UNIT II

 Unit–II : Probability Distributions for Discrete and Continuous Random Variables Probability  distributions  for  discrete  random  variables  – Bernoulli’s,  Binomial,  Geometric and Poisson distributions – applications 

Lecture Notes: Probability Distributions for Discrete and Continuous Random Variables

Introduction to Probability Distributions

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment.

• Discrete Random Variables: These take specific, countable values (e.g., number of cars on a bridge, cracks in a concrete slab).
• Continuous Random Variables: These take any value within a range (e.g., rainfall in mm, compressive strength of concrete).

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Discrete Probability Distributions

1. Bernoulli Distribution

• Definition: A distribution for a single trial with only two outcomes: success ($p$) or failure ($1-p$).

• Formula:

  $$P(X = x) = \begin{cases} p, & \text{if } x = 1 \, (\text{success}) \\ 1-p, & \text{if } x = 0 \, (\text{failure}) \end{cases}$$

  Here, $X \in \{0, 1\}$.

• Example in Civil Engineering: A soil test is conducted to check whether it is suitable ($X=1$) or unsuitable ($X=0$) for construction.

• Numerical Example:

  • Probability that a soil sample is suitable ($p$) = 0.8.
  • Probability of being unsuitable = $1-p = 0.2$.

  Outcome probabilities:

  $$P(X=1) = 0.8, \quad P(X=0) = 0.2$$

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2. Binomial Distribution

• Definition: Models the number of successes in $n$ independent Bernoulli trials.

• Formula:

  $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k = 0, 1, 2, \dots, n$$

  Where:

  • $n$: Number of trials.
  • $k$: Number of successes.
  • $p$: Probability of success.

• Example in Civil Engineering: Estimating the number of defective parts in a batch of $n$ parts manufactured.

• Numerical Example:

  • Probability of a defective part ($p$) = 0.1.
  • Batch size ($n$) = 5.
  • Calculate $P(X=2)$, the probability of 2 defective parts: $$P(X=2) = \binom{5}{2} (0.1)^2 (0.9)^3 = 10 \cdot 0.01 \cdot 0.729 = 0.0729$$

  Result: Probability of 2 defective parts is $7.29\%$.

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3. Geometric Distribution

• Definition: Models the number of trials needed to get the first success.

• Formula:

  $$P(X = k) = (1-p)^{k-1} p, \quad k = 1, 2, 3, \dots$$

  Where:

  • $k$: Number of trials.
  • $p$: Probability of success.

• Example in Civil Engineering: Number of soil tests required to find the first suitable site for construction.

• Numerical Example:

  • Probability that a site is suitable ($p$) = 0.3.
  • Find $P(X=3)$, the probability that the 3rd test is the first success: $$P(X=3) = (1-0.3)^2 \cdot 0.3 = 0.7^2 \cdot 0.3 = 0.147$$

  Result: Probability is $14.7\%$.

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4. Poisson Distribution

• Definition: Models the number of events occurring in a fixed interval of time or space.

• Formula:

  $$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots$$

  Where:

  • $\lambda$: Average rate of occurrence.
  • $k$: Number of events.

• Example in Civil Engineering: Number of vehicles crossing a bridge in an hour.

• Numerical Example:

  • Average number of vehicles per hour ($\lambda$) = 5.
  • Find $P(X=3)$, the probability of 3 vehicles crossing: $$P(X=3) = \frac{5^3 e^{-5}}{3!} = \frac{125 \cdot 0.0067}{6} \approx 0.14$$

  Result: Probability is $14\%$.

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Continuous Probability Distributions

1. Uniform Distribution

• Definition: Models variables uniformly distributed over an interval $[a, b]$.

• Formula:

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

• Example in Civil Engineering: The thickness of a concrete slab uniformly varying between 10 cm and 20 cm.

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2. Normal Distribution

• Definition: Models natural phenomena, with a symmetric bell-shaped curve.

• Formula:

  $$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$

  Where:

  • $\mu$: Mean.
  • $\sigma^2$: Variance.

