Thursday, February 6, 2025

Unit II Uniform, normal, log normal, Exponential & Gamma Distributions

Lecture Notes on Probability Distributions for Continuous Random Variables

Probability distributions for continuous random variables – Uniform, normal, log normal, exponential and Gamma distributions – Applications

Introduction

Probability distributions for continuous random variables play a fundamental role in probability and statistics. In this lecture, we will explore five essential continuous distributions:

  • Uniform Distribution
  • Normal Distribution
  • Log-normal Distribution
  • Exponential Distribution
  • Gamma Distribution

Each of these distributions has important applications in civil engineering, including material strength analysis, reliability assessment, and environmental studies.


1. Uniform Distribution

Definition

A random variable XX follows a uniform distribution on the interval (a,b)(a, b) if its probability density function (PDF) is:

f(x)={1ba,axb0,otherwisef(x) = \begin{cases} \frac{1}{b-a}, & a \leq x \leq b \\ 0, & \text{otherwise} \end{cases}

Cumulative Distribution Function (CDF)

F(x)={0,x<axaba,axb1,x>bF(x) = \begin{cases} 0, & x < a \\ \frac{x-a}{b-a}, & a \leq x \leq b \\ 1, & x > b \end{cases}

Mean and Variance

E[X]=a+b2,Var(X)=(ba)212E[X] = \frac{a+b}{2}, \quad Var(X) = \frac{(b-a)^2}{12}

Application in Civil Engineering

  • Traffic Flow Analysis: Modeling vehicle arrival times at an intersection.
  • Construction Materials: Estimating the uniform spread of loads on a beam.

Example Problem

A pedestrian waits for a bus that arrives uniformly between 10:00 and 10:30 AM. What is the probability they wait less than 10 minutes?

Solution: Given a=0a = 0 min and b=30b = 30 min, the probability of waiting less than 10 minutes is:

P(X<10)=100300=1030=13P(X < 10) = \frac{10 - 0}{30 - 0} = \frac{10}{30} = \frac{1}{3}

2. Normal Distribution

Definition

A random variable XX follows a normal distribution with mean μ\mu and variance σ2\sigma^2, denoted as XN(μ,σ2)X \sim N(\mu, \sigma^2), if its PDF is:

f(x)=12πσ2e(xμ)22σ2,<x<f(x) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \quad -\infty < x < \infty

CDF

There is no closed-form expression, but values are found using normal tables or statistical software.

Mean and Variance

E[X]=μ,Var(X)=σ2E[X] = \mu, \quad Var(X) = \sigma^2

Application in Civil Engineering

  • Material Strength: Concrete strength follows a normal distribution.
  • Bridge Load Analysis: The distribution of vehicle weights on a bridge.

Example Problem

The compressive strength of concrete follows N(30,42)N(30, 4^2). What is the probability that a randomly selected sample has strength above 35 MPa?

Solution: Standardizing:

Z=Xμσ=35304=54=1.25Z = \frac{X - \mu}{\sigma} = \frac{35 - 30}{4} = \frac{5}{4} = 1.25

From the normal table: P(Z>1.25)=10.8944=0.1056P(Z > 1.25) = 1 - 0.8944 = 0.1056. Thus, the probability is 10.56%.


3. Log-normal Distribution

Definition

A random variable XX follows a log-normal distribution if Y=ln(X)Y = \ln(X) follows a normal distribution.

PDF:

f(x)=1xσ2πe(lnxμ)22σ2,x>0f(x) = \frac{1}{x \sigma \sqrt{2\pi}} e^{-\frac{(\ln x - \mu)^2}{2\sigma^2}}, \quad x > 0

Application in Civil Engineering

  • Soil Permeability: The distribution of soil permeability measurements.
  • Building Service Life: Modeling the lifespan of materials subject to degradation.

Example Problem

If Y=ln(X)Y = \ln(X) follows N(2,0.25)N(2, 0.25), find the median of XX.

Solution: The median of a log-normal distribution is:

Median(X)=eμ=e2=7.389Median(X) = e^\mu = e^2 = 7.389

Thus, the median is 7.39.


4. Exponential Distribution

Definition

The exponential distribution models the time until an event occurs. The PDF is:

f(x)=λeλx,x>0f(x) = \lambda e^{-\lambda x}, \quad x > 0

Mean and Variance

E[X]=1λ,Var(X)=1λ2E[X] = \frac{1}{\lambda}, \quad Var(X) = \frac{1}{\lambda^2}

Application in Civil Engineering

  • Reliability Analysis: Modeling the time between failures of construction equipment.
  • Queueing Theory: Modeling the time between vehicle arrivals at toll booths.

Example Problem

A water pump fails on average once every 10 years. What is the probability it lasts more than 15 years?

Solution: Here, λ=110\lambda = \frac{1}{10}. The probability is:

P(X>15)=eλx=e1510=e1.50.2231P(X > 15) = e^{-\lambda x} = e^{-\frac{15}{10}} = e^{-1.5} \approx 0.2231

So, the probability is 22.31%.


5. Gamma Distribution

Definition

The gamma distribution is a generalization of the exponential distribution. It is given by:

f(x)=λkxk1eλxΓ(k),x>0f(x) = \frac{\lambda^k x^{k-1} e^{-\lambda x}}{\Gamma(k)}, \quad x > 0

where kk is the shape parameter and λ\lambda is the rate parameter.

Mean and Variance

E[X]=kλ,Var(X)=kλ2E[X] = \frac{k}{\lambda}, \quad Var(X) = \frac{k}{\lambda^2}

Application in Civil Engineering

  • Flood Risk Analysis: Modeling the time between extreme rainfall events.
  • Structural Load Analysis: Modeling the cumulative load effects over time.

Example Problem

If failures in a bridge structure follow a gamma distribution with k=2k = 2 and λ=1/5\lambda = 1/5, find the probability that failure occurs before 8 years.

Solution: The cumulative probability is obtained using the gamma CDF:

P(X8)=08(1/5)2xex/5Γ(2)dxP(X \leq 8) = \int_0^8 \frac{(1/5)^2 x e^{-x/5}}{\Gamma(2)} dx

Using tables or software, P(X8)0.798P(X \leq 8) \approx 0.798, or 79.8%.


Summary Table

Distribution PDF Mean Variance Civil Engineering Applications
Uniform 1ba\frac{1}{b-a} a+b2\frac{a+b}{2} (ba)212\frac{(b-a)^2}{12} Traffic flow, material loads
Normal 12πσ2e(xμ)22σ2\frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} μ\mu σ2\sigma^2 Concrete strength, vehicle loads
Log-normal 1xσ2πe(lnxμ)22σ2\frac{1}{x \sigma \sqrt{2\pi}} e^{-\frac{(\ln x - \mu)^2}{2\sigma^2}} eμ+σ2/2e^{\mu + \sigma^2/2} (eσ21)e2μ+σ2(e^{\sigma^2} -1)e^{2\mu + \sigma^2} Soil permeability, material lifespan
Exponential λeλx\lambda e^{-\lambda x} 1/λ1/\lambda 1/λ21/\lambda^2 Equipment failure, traffic delays
Gamma λkxk1eλxΓ(k)\frac{\lambda^k x^{k-1} e^{-\lambda x}}{\Gamma(k)} k/λk/\lambda k/λ2k/\lambda^2 Flood risk, structural loads

Conclusion

Understanding these distributions helps civil engineers make informed decisions in infrastructure design, reliability analysis, and risk management.

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