Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

A company monitors the total impurities in incoming batches of chemicals. The PDF for total impurity \(X\) in a batch, measured in parts per million (PPM), has PDF \(f(x)=\) \(k x^{2}(200-x)^{8}, 0 \leq x \leq 200 .\) (a) Find the value of \(k\) that makes this a valid PDF. (b) The company does not accept batches whose total impurity is 100 or above. What is the probability that a batch is not accepted? (c) Find the expected value of the total impurity in PPM. (d) Find the CDF \(F(x)\). (e) Let \(Y\) denote the total impurity in percent, rather than in PPM. What is the CDF of \(Y\) ?

Step-by-Step Solution

Answer

(a) Integrate to find \(k\). (b) Calculate \(P(X \geq 100)\). (c) Integrate for \(E[X]\). (d) Integrate for \(F(x)\). (e) Scale \(F(x)\) by \(x = 200y\).

Step 1: Determine k for Valid PDF

To find the value of \(k\), we need to ensure that the total area under the probability density function (PDF) is equal to 1. This means we must solve the integral: \[\int_{0}^{200} k x^{2}(200-x)^{8} \, dx = 1.\]Let's evaluate this: 1. First, solve the integral without the constant \(k\): - Use the substitution \(u = 200 - x\) which implies \(du = -dx\), then rewrite the integral in terms of \(u\). - Compute the new bounds: when \(x = 0\), \(u = 200\); when \(x = 200\), \(u = 0\). 2. \[\begin{align*}&\int_{0}^{200} x^{2}u^{8} \, dx = -\int_{200}^{0} (200-u)^{2}u^{8} \, (-du) \&= \int_{0}^{200} (200-u)^{2}u^{8} \, du.\end{align*}\] - Expand \((200-u)^{2} = 40000 - 400u + u^{2}\) and integrate each term separately. - Sum the results to get the final value \(I\).3. Solve for \(k\) using:\[k \times I = 1\\Rightarrow k = \frac{1}{I}.\]

Step 2: Calculate Probability Batch is Rejected

To find the probability that a batch is not accepted (i.e., \(X \geq 100\)), we need to compute:\[P(X \geq 100) = \int_{100}^{200} k x^{2}(200-x)^{8} \, dx.\]Using the value of \(k\) from Step 1, compute this integral similarly using substitution and expansion techniques as described in Step 1.

Step 3: Find Expected Value of Total Impurity

The expected value \(E[X]\) is calculated using the integral:\[E[X] = \int_{0}^{200} x \cdot f(x) \, dx = \int_{0}^{200} k x^{3}(200-x)^{8} \, dx.\]Evaluate this integral through substitution, similar steps as before, and using the result of the integration to find the expected value.

Step 4: Derive Cumulative Distribution Function (CDF) F(x)

The CDF \(F(x)\) is found by integrating the PDF from the lower bound to \(x\):\[F(x) = \int_{0}^{x} k t^{2}(200-t)^{8} \, dt.\]Evaluate this integral by substitution method or numerical approximation to express \(F(x)\) in terms of \(x\).

Step 5: Convert CDF to Percentage

To convert the CDF of impurities in PPM to a percentage, note that if \(X\) is impurity in PPM, then \(Y = \frac{X}{200}\) gives impurity as a percentage. Thus, the CDF of \(Y\), denoted as \(F_Y(y)\), is obtained by substituting \(y = \frac{x}{200}\) and scaling the function \(F(x)\). Therefore, find: \[F_Y(y) = F(200y).\]

Key Concepts

Expected Value

The expected value is a fundamental concept in probability and statistics. It represents the average outcome one can expect from a random variable after a large number of trials. In this exercise, we want to determine the expected value of total impurity in parts per million (PPM). Given a probability density function (PDF), denoted as \(f(x)\), the expected value \(E[X]\) can be calculated through integration. Specifically, it involves the integral of the variable \(x\) multiplied by the PDF over its entire range of definition.
For this particular problem, the expected value of the total impurity is computed using the formula:

• \[E[X] = \int_{0}^{200} x \cdot f(x) \, dx = \int_{0}^{200} k x^{3}(200-x)^{8} \, dx.\]

In this integral, we multiply each possible impurity level by its probability and sum these contributions. This weighted average provides insight into what **typical** impurity level can be expected in a batch.
We substitute the value of \(k\) determined from normalizing the PDF to find the expected value. Using methods of substitution helps unravel the complexity of these integrals, allowing us to achieve the desired calculation.

Cumulative Distribution Function

The cumulative distribution function (CDF) is a critical concept in understanding the distribution of a random variable. It provides the probability that a random variable will take a value less than or equal to a certain level.
In this exercise, we derive the CDF \(F(x)\) for the total impurity in PPM by integrating the PDF from the lower boundary up to a point \(x\). Mathematically, it is expressed as:

• \[F(x) = \int_{0}^{x} k t^{2}(200-t)^{8} \, dt.\]

This integral yields a function \(F(x)\) that describes the likelihood of the impurity being less than any given value \(x\). Thus, the CDF helps in determining not only the probability for simple inequalities but also provides a complete picture of the probabilities for all values up to \(x\).
A noteworthy aspect is that a CDF always lies between 0 and 1, reflecting probabilities. By understanding the CDF, one can easily derive various probabilistic inquiries, such as finding the median or any percentile of the distribution.

Integration

Integration is a fundamental mathematical tool widely used in probability and statistics, particularly in dealing with continuous random variables and their distributions. In problems involving PDFs, integration is necessary to find total probabilities and other related measures like expected values and cumulative distribution functions.
In this exercise, integration transforms the PDF into a meaningful quantity, such as the expected value or CDF, by accumulating "infinitesimal" probability slices over a given interval. Solving integrals often requires techniques such as:

• Variable substitution - changing variables to simplify expressions.
• Expanding polynomials - when necessary, breaking down complicated expressions.

For instance, calculating the expected value involves integrating the function multiplied by \(x\), while finding the probability of a batch with impurities above a threshold necessitates integrating over a specific range.
These integrations help translate complex mathematical concepts into practical tools, enabling statisticians and students to predict outcomes or understand distribution behaviors.

Substitution Method

The substitution method is a powerful technique used to simplify the process of integration, especially when dealing with complex polynomials as seen in this exercise. It is analogous to the chain rule in differentiation.
When attempting to solve an integral like \(\int x^{2}(200-x)^{8} \, dx\), you may use substitution to make the expression more manageable. Suppose you set a new variable, such as \(u = 200 - x\), which transforms the integral into an easier form \(du = -dx\). Consequently, this simplifies the polynomial, turning an otherwise daunting integration task into a tractable one.

• The bounds of integration also change accordingly: if \(x=0\), then \(u=200\), and if \(x=200\), then \(u=0\).

This technique doesn't just reframe the problem but also aids in cleansing the calculation from potential algebraic clutter, thus enabling effortless computation. Implementing substitution allows the integrals to be calculated either directly or by further polynomial simplifications, paving the way for solving the problem at hand.