Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

A PDF for a continuous random variable \(X\) is given. Use the \(P D F\) to find (a) \(P(X \geq 2)\), (b) \(E(X)\), and (c) the CDF: $$ f(x)= \begin{cases}(8-x) / 32, & \text { if } 0 \leq x \leq 8 \\ 0, & \text { otherwise }\end{cases} $$

Step-by-Step Solution

Answer

(a) \(P(X \geq 2) = \frac{7}{16}\), (b) \(E(X) = 4\), (c) CDF: \(F(x) = \frac{x(16-x)}{64}\) for \(0 \leq x \leq 8\).

Step 1: Understanding the PDF

The probability density function (PDF) given is \( f(x) = \frac{8-x}{32} \) for \( 0 \leq x \leq 8 \) and zero otherwise. This formula represents the density of the continuous random variable \( X \).

Step 2: Finding \( P(X \geq 2) \)

To calculate \( P(X \geq 2) \), compute \( \int_2^8 \frac{8-x}{32} \, dx \). This integral represents the probability that \( X \) takes a value greater than or equal to 2.

Step 3: Evaluating the Integral

Evaluate the integral \( \int_2^8 \frac{8-x}{32} \, dx \). Calculate it as \(-\frac{1}{64} (8-x)^2 \) from 2 to 8, which yields: \( 1 - \frac{6^2}{64} = 1 - \frac{36}{64} = \frac{28}{64} = \frac{7}{16}\).

Step 4: Calculating the Expectation \( E(X) \)

To find the expected value \( E(X) \), compute \( \int_0^8 x \cdot \frac{8-x}{32} \, dx \). This represents the average value of the random variable \( X \).

Step 5: Integrating for \( E(X) \)

Calculate the integral \( \int_0^8 x \cdot \frac{8-x}{32} \, dx \). It equals to \( \frac{1}{32} \int_0^8 (8x - x^2) \, dx \). The value is \( \frac{1}{32} \left[ 4x^2 - \frac{x^3}{3} \right]_0^8 \), resulting in \( 4 \).

Step 6: Constructing the CDF

The cumulative distribution function (CDF) \( F(x) \) for \( 0 \leq x \leq 8 \) is \( F(x) = \int_0^x \frac{8-t}{32} \, dt \), simplified to \( \frac{1}{32} \left[ 8t - \frac{t^2}{2} \right]_0^x \).

Step 7: Solution of the CDF

By evaluating, \( F(x) = \frac{1}{32} \left[ 8x - \frac{x^2}{2}\right] = \frac{x(16-x)}{64} \) for \( 0 \leq x \leq 8 \). For \( x < 0 \), \( F(x) = 0 \) and for \( x > 8 \), \( F(x) = 1 \).

Key Concepts

Probability Density Function (PDF)

The Probability Density Function, often abbreviated as PDF, is a crucial concept in continuous probability theory. It describes how the probabilities are distributed over the values of a continuous random variable. In simple terms, the PDF gives us the relative likelihood of a random variable to take on a particular value. However, it's important to remember that for continuous random variables, the probability of the variable being exactly one value is zero.
Instead, probabilities are obtained by integrating the PDF over a range of values.The function provided in the exercise is a piecewise PDF:\[ f(x) = \begin{cases} \frac{8-x}{32}, & \text{if } 0 \leq x \leq 8 \ 0, & \text{otherwise} \end{cases}\]This function is valid between 0 and 8, indicating where the random variable X has non-zero probability.- For values of X outside this range, the PDF is zero, signifying no probability.- Within the defined range, the PDF \(f(x)\) shows the relative probability of X assuming values within these limits.Integrating this function over any subset of the range [0, 8] gives the probability that X falls within that subset. The given PDF is linear, decreasing from 0.25 at x = 0 to 0 at x = 8, reflecting how the likelihood of X reduces as we move from left to right.

Cumulative Distribution Function (CDF)

The Cumulative Distribution Function, or CDF, provides a cumulative probability up to a certain value. It is a function that tells us the probability that a random variable will take a value less than or equal to X.
The CDF is derived by integrating the PDF from the lower limit of the range (in this case, 0) to the desired value of X.For the presented function, the CDF is:\[F(x) = \begin{cases} 0, & \text{if } x < 0 \ \frac{x(16-x)}{64}, & \text{if } 0 \leq x \leq 8 \ 1, & \text{if } x > 8 \end{cases}\]This shows that:- For values less than 0, \(F(x)\) is 0 because probability starts accumulating only from x = 0.- Between 0 and 8, it calculates the accumulated probability from 0 up to x.- For x greater than 8, the CDF plateaus at 1, indicating the entirety of the distribution's probability has been accounted for.In essence, the CDF offers a comprehensive snapshot of the probability distribution up to any given point, summing up probabilities from the beginning of the distribution to that point.

Expected Value (E(X))

The Expected Value, denoted as \(E(X)\), represents the long-run average or mean value of the random variable X after many repetitions of the experiment. It's a fundamental measure in probability and statistics, often considered the 'center of mass' of the probability distribution.
To find \(E(X)\), one needs to integrate the product of X and its PDF over all possible values of X.For the given PDF in the exercise, the expected value is calculated using the integral:\[E(X) = \int_0^8 x \cdot \frac{8-x}{32} \, dx\]This computes to:\[E(X) = \frac{1}{32} \int_0^8 (8x - x^2) \, dx \]Solving this integral, it turns out:\[E(X) = \frac{1}{32} \left[ 4x^2 - \frac{x^3}{3} \right]_0^8 = 4\]Thus, the expected value, or the mean, of this random variable is 4.- This tells us that, on average, the outcomes are centered around the value 4.- Note that the expected value might not be a value that X can actually take; it signifies an average.Understanding \(E(X)\) is crucial for statistical analyses as it is one of the key values used to summarize distributions.