Log Normal Distribution Calculator With Examples

Log-normal distribution calculator

Log-normal Distribution Calculator is used to find mean, variance and probabilities of various type of events for Log-normal distribution with parameter $\mu$ and $\sigma^2$.

Log-Normal Probability Calculator
First Parameter ($\mu$)
Second parameter ($\sigma$)
P(X< A)
P(X > B)
P(A< X < B) and
Outside A and B and
Results
Mean :
Variance :
Required Probability :

How to calculate probabilities of Log-normal Distribution?

Step 1 - Enter the first parameter $\mu$

Step 2 - Enter the second parameter $\sigma$

Step 3 - Select the probability type

Step 4 - Click on "Calculate" button to get Log-normal distribution probabilities

Step 5 - Gives the output of mean and variance of log-normal distribution

Step 6 - Gives the output of required probability for log-normal distribution

Definition of Log-normal Distribution

The continuous random variable $X$ has a Log Normal Distribution if the random variable $Y=\ln (X)$ has a normal distribution with mean $\mu$ and standard deviation $\sigma$. The probability density function of $X$ is

$$ \begin{align*} f(x;\mu,\sigma) &= \begin{cases} \frac{1}{\sqrt{2\pi}\sigma x}e^{-\frac{1}{2\sigma^2}(\ln x -\mu)^2}, & x\geq 0; \\ 0, & x < 0. \end{cases} \end{align*} $$

  • $\mu$ is location parameter
  • $\sigma$ is scale parameter

Normal distribution is not suitable when the data are highly skewed or data contains outliers. In such a situation, log-normal distribution is often a good choice.
The log-normal distribution is derived from a normal distribution as follows:
If $Y\sim N(\mu,\sigma^2)$ then $X= e^Y$ follows a log-normal distribution with parameter $\mu$ and $\sigma^2$.

Mean of Log-normal distribution

The mean of Log-normal distribution is $E(X) = e^{\mu+\sigma^2/2}$.

Variance of Log-normal distribution

The variance of Log-normal distribution is $V(X) =e^{2\mu+\sigma^2}\big(e^{\sigma^2}-1\big)$.

Log-normal Distribution Example

The life-time (in days) of certain electrionic component that operates in a high-temperature environment is log-normally distributed with $\mu=1.2$ and $\sigma=0.5$.

a. Find mean and variance of lifetime of electronic component.

b. Find the probability that the component works till 4 days.

c. Find the probability that the component works more than 5 days.

d. Find the probability that the component works between 3 and 5 days.

Solution

Let $X$ denote the life-time (in days) of certain electronic components that operates in a high-temperature environment. Given that $X\sim LN(1.2, 0.5^2)$. That is $\mu = 1.2$ and $\sigma = 0.5$.

Then $\ln(X)\sim N(1.2,0.25)$ distribution.

a. The mean of Log-normal distribution is

$$ \begin{aligned} E(X) &= e^{\mu+\sigma^2/2}\\ &= e^{1.2 + 0.5^2/2}\\ &= e^{1.325}\\ &= 3.7622 \end{aligned} $$

and the variance of log-normal distribution is

$$ \begin{aligned} V(X) &= e^{2\mu+\sigma^2}\big(e^{\sigma^2}-1\big)\\ &= e^{2*1.2 + 0.5^2}\big(e^{0.5^2}-1\big)\\ &= e^{2.65}\big(e^{0.25}-1\big)\\ &= 14.154\big(0.284\big)\\ &= 4.0197 \end{aligned} $$

b. The probability that the component works till 4 days is $P(X<4)$.

The $Z$ score that corresponds to $4$ is

$$ \begin{aligned} z&=\dfrac{\ln(X)-\mu}{\sigma}\\ &=\dfrac{\ln(4)-1.2}{0.5}\\ &\approx0.37 \end{aligned} $$
Thus the probability that the component works till 4 days is

$$ \begin{aligned} P(X < 4) &=P(\ln(X) < \ln(4))\\ &=P(Z < 0.37)\\ &=0.6443 \end{aligned} $$

c. The probability that the component works more than 5 days is $P(X>5)$.

The $Z$ score that corresponds to $5$ is

$$ \begin{aligned} z&=\dfrac{\ln(X)-\mu}{\sigma}\\ &=\dfrac{\ln(5)-1.2}{0.5}\\ &\approx0.82 \end{aligned} $$

The probability that the component works more than 5 days is

$$ \begin{aligned} P(X > 5) &=1-P(X < 5)\\ &= 1-P(\ln X < \ln (5))\\ &= 1-P(Z < 0.82)\\ &=1-0.7939\\ &=0.2061 \end{aligned} $$

d. The probability that the component works between 3 and 5 days is $P(3 < X < 5)$.

The Z score that corresponds to $3$ and $5$ are respectively

$$ \begin{aligned} z_1&=\dfrac{\ln(X)-\mu}{\sigma}\\ &=\dfrac{\ln(3)-1.2}{0.5}\\ &\approx-0.2 \end{aligned} $$

and

$$ \begin{aligned} z_2&=\dfrac{\ln(X)-\mu}{\sigma}\\ &=\dfrac{\ln(5)-1.2}{0.5}\\ &\approx0.82 \end{aligned} $$

The probability that the component works between 3 and 5 days is

$$ \begin{aligned} P(3 \leq X\leq 5) &=P(\ln (3) \leq \ln X\leq \ln(5))\\ &=P(-0.2\leq Z\leq 0.82)\\ &= P(Z < 0.82) -P( Z < -0.2)\\ &=0.7939-0.4207\\ &= 0.3732 \end{aligned} $$

Conclusion

In this tutorial, you learned about how to calculate probabilities of Log-normal distribution. You also learned about how to solve numerical problems based on Log-normal distribution.

To read more about the step by step tutorial on Log-normal distribution refer the link Log-normal Distribution. This tutorial will help you to understand Log-normal distribution and you will learn how to derive mean, variance, moments Log-normal distribution and other properties of Log-normal distribution.

To learn more about other probability distributions, please refer to the following tutorial:

Probability distributions

Let me know in the comments if you have any questions on Log-normal Distribution Examples and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

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