/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Suppose that \(X\) has a beta di... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(X\) has a beta distribution with parameters \(\alpha=2.5\) and \(\beta=2.5 .\) Sketch an approximate graph of the probability density function. Is the density symmetric?

Short Answer

Expert verified
The density is symmetric around 0.5 as \(\alpha = \beta = 2.5\).

Step by step solution

01

Understanding the Beta Distribution Definition

The beta distribution is defined on the interval \([0,1]\) and is characterized by two shape parameters, \(\alpha\) and \(\beta\). The probability density function (PDF) of the beta distribution is given by:\[ f(x; \alpha, \beta) = \frac{x^{\alpha-1} (1-x)^{\beta-1}}{B(\alpha, \beta)},\]where \(B(\alpha, \beta)\) is the beta function, serving as a normalization constant to ensure the total area under the curve is 1.
02

Calculating the Symmetry

The beta distribution is symmetric when \(\alpha = \beta\). Here, both \(\alpha\) and \(\beta\) are 2.5, signaling that the distribution is symmetric around 0.5.
03

Visualizing the Probability Density Function

For a beta distribution with \(\alpha = \beta = 2.5\), sketch a graph on a coordinate plane. On the y-axis, plot the PDF value and on the x-axis, plot the values from 0 to 1. Since the distribution is symmetric, the peak of the graph should occur at \(x = 0.5\), with the graph equally tapering on both sides towards the ends 0 and 1.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding the Probability Density Function
The probability density function (PDF) is a fundamental concept used to define the behavior of continuous random variables, such as those described by the beta distribution. In the context of the beta distribution, the PDF tells us how the values of this distribution are spread across the interval ([0,1]). The formula is as follows:

\[ f(x; \alpha, \beta) = \frac{x^{\alpha-1} (1-x)^{\beta-1}}{B(\alpha, \beta)}\]

What makes the PDF particularly interesting is that it is influenced by the "shape parameters" \(\alpha\) and \(\beta\). These parameters adjust the steepness and peak (or peaks) of the curve. Furthermore, the function \(B(\alpha, \beta)\), known as the beta function, works to ensure that the total area under the curve created by the PDF along the interval ([0,1]) is exactly one, fulfilling the primary rule of probability.

To visualize this, graphing the PDF with different \(\alpha\) and \(\beta\) values will show different shapes. Many of them might have a mode, which is the value that appears most frequently. The PDF not only helps in determining probabilities for a given range but also in understanding how data behaves in various scenarios, crucial for both theoretical and applied statistics.
What is a Symmetric Distribution?
A symmetric distribution is a type of distribution where the left and right sides of the distribution are mirror images of each other. This symmetry means that the distribution has balanced tails, and it is most often centered around a point where the peak occurs, known as the median or mean.

In the case of the beta distribution, symmetry happens specifically when the shape parameters \(\alpha\) and \(\beta\) are equal. For the exercise you described, \(\alpha = \beta = 2.5\). This equality signifies that the distribution is perfectly symmetric around the midpoint of the interval ([0,1]), which is 0.5.

The symmetry in such distributions is visually intuitive. Imagine folding a plotted PDF graph along the line \(x = 0.5\); both sides should overlap completely. This property of symmetry can simplify calculations and assumptions in statistical analysis, making symmetric distributions particularly appealing in various research fields.
Exploring Shape Parameters in Beta Distribution
In a beta distribution, shape parameters play a crucial role in defining the "shape" or look of the probability density function. These parameters are denoted as \(\alpha\) and \(\beta\), and they can significantly modify the distribution's properties.

  • If both \(\alpha\) and \(\beta\) are greater than 1, the beta distribution form a unimodal curve, peaking somewhere within the (0,1) interval depending on the values.
  • If \(\alpha\) and \(\beta\) are equal, the distribution becomes symmetric, as previously explained. It's centered around the midpoint \(x = 0.5\) on the interval (0,1).
  • Values of \(\alpha\) and \(\beta\) less than 1 can lead to "U-shaped" distributions, indicating that the probability density is higher at the boundaries (0 and 1) rather than the middle.


Thus, the shape parameters define how the area under the curve is spread. These parameters provide versatility, allowing the beta distribution to model a wide variety of data shapes, from skewed to perfectly symmetric forms. Adjusting \(\alpha\) and \(\beta\) can help tailor the distribution to fit datasets with specific characteristics, enhancing analytical precision in statistical projects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The thickness of photoresist applied to wafers in semiconductor manufacturing at a particular location on the wafer is uniformly distributed between 0.2050 and 0.2150 micrometers. Determine the following: a. Cumulative distribution function of photoresist thickness b. Proportion of wafers that exceeds 0.2125 micrometers in photoresist thickness c. Thickness exceeded by \(10 \%\) of the wafers d. Mean and variance of photoresist thickness

Use integration by parts to show that \(\Gamma(r)=(r-1)\) \(\Gamma(r-1)\)

An article in the Journal of Geophysical Research ["Spatial and Temporal Distributions of U.S. Winds and Wind Power at 80 m Derived from Measurements" (2003, Vol. 108) ] considered wind speed at stations throughout the United States. For a station at Amarillo, Texas, the mean wind speed at \(80 \mathrm{~m}\) (the hub height of large wind turbines) was \(10.3 \mathrm{~m} / \mathrm{s}\) with a standard deviation of \(4.9 \mathrm{~m} / \mathrm{s}\). Determine the shape and scale parameters of a Weibull distribution with these properties.

A European standard value for a low-emission window glazing uses 0.59 as the proportion of solar energy that enters a room. Suppose that the distribution of the proportion of solar energy that enters a room is a beta random variable. a. Calculate the mode, mean, and variance of the distribution for \(\alpha=3\) and \(\beta=1.4\) b. Calculate the mode, mean, and variance of the distribution for \(\alpha=10\) and \(\beta=6.25 .\) c. Comment on the difference in dispersion in the distribution from parts (a) and (b).

An article in Electronic Journal of Applied Statistical Analysis ["Survival Analysis of Dialysis Patients Under Parametric and Non-Parametric Approaches" (2012, Vol. 5(2), pp. \(271-288\) ) ] modeled the survival time of dialysis patients with chronic kidney disease with a Weibull distribution. The mean and standard deviation of survival time were 16.01 and 11.66 months, respectively. Determine the following: a. Shape and scale parameters of this Weibull distribution b. Probability that survival time is more than 48 months c. Survival time exceeded with \(90 \%\) probability

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.