• Example in Civil Engineering: Variations in compressive strength of concrete.

• Numerical Example:

  • Compressive strength ($X$) is normally distributed with $\mu = 40$ MPa and $\sigma = 5$ MPa.
  • Find $P(35 \leq X \leq 45)$:

    • Standardize $X$ to $Z$:

      $$Z = \frac{X - \mu}{\sigma}$$

      For $X=35$, $Z = \frac{35-40}{5} = -1$.
      For $X=45$, $Z = \frac{45-40}{5} = 1$.

    • Use standard normal tables:

      $$P(-1 \leq Z \leq 1) = P(Z \leq 1) - P(Z \leq -1)$$ $$P(Z \leq 1) = 0.8413, \quad P(Z \leq -1) = 0.1587$$ $$P(-1 \leq Z \leq 1) = 0.8413 - 0.1587 = 0.6826$$

  Result: Probability is $68.26\%$.

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Applications in Civil Engineering

1. Bernoulli Distribution: Checking soil suitability in a binary pass/fail test.
2. Binomial Distribution: Predicting the number of defective items in a batch.
3. Geometric Distribution: Determining the number of tests to find a viable construction site.
4. Poisson Distribution: Estimating vehicle counts or defects over a time period.
5. Normal Distribution: Modeling variability in material properties like tensile strength or elasticity.

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Conclusion

Understanding probability distributions is critical in civil engineering for risk assessment, design, and decision-making under uncertainty. These distributions allow engineers to quantify variability and make informed decisions.

Probability  distributions  for  continuous  random  variables  – Uniform,  normal,  log normal,  exponential  and  gamma  distributions – statistical  probability  distributions  – Students’ t, Chi – square and F – distributions – applications.

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Lecture Notes: Probability Distributions for Continuous Random Variables

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Introduction

For continuous random variables, probability distributions describe the likelihood of a variable taking a value within a certain range. Unlike discrete variables, the probability for a specific value is zero, and we use probability density functions (PDFs) to calculate probabilities over intervals.

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Continuous Probability Distributions

1. Uniform Distribution

• Definition: The variable is equally likely to occur anywhere within a specified range $[a, b]$.

• PDF:

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

• Mean and Variance:

  $$\mu = \frac{a+b}{2}, \quad \sigma^2 = \frac{(b-a)^2}{12}$$

• Application in Civil Engineering: Thickness of asphalt layers laid by machines within a tolerance range.

• Numerical Example:
  Thickness varies uniformly between $a = 10 \, \text{cm}$ and $b = 20 \, \text{cm}$. Find $P(12 \leq X \leq 15)$.

  $$P(12 \leq X \leq 15) = \int_{12}^{15} \frac{1}{20-10} dx = \frac{1}{10} \cdot (15 - 12) = 0.3$$

  Result: Probability is $30\%$.

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2. Normal Distribution

• Definition: A bell-shaped distribution often used for natural phenomena.

• PDF:

  $$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$

• Mean and Variance: $\mu$, $\sigma^2$ (parameters of the distribution).

• Application in Civil Engineering: Variations in concrete strength or traffic flow rates.

• Numerical Example:
  Compressive strength of concrete is normally distributed with $\mu = 30 \, \text{MPa}$ and $\sigma = 5 \, \text{MPa}$. Find $P(25 \leq X \leq 35)$.

  • Standardize $Z = \frac{X - \mu}{\sigma}$. For $X = 25$, $Z = \frac{25-30}{5} = -1$; for $X = 35$, $Z = \frac{35-30}{5} = 1$.
  • Use standard normal tables: $$P(-1 \leq Z \leq 1) = 0.6826$$

  Result: Probability is $68.26\%$.

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3. Log-Normal Distribution

• Definition: The logarithm of the variable follows a normal distribution. Used for non-negative, skewed data.

• PDF:

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

• Application in Civil Engineering: Modeling particle sizes in soil or material lifetimes.

