σ are the parameters of ln(X), not of X directly. The mean of X is exp(μ + σ²/2) and its variance is [exp(σ²) − 1]·exp(2μ + σ²).

Formula

$ f(x;\mu,\sigma)=\frac{1}{x\sigma\sqrt{2\pi}}e^{-\frac{(\ln x-\mu)^2}{2\sigma^2}},\quad x>0 $

Parameters

μ (mu): mean of the natural logarithm of the variable. Can be any real number.
σ (sigma): standard deviation of the natural logarithm. Must be σ > 0.
x: positive value of the original variable (x > 0).

Worked example

Situation: The resolution time for incidents in a computer system (in minutes) follows a log-normal distribution with parameters $\mu = 2$ and $\sigma = 0.5$ (in natural logarithm). These parameters describe the behavior of the logarithm of the time, not the time directly.

Question 1: What is the probability that an incident is resolved in 10 minutes or less, $P(X \leq 10)$?

Solution: We standardize using the relation $\ln(X) \sim \mathcal{N}(\mu, \sigma^2)$: \[ P(X \leq 10) = P\!\left(Z \leq \frac{\ln 10 - 2}{0.5}\right) = P\!\left(Z \leq \frac{2.3026 - 2}{0.5}\right) = P(Z \leq 0.605) \approx 0.727 \] There is approximately a 72.7% probability of resolving the incident in 10 minutes or less.

Question 2: What is the median resolution time?

Solution: The median of the log-normal is: \[ \text{median} = e^\mu = e^2 \approx 7.389 \text{ minutes} \] The median is 7.39 minutes. The arithmetic mean, on the other hand, is higher: \[ E[X] = e^{\mu + \sigma^2/2} = e^{2 + 0.125} = e^{2.125} \approx 8.37 \text{ minutes} \]

Interpretation: The median (7.39 min) is lower than the mean (8.37 min), which reflects the typical right skew of the log-normal: most incidents are resolved relatively quickly, but some drag on much longer and pull the mean up. To size a technical support team, it is preferable to use high percentiles (e.g., the 90th percentile) as a service-level target.

How to interpret the result

The calculator offers three types of output. The PDF, $ f(x) $, gives the probability density at the point $ x > 0 $. The log-normal distribution is always positive and has a long right tail: most values cluster relatively close to the origin, but there is a non-negligible probability of very large values. The peak of the curve (the mode) is at $ e^{\mu - \sigma^2} $, which is lower than the median $ e^\mu $ and the mean $ e^{\mu + \sigma^2/2} $; this ordering is typical of distributions with positive skew. In the chart, the green area represents the probability of the selected range; the horizontal axis starts at zero.

The CDF, $ P(X \leq x) $, accumulates the probability up to the value $ x $ and is equivalent to the CDF of the standard normal evaluated at $ (\ln x - \mu)/\sigma $. For example, if income follows a log-normal with $ \mu = 10.8 $ and $ \sigma = 0.4 $, then $ P(X \leq 40{,}000) $ gives the proportion of the population earning $40,000 or less. The right tail $ P(X > x) $ measures the fraction of the distribution with extremely high values, which is relevant in risk analysis and finance, where tail events are the most costly.

The percentile or quantile result returns $ x = e^{\mu + z_p \sigma} $, where $ z_p $ is the quantile of the standard normal for the probability $ p $. The median (50th percentile) is exactly $ e^\mu $, always lower than the arithmetic mean. The 90th or 95th percentile is especially useful in value-at-risk (VaR) analysis, buffer sizing and system dimensioning where extreme values are the ones that drive the design.

Frequently asked questions

How is it related to the normal? If X is lognormal, then ln X follows a normal distribution. Its parameters μ and σ are the mean and standard deviation of the logarithm, not of the original variable.
Why is it right-skewed? Because it arises from multiplying many independent positive factors (percentage growths), which produces occasional very large values. As a result the mean exceeds the median, and the median e^μ often describes the typical case better.
What phenomena does it model well? Positive quantities with a long right tail: incomes and salaries, prices, response times, particle or city sizes, and disability durations in insurance.

Reference: Log-normal distribution — Wikipedia