Problem 32 - Calculus Single Variable

Question

In each of Exercises \(29-36,\) calculate the mean of the random variable whose probability density function is given. $$ f(x)=6 x(1-x) \quad I=[0,1] $$

Step-by-Step Solution

Answer

The mean is \( \frac{1}{2} \).

Step 1: Understand the Problem

The problem involves finding the mean of a random variable over a given interval [0,1] with the probability density function (PDF) \( f(x) = 6x(1-x) \). The mean is defined as the expected value (\( E[X] \)) of the random variable \( X \).

Step 2: Set Up the Integral for the Expected Value

To find the mean (expected value) of \( X \), use the formula \( E[X] = \int_{a}^{b} x \cdot f(x) \, dx \), where \( a \) and \( b \) are the limits of the interval. Substitute \( f(x) = 6x(1-x) \) and \( [a, b] = [0, 1] \), so the integral becomes \( E[X] = \int_{0}^{1} x \cdot 6x(1-x) \, dx \).

Step 3: Simplify the Integral Expression

Substitute \( f(x) \) into the integral, giving \( E[X] = \int_{0}^{1} 6x^2(1-x) \, dx \). Expand the expression inside the integral: \( 6x^2(1-x) = 6x^2 - 6x^3 \). This simplifies to \( \int_{0}^{1} (6x^2 - 6x^3) \, dx \).

Step 4: Calculate the Integral

Integrate the simplified expression with respect to \( x \). First, integrate \( 6x^2 \) to get \( 2x^3 \), and \( -6x^3 \) to get \( -\frac{3}{2}x^4 \). Evaluate the integral from 0 to 1: \[ 2x^3 - \frac{3}{2}x^4 \Bigg|_0^1 = \left(2(1)^3 - \frac{3}{2}(1)^4\right) - \left(2(0)^3 - \frac{3}{2}(0)^4\right)\] This results in \( 2 - \frac{3}{2} = \frac{1}{2} \).

Step 5: Conclude the Solution

The expected value, or the mean, of the random variable \( X \) over the interval [0,1] with the given probability density function is \( \frac{1}{2} \).

Key Concepts

Probability Density Function

A Probability Density Function (PDF) is essential when dealing with continuous random variables. It defines the likelihood of a random variable falling within a particular range of values, in this case, the interval [0,1].
The function itself is not the probability but rather a tool to calculate probability over an interval.
For the function to be a proper PDF, it must satisfy two conditions:

• The function is non-negative, meaning that for all values in its domain, the output should be equal to or greater than zero.
• The integral of the PDF over its domain must equal 1, ensuring the total probability across the interval sums up to 1.

In this specific case, the PDF is given by \( f(x) = 6x(1-x) \), which is non-negative and integrates to 1 over the interval [0,1], making it a valid Probability Density Function.

Expected Value

The expected value of a random variable is a measure of its central tendency—it's essentially the mean of the random variable, representing the long-term average if the variable were to be sampled repeatedly.
To compute it for a continuous random variable, you'll use the Probability Density Function. The expected value formula is\[ E[X] = \int_{a}^{b} x \cdot f(x) \ dx \]
where \( a \) and \( b \) are the endpoints of the interval over which the PDF is defined.
In this exercise, the expected value is calculated over \([0, 1]\) for the PDF \( f(x) = 6x(1-x) \). Substituting these into the formula gives the integral for the expected value as \( E[X] = \int_{0}^{1} x \cdot 6x(1-x) \, dx \), which simplifies and calculates to \( \frac{1}{2} \).
Thus, the expected value or mean of the random variable is \( \frac{1}{2} \), indicating that the average expected position of \( x \) on the interval \([0, 1]\) is at \( x = 0.5 \).

Integration

Integration plays a crucial role in calculating probabilities and expected values for continuous random variables.
It's the process of finding the integral of a function over a given interval, representing the sum of infinite, infinitesimally small values.
For the mean of a random variable, integration helps in adding up the effect of all possible values of the variable, weighted by their corresponding probabilities.
In this exercise, the integral \( \int_{0}^{1} x \cdot 6x(1-x) \, dx \) needed to be evaluated to determine the expected value. Breaking it down involved simplifying \( 6x^2(1-x) \) into \( 6x^2 - 6x^3 \), and then integrating each term separately:

• Integrate \( 6x^2 \) to get \( 2x^3 \).
• Integrate \( -6x^3 \) to get \( -\frac{3}{2}x^4 \).

Solving these integrals over the defined interval, from 0 to 1, yields the expected value \( \frac{1}{2} \). This crucial calculus tool allows us to quantify expectations and average outcomes effectively.

Interval [0,1]

The interval [0,1] is used frequently in probability and statistics, often serving as a standard domain for many probability density functions.
This interval is particularly significant because it represents a closed range from 0 to 1, including both endpoint values.
When the PDF is defined on [0,1], it implies that the random variable's values are restricted to lie within this range.
For our specific PDF \( f(x) = 6x(1-x) \), this interval means that we are considering all potential outcomes for the random variable 'x' between and including 0 and 1.
Understanding this interval is crucial when setting up integrals for calculating expectations or probabilities, as it defines the bounds within which the function is evaluated.