• Numerical Example:
  If the log of concrete strength follows $N(\mu = 3, \sigma^2 = 0.25)$, find the probability that strength exceeds $25$.

  • Convert: $\ln(25) = 3.2189$.
  • Standardize: $Z = \frac{3.2189 - 3}{\sqrt{0.25}} = 0.438$.
  • Use normal table: $P(Z > 0.438) = 1 - 0.668 \approx 0.332$.
    Result: Probability is $33.2\%$.

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4. Exponential Distribution

• Definition: Models the time until an event occurs (e.g., failure rates).

• PDF:

  $$f(x) = \lambda e^{-\lambda x}, \quad x \geq 0$$

• Mean and Variance: $\mu = \frac{1}{\lambda}, \quad \sigma^2 = \frac{1}{\lambda^2}$.

• Application in Civil Engineering: Time until machinery failure or time between vehicle arrivals.

• Numerical Example:
  The mean time between pump failures is $\mu = 5$ hours. Find the probability that the pump operates for more than 8 hours.

  • $\lambda = \frac{1}{\mu} = 0.2$.
  $$P(X > 8) = \int_8^\infty 0.2 e^{-0.2x} dx = e^{-0.2 \cdot 8} = e^{-1.6} \approx 0.201$$

  Result: Probability is $20.1\%$.

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5. Gamma Distribution

• Definition: Models the time for multiple events to occur.

• PDF:

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

  Where $\Gamma(k)$ is the gamma function.

• Mean and Variance: $\mu = \frac{k}{\lambda}, \quad \sigma^2 = \frac{k}{\lambda^2}$.

• Application in Civil Engineering: Total rainfall over a season or time to complete multiple stages of a project.

• Numerical Example:
  A storm produces rainfall where $k = 3$, $\lambda = 2$. Find the probability that total rainfall exceeds $5$ units.

  • Use the incomplete gamma function: $$P(X > 5) = 1 - \int_0^5 \frac{2^3 x^2 e^{-2x}}{\Gamma(3)} dx$$ Using software or tables: $$P(X > 5) \approx 0.184$$

  Result: Probability is $18.4\%$.

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Statistical Probability Distributions

1. Student’s $t$-Distribution

• Definition: Models small-sample means when the population standard deviation is unknown.
• Application: Estimating mean material properties with limited data.

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2. Chi-Square Distribution

• Definition: Models the sum of squared standard normal variables.
• Application: Analyzing variance or goodness-of-fit tests.

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3. $F$-Distribution

• Definition: Ratio of two independent chi-square distributions.
• Application: Comparing variances between two groups (e.g., soil strengths under different treatments).

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Summary of Applications

Distribution	Applications
Uniform	Material tolerances (e.g., thickness).
Normal	Variations in material properties.
Log-Normal	Particle sizes, material lifetimes.
Exponential	Time to failure, inter-arrival times.
Gamma	Total rainfall, project completion times.
Student’s $t$	Small-sample means for material properties.
Chi-Square	Variance analysis, goodness-of-fit.
$F$	Comparing variability across groups.

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These notes cover definitions, formulas, real-life examples, and numerical problems for key continuous probability distributions and their applications in civil engineering. Let me know if you'd like further clarifications or additional examples!

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Numerical Examples for Bernoulli, Binomial, Geometric, and Poisson Distributions

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1. Bernoulli Distribution

Scenario:

A civil engineer tests soil samples for suitability for construction. The probability of a soil sample being suitable ($X = 1$) is $p = 0.7$, and the probability of being unsuitable ($X = 0$) is $1-p = 0.3$.

Objective:

Find the probability that:

1. A soil sample is suitable ($P(X=1)$).
2. A soil sample is unsuitable ($P(X=0)$).

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Solution:

1. Understand the Bernoulli Distribution Formula:

   $$P(X = x) = \begin{cases} p, & \text{if } x = 1 \\ 1-p, & \text{if } x = 0 \end{cases}$$

2. Calculate Probabilities:

   • For $X = 1$ (suitable):

     $$P(X = 1) = p = 0.7$$

   • For $X = 0$ (unsuitable):

     $$P(X = 0) = 1-p = 0.3$$

Final Answer:

• Probability of suitability ($X=1$) is 0.7 (70%).
• Probability of unsuitability ($X=0$) is 0.3 (30%).

