Question

A PDF for a continuous random variable \(X\) is given. Use the PDF to find (a) \(P(X \geq 2),(b) E(X)\), and \((c)\) the \(\mathrm{CDF}\). $$ f(x)=\left\\{\begin{array}{ll} \frac{3}{64} x^{2}(4-x), & \text { if } 0 \leq x \leq 4 \\ 0, & \text { otherwise } \end{array}\right. $$

Short answer

(a) \(P(X \geq 2) = 0.265625\), (b) \(E(X) = 2.4\), (c) CDF: \(F(x) = \frac{3}{64}(4x^3/3 - x^4/4)\).

Step-by-step solution

Understanding the PDF

The probability density function (PDF) is given as \( f(x) = \frac{3}{64} x^2 (4-x) \) for \( 0 \leq x \leq 4 \). This function is zero otherwise. This means the random variable \(X\) can take any continuous value in the interval from 0 to 4.

Calculating \(P(X \geq 2)\)

To find \( P(X \geq 2) \), we compute the integral of the PDF from 2 to 4.\[P(X \geq 2) = \int_{2}^{4} \frac{3}{64} x^2 (4-x) \, dx\]Calculate this integral to determine the probability. First, expand and compute the antiderivative of \( \frac{3}{64} x^2 (4-x) \).

Evaluate the Integral

Expand the integrand: \( \frac{3}{64} x^2 (4-x) = \frac{3}{64} (4x^2 - x^3) \).Find the antiderivative:\[\int \frac{3}{64} (4x^2 - x^3) \, dx = \frac{3}{64} \left( \frac{4}{3} x^3 - \frac{1}{4} x^4 \right)\]Evaluate from 2 to 4:\[\left[ \frac{1}{16} x^3 - \frac{3}{256} x^4 \right]_{2}^{4} = \left(\frac{1}{16}(64) - \frac{3}{256}(256)\right) - \left(\frac{1}{16}(8) - \frac{3}{256}(16)\right)\]Calculate the final values to find \( P(X \geq 2) \).

Calculating \(E(X)\) (Expected Value)

The expected value \( E(X) \) is found by integrating \( x \times f(x) \) over the interval from 0 to 4:\[E(X) = \int_{0}^{4} x \times \frac{3}{64} x^2 (4-x) \, dx\]This becomes:\[E(X) = \frac{3}{64} \int_{0}^{4} x^3 (4-x) \, dx = \frac{3}{64} \left( \int_{0}^{4} 4x^3 \, dx - \int_{0}^{4} x^4 \, dx \right)\]Calculate the antiderivatives and evaluate from 0 to 4.

Find the Antiderivatives and Evaluate for \(E(X)\)

Compute the antiderivatives:\[\frac{3}{64} \left( 4 \times \frac{x^4}{4} - \frac{x^5}{5} \right)\]Evaluate from 0 to 4 to find:\[\left[ x^4 - \frac{x^5}{5} \right]_{0}^{4}\]Calculate and simplify to find \( E(X) \).

Finding the CDF

The cumulative distribution function (CDF) \( F(x) \) is the integral of \( f(x) \) from 0 to \( x \).For \( 0 \leq x \leq 4 \),\[F(x) = \int_{0}^{x} \frac{3}{64} t^2 (4-t) \, dt\]Calculate the antiderivative, then substitute to find \( F(x) \) as a function of \( x \).

Calculate Antiderivative for CDF and Simplify

Integrate \( \frac{3}{64} t^2 (4-t) \, dt \) to find \( F(x) \):\[\int \frac{3}{64} t^2 (4-t) \, dt = \frac{3}{64} \left( 4 \frac{t^3}{3} - \frac{t^4}{4} \right)\]Evaluate from 0 to \( x \):\[ F(x) = \frac{3}{64} \left( \frac{4}{3} x^3 - \frac{x^4}{4} \right) \]Simplify to find a final expression for the CDF in the range of \( 0 \leq x \leq 4 \).

Key concepts

Probability Density Function

In continuous probability distributions, a Probability Density Function (PDF) is a function that describes the likelihood of a random variable taking on a particular value within a given range. The PDF for a random variable is denoted as \( f(x) \) and is defined in such a way that the area under the curve of \( f(x) \) over a specific interval represents the probability of the variable falling within that interval.

For a given random variable \( X \), the PDF \( f(x) = \frac{3}{64}x^2(4-x) \) applies within the range \( 0 \leq x \leq 4 \). Outside this range, the PDF is zero, reflecting that \( X \) cannot take on values outside this interval. It's important to know that, unlike probabilities, the value of a PDF can exceed 1, but the integral (area under the curve) over the entire range is always equal to 1. This ensures that the total probability of all possible outcomes of \( X \) is summed up to one.

The PDF allows us to calculate probabilities for specific intervals. For instance, finding \( P(X \geq 2) \) requires calculating the integral of the PDF from 2 to 4, which will give the total probability of \( X \) taking a value in this range.

Expected Value

The Expected Value, commonly denoted as \( E(X) \), is a measure of the central tendency of a random variable. It is essentially the "long-term average" of the random variable's possible outcomes, weighted by their probabilities. In the context of continuous probability distributions, the expected value is computed using the PDF.

For the given continuous random variable \( X \), the expected value \( E(X) \) is determined by evaluating the integral:
\[ E(X) = \int_{0}^{4} x \times f(x) \, dx = \int_{0}^{4} x \times \left( \frac{3}{64} x^2 (4-x) \right) \, dx \]
Breaking it down further, this involves computing the integral of the expression
\[ \frac{3}{64} x^3 (4-x) = \frac{3}{64} \left( 4x^3 - x^4 \right) \] from 0 to 4.
The expected value provides a single number summarizing the likelihood of \( X \)'s outcomes. It is valuable in many fields, including insurance, economics, and risk management, as a predictor of future trends based on historical data.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a fundamental concept in probability that describes the probability that a random variable is less than or equal to a particular value. For a continuous random variable with a known probability density function \( f(x) \), the CDF, denoted as \( F(x) \), is obtained by integrating the PDF from the lower limit of \( 0 \) to the value \( x \).

In mathematical terms, the CDF is expressed as:
\[ F(x) = \int_{0}^{x} f(t) \, dt = \int_{0}^{x} \frac{3}{64} t^2 (4-t) \, dt \]
For the range \( 0 \leq x \leq 4 \), integrating \( f(x) \) gives us a function showing the accumulated probability up to any point \( x \). This cumulative function progressively increases as \( x \) rises from 0 to 4, ultimately reaching a value of 1, reflecting the total probability of all possible outcomes.

The CDF is a useful tool because it provides a straightforward way to compute probabilities for ranges and specific points, especially in cases where direct integration of the PDF might be complex. Additionally, by understanding the shape of the CDF, one can quickly infer where most of the distribution's probability mass is concentrated.