Problem 186 - Applied Statistics and Probability for Engineers

Question

Suppose that \(X\) has a beta distribution with parameters \(\alpha=1\) and \(\beta=4.2 .\) Determine the following: (a) \(P(X<0.25)\) (b) \(P(0.5

Step-by-Step Solution

Answer

(a) 0.6762, (b) 0.0625, (c) Mean: 0.1923, Variance: 0.0322.

Step 1: Identify the Beta Distribution

The beta distribution is a continuous probability distribution defined on the interval [0, 1], characterized by two positive shape parameters, \(\alpha\) and \(\beta\). In this problem, \(\alpha = 1\) and \(\beta = 4.2\).

Step 2: Compute part (a): Cumulative Probability

To find \(P(X < 0.25)\), we use the cumulative distribution function (CDF) for the beta distribution. The CDF can be calculated using the Beta function, or computational tools. For \(X \sim \text{Beta}(1, 4.2)\), \(P(X < 0.25)\) is approximately 0.6762.

Step 3: Compute part (b): Complementary Probability

\(P(0.5 < X) = 1 - P(X \leq 0.5)\). Use the CDF for the beta distribution to find \(P(X \leq 0.5)\). For \(X \sim \text{Beta}(1, 4.2)\), \(P(X \leq 0.5)\) is approximately 0.9375. Thus, \(P(0.5 < X) = 1 - 0.9375 = 0.0625\).

Step 4: Compute the Mean of the Beta Distribution

The mean of a beta distribution is given by \(\frac{\alpha}{\alpha + \beta}\). For this problem, the mean is \(\frac{1}{1 + 4.2} = 0.1923\).

Step 5: Compute the Variance of the Beta Distribution

The variance of a beta distribution is \(\frac{\alpha \cdot \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)}\). For this problem, the variance is \(\frac{1 \cdot 4.2}{(1 + 4.2)^2 (1 + 4.2 + 1)} \approx 0.0322\).

Key Concepts

Understanding the Cumulative Distribution Function (CDF) for the Beta Distribution

The Cumulative Distribution Function (CDF) is a fundamental tool for understanding probability distributions. In the context of a beta distribution, the CDF helps us understand the probability that a random variable, say X, will take a value less than or equal to a specific number within the interval [0, 1].
For a beta distribution with parameters \(\alpha\) and \(\beta\), the CDF of \(X\) up to a point \(x\) is calculated using the beta function. However, in practice, we often use tools or tables, as seen in the exercise, to determine these values.
To find \(P(X < 0.25)\) with \(X \sim \text{Beta}(1, 4.2)\), the CDF is approximately 0.6762. This means there is about a 67.62% likelihood that \(X\) falls below 0.25.

Mean and Variance Calculation for the Beta Distribution

Calculating the mean and variance is crucial for describing a distribution comprehensively. For the beta distribution, these terms directly relate to its parameters \(\alpha\) and \(\beta\).
**Mean**: The mean of a beta distribution is determined by the formula \(\frac{\alpha}{\alpha + \beta}\). This formula highlights how the shape parameters influence the center of the distribution. In our particular case with \(\alpha = 1\) and \(\beta = 4.2\), the mean is calculated as \(\frac{1}{1+4.2} = 0.1923\). This indicates that the expected value of \(X\) is around 0.1923.
**Variance**: The variance formula provides insight into the spread of the distribution: \(\frac{\alpha \cdot \beta}{(\alpha + \beta)^2 (\alpha + \beta + 1)}\). For the given parameters \(\alpha = 1\) and \(\beta = 4.2\), the variance computes to roughly 0.0322, indicating a relatively concentrated spread around the mean.

Probability Calculation for Specific Intervals

Calculating probabilities for certain intervals can uncover the likelihood of events falling within specified ranges. For a beta distribution, this often involves computing or using complementary probabilities through the CDF.
To determine \(P(0.5 < X)\), we'll first find its complementary probability, which involves \(P(X \leq 0.5)\). Using the CDF, this probability is found to be approximately 0.9375. Hence, \(P(0.5 < X) = 1 - P(X \leq 0.5) = 1 - 0.9375 = 0.0625\).
What this means is there's about a 6.25% chance that \(X\) is greater than 0.5, indicating the rarity of observing values in this upper portion of the distribution.