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2. Binomial Distribution

Scenario:

In a batch of 10 soil samples, the probability of a sample being suitable for construction is $p = 0.7$. The engineer wants to calculate the probability that exactly 6 out of the 10 samples are suitable.

Objective:

Find $P(X = 6)$ using the Binomial Distribution.

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Solution:

1. Understand the Binomial Distribution Formula:

   $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$

   Where:

   • $n = 10$ (number of trials).
   • $k = 6$ (number of successes).
   • $p = 0.7$ (probability of success).
   • $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

2. Calculate the Binomial Coefficient:

   $$\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10 \cdot 9 \cdot 8 \cdot 7}{4 \cdot 3 \cdot 2 \cdot 1} = 210$$

3. Calculate $P(X = 6)$:

   $$P(X = 6) = \binom{10}{6} (0.7)^6 (1-0.7)^4$$
   • Calculate powers: $$(0.7)^6 = 0.117649, \quad (0.3)^4 = 0.0081$$
   • Substitute values: $$P(X = 6) = 210 \cdot 0.117649 \cdot 0.0081 \approx 0.2001$$

Final Answer:

The probability that exactly 6 out of 10 samples are suitable is approximately 0.2001 (20.01%).

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3. Geometric Distribution

Scenario:

A civil engineer is testing sites for water availability. The probability that a site is water-abundant ($X=1$) is $p = 0.4$. The engineer wants to calculate the probability that the first water-abundant site is found on the 4th test.

Objective:

Find $P(X = 4)$.

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Solution:

1. Understand the Geometric Distribution Formula:

   $$P(X = k) = (1-p)^{k-1} p$$

   Where:

   • $p = 0.4$ (probability of success).
   • $k = 4$ (first success on the 4th trial).

2. Calculate $P(X = 4)$:

   $$P(X = 4) = (1-0.4)^{4-1} \cdot 0.4$$
   • Calculate $(1-p)^{k-1}$: $$(1-0.4)^3 = (0.6)^3 = 0.216$$
   • Multiply by $p$: $$P(X = 4) = 0.216 \cdot 0.4 = 0.0864$$

Final Answer:

The probability that the first water-abundant site is found on the 4th test is 0.0864 (8.64%).

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4. Poisson Distribution

Scenario:

The number of vehicles crossing a rural bridge follows a Poisson distribution with an average rate of 3 vehicles per hour ($\lambda = 3$). The engineer wants to find the probability that exactly 5 vehicles cross the bridge in an hour.

Objective:

Find $P(X = 5)$.

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Solution:

1. Understand the Poisson Distribution Formula:

   $$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$$

   Where:

   • $\lambda = 3$ (mean rate).
   • $k = 5$ (number of occurrences).

2. Calculate $P(X = 5)$:

   • $\lambda^k = 3^5 = 243$.
   • $e^{-\lambda} \approx e^{-3} = 0.0498$.
   • $k! = 5! = 120$.
   $$P(X = 5) = \frac{3^5 \cdot e^{-3}}{5!} = \frac{243 \cdot 0.0498}{120} \approx 0.1008$$

Final Answer:

The probability of exactly 5 vehicles crossing the bridge in an hour is approximately 0.1008 (10.08%).

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Summary of Examples

1. Bernoulli Distribution: Probability of a soil sample being suitable or unsuitable.
2. Binomial Distribution: Number of suitable soil samples in a batch.
3. Geometric Distribution: Number of trials to find the first water-abundant site.
4. Poisson Distribution: Number of vehicles crossing a bridge in a fixed time.

These step-by-step examples highlight how these distributions are applied to real-life civil engineering problems. Let me know if you need additional clarifications or examples!