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Chapter 4: Problem 191

Suppose that the construction of a solar power station is initiated. The project's completion time has not been set due to uncertainties in financial resources. The completion time for the first phase is modeled with a beta distribution and the minimum, most likely (mode), and maximum completion times for the first phase are \(1.0,1.25,\) and 2.0 years, respectively. Also, the mean time is assumed to equal \(\mu=1+4(1.25)+2) / 6=1.333 .\) Determine the following in parts (a) and (b): (a) Parameters \(\alpha\) and \(\beta\) of the beta distribution. (b) Standard deviation of the distribution. (c) Sketch the probability density function.

Short Answer

Expert verified
Parameters are \(\alpha = 3\), \(\beta = 2\), and standard deviation is 0.2.

Step by step solution

01

Understand the Beta Distribution

A beta distribution is defined by two parameters, \(\alpha\) and \(\beta\). It is often used to model random variables limited to intervals of finite length in a wide variety of disciplines.
02

Set Up the Mean Formula

The mean of a beta distribution is given by \(\mu = \frac{\alpha}{\alpha + \beta}\). We are given that \(\mu = 1.333\).
03

Use the Mode Formula

The mode of a beta distribution is given by \(\frac{\alpha - 1}{\alpha + \beta - 2}\), and we know the mode is 1.25.
04

Apply the Parameter Conditions

Since we have both the mean and mode formulas, we need to use these formulas alongside boundary conditions to solve for \(\alpha\) and \(\beta\). Be aware that the range for completion times is \([1.0, 2.0]\).
05

Calculate \(\alpha\) and \(\beta\)

By using the boundary conditions and given equations:1. \(\mu = \frac{\alpha}{\alpha + \beta} = 1.333\) leading to \(1.333\alpha + 1.333\beta = \alpha\).2. \(1.25 = \frac{\alpha - 1}{\alpha + \beta - 2}\) leading to \(\alpha - 1 = 1.25(\alpha + \beta - 2)\).By solving these simultaneous equations, we find \(\alpha=3\) and \(\beta=2\).
06

Calculate the Standard Deviation

The standard deviation \(\sigma\) for a beta distribution is calculated using \(\sigma = \sqrt{\frac{\alpha\beta}{(\alpha + \beta)^2(\alpha + \beta + 1)}} = \sqrt{\frac{3 \times 2}{(3 + 2)^2(3 + 2 + 1)}} = \sqrt{\frac{6}{5^2 \times 6}} = \sqrt{\frac{6}{150}} = \sqrt{0.04}\). Hence, \(\sigma = 0.2\).
07

Sketch the Probability Density Function

Sketching the graph involves plotting the beta distribution over the interval \([1.0, 2.0]\) with calculated \(\alpha\) and \(\beta\) values. The shape will indicate that it is right-skewed as \(\alpha > \beta\). This graph depicts how the probability density changes for each completion time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function
The Probability Density Function (PDF) is a core concept in understanding how probabilities are distributed across a continuous random variable. For the beta distribution in our exercise, the PDF defines the likelihood of different completion times for the solar power station phase, ranging from 1.0 to 2.0 years.

The PDF graphically displays how probabilities accumulate. It allows us to visualize the distribution's shape. In a beta distribution, this shape can vary extensively based on the values of the parameters \( \alpha \) and \( \beta \). With \( \alpha = 3 \) and \( \beta = 2 \), our distribution is right-skewed. This skewness indicates that shorter completion times are more probable.
  • The PDF must integrate to 1 over the defined interval, ensuring the total probability sums up correctly.
  • Being right-skewed reflects that unusual delays beyond the most likely time are less probable but possible.
Understanding the PDF gives insights into the likelihood of different outcomes in our modeled interval.
Standard Deviation
Standard deviation is a statistical tool that helps us understand the spread of observations around their mean in a distribution. It measures variability. In a beta distribution, it is critical for understanding project completion variability.

Through our calculations, we determined that the standard deviation \( \sigma \) for the given beta distribution is 0.2 years. A smaller standard deviation suggests that most completion times are closely packed around the mean of 1.333 years.
  • A standard deviation tells us the typical distance from the average completion time.
  • It can inform project managers about the reliability and predictability of their schedules.
The precise calculation of \( \sigma \) is crucial for risk assessment, ensuring planning accounts for potential delays and variability.
Random Variables
Random variables are fundamental to probability and statistics, representing outcomes of uncertain events numerically. In this exercise, the completion time of a solar project phase acts as a random variable. It can take any real value in its defined range, thus being continuous rather than discrete.

Since it's modeled by a beta distribution, this completion time is influenced by parameters \( \alpha \) and \( \beta \), which govern its distribution shape.
  • The range for our random variable is set between 1.0 and 2.0 years, controlling possible outcomes.
  • It introduces the element of uncertainty from unpredictable financial factors.
Utilizing random variables helps quantify uncertainty, facilitating data-informed decisions.
Parameters of Distribution
Parameters \((\alpha, \beta)\) determine the unique characteristics of the beta distribution. They shape the density function and control its skewness and variability.

In our example, \( \alpha = 3 \) and \( \beta = 2 \), meaning the distribution is right-skewed. This occurs because \( \alpha > \beta \), suggesting a stronger pull towards shorter completion times around the mode of 1.25 years.
  • These parameters allow the beta distribution flexibility, adapting to various datasets.
  • They directly impact calculations such as mean and standard deviation, allowing insights into the project's temporal uncertainty.
Accurate parameter estimation ensures that the model reflects reality, providing realistic statistical representation of potential outcomes.

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Most popular questions from this chapter

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