## Basic exercises for lognormal distribution

This post presents exercises on the lognormal distribution. These exercises are to reinforce the basic properties discussed in this companion blog post.

Additional resources: another discussion of lognormal, a concise summary of lognormal and a problem set on lognormal.

_____________________________________________________________________________________

Exercises

Exercise 1
Let $X$ be a normal random variable with mean 6.5 and standard deviation 0.8. Consider the random variable $Y=e^X$. what is the probability $P(800 \le Y \le 1000)$?

$\text{ }$

Exercise 2
Suppose $Y$ follows a lognormal distribution with parameters $\mu=1$ and $\sigma=1$. Let $Y_1=1.25 Y$. Determine the following:

• The probability that $Y_1$ exceed 1.
• The 40th percentile of $Y_1$.
• The 80th percentile of $Y_1$.

$\text{ }$

Exercise 3
Let $Y$ follows a lognormal distribution with parameters $\mu=4$ and $\sigma=0.9$. Compute the mean, second moment, variance, third moment and the fourth moment.

$\text{ }$

Exercise 4
Let $Y$ be the same lognormal distribution as in Exercise 3. Use the results in Exercise 3 to compute the coefficient of variation, coefficient of skewness and the kurtosis.

$\text{ }$

Exercise 5
Given the following facts about a lognormal distribution:

• The lower quartile (i.e. 25% percentile) is 1000.
• The upper quartile (i.e. 75% percentile) is 4000.

Determine the mean and variance of the given lognormal distribution.

$\text{ }$

Exercise 6
Suppose that a random variable $Y$ follows a lognormal distribution with mean 149.157 and variance 223.5945. Determine the probability $P(Y>150)$.

$\text{ }$

Exercise 7
Suppose that a random variable $Y$ follows a lognormal distribution with mean 1200 and median 1000. Determine the probability $P(Y>1300)$.

$\text{ }$

Exercise 8
Customers of a very popular restaurant usually have to wait in line for a table. Suppose that the wait time $Y$ (in minutes) for a table follows a lognormal distribution with parameters $\mu=3.5$ and $\sigma=0.10$. Concerned about long wait time, the restaurant owner improves the wait time by expanding the facility and hiring more staff. As a result, the wait time for a table is cut by half. After the restaurant expansion,

• what is the probability distribution of the wait time for a table?
• what is the probability that a customer will have to wait more than 20 minutes for a table?

$\text{ }$
_____________________________________________________________________________________

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$
_____________________________________________________________________________________

Exercise 1

• 0.1040

$\text{ }$

Exercise 2

• 0.8888
• 1.4669
• 7.8707

$\text{ }$

Exercise 3

• $E(Y)=e^{4.405}$
• $E(Y^2)=e^{9.62}$
• $E(Y^3)=e^{15.645}$
• $E(Y^4)=e^{22.48}$
• $Var(Y)=(e^{0.81}-1) \ e^{8.81}$

$\text{ }$

Exercise 4

• $\displaystyle \text{CV}=$ 1.117098
• $\displaystyle \gamma_1=$ 4.74533
• $\displaystyle \beta_2=$ 60.41075686

$\text{ }$

Exercise 5

• $\displaystyle E(Y)=$ 3415.391045
• $\displaystyle E(Y^2)=$ 34017449.61
• $\displaystyle Var(Y)=$ 22352553.62

$\text{ }$

Exercise 6

• 0.4562

$\text{ }$

Exercise 7

• 0.3336

$\text{ }$

Exercise 8

• longnormal with $\mu=3.5+\log(0.5)$ and $\sigma=0.1$ where $\log$ is the natural logarithm.
• 0.0294

_____________________________________________________________________________________

$\copyright \ 2015 \text{ by Dan Ma}$
Revised October 18, 2018

## Two Practice Problems on the Standard Normal Distribution

This post presents two practice problems with calculation involving the standard normal distribution.

Problems
Let $Z$ be a standard normal random variable.

1. Evaluate $\displaystyle E(\lvert Z \lvert)$
2. Evaluate $\displaystyle E(Z^2)$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

We show that $\displaystyle E(\lvert Z \lvert)=\sqrt{\frac{2}{\pi}}$. Problem 2 is left as an exercise.

Let $X=\lvert Z \lvert$. The cumulative distribution function of $X$ is $F(x)=P(X \le x)$. We have the following.

\displaystyle \begin{aligned}(1) \ \ \ \ \ F(x)&= P(X \le x) \\&\text{ } \\&= P(\lvert Z \lvert \le x) \\&\text{ } \\&=P(-x \le Z \le x) \\&\text{ } \\&=\int_{-x}^x \frac{1}{\sqrt{2 \pi}} \ e^{-\frac{t^2}{2}} \ dt \\&\text{ } \\&=2\int_{0}^x \frac{1}{\sqrt{2 \pi}} \ e^{-\frac{t^2}{2}} \ dt \\&\text{ } \end{aligned}

Upon differentiation of this cdf, we have the probability density function (pdf) of $X$.

$\displaystyle (2) \ \ \ \ \ f(x)=\sqrt{\frac{2}{\pi}} \ e^{-\frac{x^2}{2}}$

The following is the calculation for $E(X)$.

$\displaystyle (3) \ \ \ \ \ E(X)=\sqrt{\frac{2}{\pi}} \ \int_0^\infty \ x e^{-\frac{x^2}{2}}=\sqrt{\frac{2}{\pi}}$