Problem 18 - Calculus Single Variable

Question

In each of Exercises \(17-22,\) the probability density function \(f\) of a random variable \(X\) with range \(I=[a, b]\) is given. Calculate \(P(\alpha \leq X \leq \beta)\) for the given subinterval \(J=[\alpha, \beta]\) of \(I .\) $$ f(x)=12 x^{2}(1-x) \quad I=[0,1] \quad J=[0,1 / 2] $$

Step-by-Step Solution

Answer

The probability is \( \frac{5}{16} \).

Step 1: Verify PDF and Define Objective

First, we confirm that the function given is a valid probability density function. For a function to be a PDF, it must satisfy two conditions: \( f(x) \geq 0 \) for all \( x \) in its range, and \( \int_{a}^{b} f(x) \, dx = 1 \). Next, identify the objective, which is to find the probability \( P(\alpha \leq X \leq \beta) \). Here, we have \( \alpha = 0 \) and \( \beta = \frac{1}{2} \).

Step 2: Calculate the Normalization Constant

We need to check if the integral of \( f(x) \) from \( 0 \) to \( 1 \) is 1 to assure that it is normalized. Compute \( \int_{0}^{1} f(x) \, dx \):\[\int_{0}^{1} 12x^2(1-x) \, dx = \int_{0}^{1} 12x^2 - 12x^3 \, dx.\]Evaluate the integral to ensure normalization:\[\left[ 12\frac{x^3}{3} - 12\frac{x^4}{4} \right]_{0}^{1} = 4 - 3 = 1.\]The integral equals 1, confirming \( f(x) \) is properly normalized.

Step 3: Setup the Integral for Probability

Define the integral to find the probability over the interval \( J=[0,\frac{1}{2}] \). Setup the integral:\[P\left(0 \leq X \leq \frac{1}{2}\right) = \int_{0}^{1/2} 12x^2(1-x) \, dx.\]

Step 4: Evaluate the Integral

Compute \( \int_{0}^{1/2} 12x^2 - 12x^3 \, dx \):\[\left[ 12\frac{x^3}{3} - 12\frac{x^4}{4} \right]_{0}^{1/2} = \left[ 4x^3 - 3x^4 \right]_{0}^{1/2}.\]Plug in the limits of integration:\[4\left( \frac{1}{2} \right)^3 - 3\left( \frac{1}{2} \right)^4 = \frac{1}{2} - \frac{3}{16} = \frac{8}{16} - \frac{3}{16} = \frac{5}{16}.\]

Step 5: Conclusion

The probability that the random variable \( X \) falls within the interval \( [0, 0.5] \) is \( \frac{5}{16} \).

Key Concepts

Random Variable

A random variable is a concept used in probability and statistics to quantify the outcomes of random phenomena. It is essentially a variable that can take on different values, each associated with a probability. In simple terms, you can think of a random variable as a way to map outcomes of a random process to numerical values. This makes it easy to perform arithmetic and statistical calculations.

• Types of Random Variables: There are two main types of random variables: discrete and continuous. Discrete random variables have countable outcomes, like the roll of a dice, while continuous random variables have outcomes that can take on any value within a range.
• Properties: Each random variable has a probability distribution that provides the probabilities of occurring for all possible outcomes.
• Example in Context: In the given exercise, the function represents a continuous random variable, where the outcomes range between 0 and 1.

Integral Calculus

Integral calculus is a branch of calculus dealing with integrals and their properties. It focuses on understanding areas under curves, which is essential for finding probabilities involving continuous random variables.

• Definite Integrals: These are used to calculate the area under a curve over a specified interval. In probability, this area corresponds to the probability that a random variable falls within a certain range.
• Application: To find the probability of a random variable, we often need to compute the integral of its probability density function (PDF) over the desired range.
• Example in Context: In the exercise, integral calculus is used to compute the normalization constant and find the specific probability of the interval [0, 0.5].

Probability

Probability measures how likely an event is to occur. It helps quantify the uncertainties associated with random processes.

• Basics: Probabilities range from 0 to 1. A probability of 0 means the event will not happen, while a probability of 1 means it will certainly happen.
• In Continuous Settings: When dealing with continuous random variables, probabilities are found using the area under the curve of the probability density function.
• Example in Context: For the exercise, the probability that the random variable X falls within [0, 0.5] was found to be \( \frac{5}{16} \).

Normalization

Normalization in the context of probability density functions (PDFs) ensures that the total probability across all possible outcomes equals 1. This is a crucial step to validate that a function is a proper PDF.

• Purpose: It confirms that the function accurately describes a completely random variable that accounts for all possible events.
• Method: This is achieved by integrating the PDF over its entire range and ensuring the result equals 1.
• Example in Context: In the solution, the function \( f(x) = 12x^2(1-x) \) was confirmed to be normalized because \( \int_{0}^{1} f(x) \, dx = 1 \).