Log-normal distribution
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.
If $Y\sim N(\mu,\sigma^2)$ then $X= e^Y$ follows a log-normal distribution with parameter $\mu$ and $\sigma^2$.
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
\[ \begin{equation*} f(x;\mu,\sigma) =\left\{ \begin{array}{ll} \frac{1}{\sqrt{2\pi}\sigma x}e^{-\frac{1}{2\sigma^2}(\ln x -\mu)^2}, & \hbox{$x\geq 0$;} \\ 0, & \hbox{$x < 0$.} \end{array} \right. \end{equation*} \]
$\mu$is location parameter$\sigma$is scale parameter
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
$$ \begin{aligned} P(3 \leq X\leq 5) &=P(\ln (3) \leq \ln X\leq \ln(5))\\ &=P(-0.2\leq Z\leq 0.82)\\ &= P(Z<0.82) -P(Z< -0.2)\\ &=0.7939-0.4207\\ &= 0.3732 \end{aligned} $$
Hope this example helps you understand how to solve the problems on log-normal distribution.
