Problem 13 - 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: $$ f(x)= \begin{cases}\frac{3}{64} x^{2}(4-x), & \text { if } 0 \leq x \leq 4 \\\ 0, & \text { otherwise }\end{cases} $$

Step-by-Step Solution

Answer

(a) \(P(X \geq 2)\) results from integrating from 2 to 4. (b) Solve \(E(X)\) by integrating \(xf(x)\) from 0 to 4. (c) The CDF \(F(x)\) integrates the PDF from 0 to \(x\).

Step 1: Understand the Given PDF

The given piecewise function defines the probability density function (PDF) for the random variable \(X\). It is defined as \(f(x) = \frac{3}{64} x^{2}(4-x)\) for \(0 \leq x \leq 4\) and \(0\) otherwise.

Step 2: Calculate P(X ≥ 2)

To find \(P(X \geq 2)\), we need to integrate \(f(x)\) from \(x = 2\) to \(x = 4\). The integral is: \[P(X \geq 2) = \int_{2}^{4} \frac{3}{64} x^{2}(4-x) \, dx\]Calculating the integral:1. Expand the integrand: \(x^2(4-x) = 4x^2 - x^3\).2. The integral becomes \(\int_{2}^{4} \frac{3}{64} (4x^2 - x^3) \, dx\).3. Separate the terms: \(\frac{3}{64} \int_{2}^{4} (4x^2 - x^3) \, dx\).4. Solve the integrals: - \(4\int x^2 \, dx = \frac{4x^3}{3}\) - \(\int x^3 \, dx = \frac{x^4}{4}\).5. Substitute the limits into the result to find the probability.

Step 3: Calculate E(X)

To find the expected value \(E(X)\), calculate:\[E(X) = \int_{0}^{4} x \cdot f(x) \, dx = \int_{0}^{4} x \cdot \frac{3}{64} x^{2}(4-x) \, dx\]1. Simplify the expression: \(x \cdot x^2 \cdot (4 - x) = x^3(4 - x)\).2. Expand: \(x^3(4-x) = 4x^3 - x^4\).3. The integral becomes \(\int_{0}^{4} \frac{3}{64} (4x^3 - x^4) \, dx\).4. Solve the integrals: - \(4\int x^3 \, dx = \frac{4x^4}{4}\) - \(\int x^4 \, dx = \frac{x^5}{5}\).5. Substitute the bounds to find \(E(X)\).

Step 4: Calculate the CDF

The cumulative distribution function (CDF), \(F(x)\), is found by integrating the PDF from 0 to \(x\):For \(x < 0\), \(F(x) = 0\).For \(0 \leq x \leq 4\):\[F(x) = \int_{0}^{x} \frac{3}{64} t^{2}(4-t) \, dt\]1. Replace \(t\) for \(x\) and integrate:2. Expand: \(t^2(4-t) = 4t^2 - t^3\).3. Integrate: \(\int_{0}^{x} \frac{3}{64} (4t^2 - t^3) \, dt\).4. Find \(\frac{3}{64}\) of the simplified integral with bounds substituted to compute \(F(x)\).For \(x > 4\), \(F(x) = 1\) since the entirety of the probability is accumulated.

Key Concepts

Continuous Random Variable

A continuous random variable is a type of variable that can take on an infinite number of possible values. Unlike discrete random variables, which have a countable number of possible values, continuous variables are often represented by intervals on the real number line. When discussing a random variable like "X" in probability, it typically refers to quantities that can occur continuously over a specified interval. An example is the temperature in a day, which can take any value within a range, such as 0 to 100 degrees. Continuous random variables are associated with a probability density function (PDF) that defines the likelihood of the variable taking on a particular value. The important feature of a PDF is that the area under the curve over an interval gives the probability that the random variable falls within that interval. The total area under the PDF over the entire interval is 1, reflecting the certainty that the variable will take on some value within the interval.

Expected Value

The expected value of a continuous random variable is essentially the "mean" or average value that the variable assumes. Mathematically, it represents the center of the distribution for that variable. To find the expected value, denoted as \(E(X)\), you compute the integral of the product of the variable and its probability density function over the range of the variable. In other words:\[E(X) = \int_{a}^{b} x \, f(x) \, dx\]where \(f(x)\) is the probability density function of the random variable, and \([a, b]\) represents the interval over which the variable is defined. In this case, it is the integral from 0 to 4. Finding \(E(X)\) means calculating this integral, which may involve algebraic manipulations and applying limits to find an eventual result.

Cumulative Distribution Function

The cumulative distribution function (CDF) provides a way to describe the probability that a continuous random variable is less than or equal to a certain value. This function is cumulative because it "accumulates" probability. Importantly, the CDF, denoted as \(F(x)\), is obtained by integrating the PDF from the lower bound of the distribution up to some value \(x\):\[F(x) = \int_{a}^{x} f(t) \, dt\]The lower limit \(a\) is typically the smallest value within the interval of the random variable. The result of the CDF is a function that increases from 0 to 1 as \(x\) runs from the lower limit of the support to the upper limit. For values of \(x\) that are less than the support interval's lower bound, the CDF is 0. For values beyond the upper bound, it reaches 1. The CDF thus allows us to easily compute probabilities for intervals.

Integration in Probability

Integration plays a crucial role in probability, especially when dealing with continuous random variables. By definition, probability density functions (PDFs) for continuous variables must be integrated across intervals to find probabilities. For example, when given a PDF, calculating the probability that the variable is within a particular range means integrating the function across that range.To handle these computations, we use definite integrals. In the given problem, integrating the piecewise defined PDF for values of \(x\) between 0 and 4 yields probabilities for different scenarios:

• To find the probability of \(P(X \geq 2)\), integrate the PDF from 2 to 4.
• The expected value involves integrating the product \(x \cdot f(x)\) over the entire interval from 0 to 4.
• For the cumulative distribution, integrate from the lower bound of the interval to \(x\) to find the total probability up to \(x\).

In all these cases, integration transforms the PDF into valuable probability information, reflecting the range and behavior of the random variable. Consequently, a solid understanding of integration techniques is key to working effectively with continuous random variables in probability.