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 \(\operatorname{CDF} F(x)\). (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) \( k = \frac{9}{78400} \). (b) Probability is 0.9998. (c) Expected value is 18.18 PPM. (d) CDF is an integrated form of \( f(x) \). (e) CDF of \( Y \) is \( F(100y) \).

Step 1: Determine the Value of k for PDF Validity

To determine the constant \( k \) so that \( f(x) = kx^2(200-x)^8 \) is a valid probability density function (PDF), we need the total area under the PDF from \( x = 0 \) to \( x = 200 \) to equal 1. Thus, we solve the integral \( \int_{0}^{200} k x^2 (200-x)^8 \, dx = 1 \). Evaluating this integral gives a numerical result of \( \frac{78400}{9} \), leading to \( k = \frac{9}{78400} \) after normalizing to equal 1.

Step 2: Calculate the Probability Batch is Not Accepted

The company does not accept batches with impurity levels \( x \geq 100 \). We find this probability by integrating the PDF from 100 to 200: \( P(X \geq 100) = \int_{100}^{200} f(x) \, dx \). Solving this gives approximately 0.9998.

Step 3: Find Expected Value of Total Impurity

The expected value of \( X \) is calculated using \( E(X) = \int_{0}^{200} x f(x) \, dx \). Evaluate \( E(X) = \int_{0}^{200} x \left( \frac{9}{78400} \cdot x^2 (200-x)^8 \right) \, dx \), which gives approximately 18.18 PPM.

Step 4: Derive the CDF F(x) of the PDF

The CDF \( F(x) \) is found by integrating the PDF from 0 to \( x \), giving \( F(x) = \int_{0}^{x} f(t) \, dt = \frac{9}{78400} \int_{0}^{x} t^2 (200-t)^8 \, dt \). This requires symbolic integration techniques to express \( F(x) \) in terms of \( x \).

Step 5: Determine the CDF of Impurity in Percent, Y

The impurity in percent is \( Y = \frac{X}{100} \). To find \( F_Y(y) \), note that \( Y = \frac{X}{100} \implies X = 100Y \). Using a change of variable, \( F_Y(y) = P(Y \leq y) = P(X \leq 100y) = F(100y) \). Thus, \( F_Y(y) = \frac{9}{78400} \int_{0}^{100y} t^2 (200-t)^8 \, dt \).

Key Concepts

Expected Value

The expected value, often denoted as \( E(X) \), is a fundamental concept in probability and statistics. It provides a measure of the central tendency of a random variable, essentially giving us the 'average' outcome if an experiment were repeated many times under the same conditions. Think of it as the weighted average of all possible values a random variable can take, where the weights are the probabilities of each outcome.

For continuous random variables, like the total impurity in a batch of chemicals, the expected value is calculated by taking the integral of the product of the variable and its probability density function over all possible values of the variable:

• Formula: \( E(X) = \int_{a}^{b} x f(x) \, dx \)
• In our case for \( X \), the impurity level, \( E(X) = \int_{0}^{200} x \left( \frac{9}{78400} \cdot x^2 (200-x)^8 \right) \, dx \)

Evaluating this integral gives an expected value of approximately 18.18 PPM, indicating that, on average, we can expect about 18.18 parts per million of impurity in a rejected batch.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a valuable tool in understanding probabilities associated with continuous random variables. It describes the probability that a random variable will take a value less than or equal to a certain point. In simpler terms, it accumulates the probabilities from the start up to a specific value, making it easier to visualize and calculate probabilities for a range of values.

For a random variable \( X \) with a probability density function \( f(x) \), the CDF, denoted as \( F(x) \), is defined as:

• \( F(x) = \int_{a}^{x} f(t) \, dt \)
• In our specific problem, \( F(x) = \int_{0}^{x} \left( \frac{9}{78400} t^2 (200-t)^8 \right) \, dt \)

When evaluating this formula, the CDF helps in determining probabilities such as the likelihood a batch is accepted or rejected by showing how probabilities accumulate up to the threshold of 100 PPM.

Change of Variable

The change of variable technique is used in calculus to transform integrals into a different form, making them easier to evaluate. This concept is particularly useful in probability when we want to switch from one variable of interest to another, such as transforming PPM (parts per million) into percentages.

When the impurity is given in PPM, we denote it by \( X \). To convert this into percentages, we define a new variable, \( Y \), where \( Y = \frac{X}{100} \). This transformation allows us to find the cumulative distribution function (CDF) of \( Y \) based on the known CDF of \( X \).

• For the CDF of \( Y \), we find \( F_Y(y) = P(Y \leq y) = P(X \leq 100y) = F(100y) \)
• This requires substituting \( 100y \) into the CDF of \( X \)

This approach is adaptable to other similar context changes, making it a highly versatile tool in both integral calculus and probability.

Integral Calculus

Integral calculus is a major branch of mathematics that deals with the accumulation of quantities, such as areas under curves, and is fundamental to understanding probability distributions. In the context of probability density functions (PDFs), integral calculus plays a critical role in validating functions and calculating various properties like cumulative distribution functions (CDFs) and expected values.

The key operations involve:

• Finding the total area under a probability curve (as seen in finding \( k \) for a valid PDF)
• Evaluating the CDF by integrating the PDF over a certain range
• Calculating expected values by integrating the product of the variable and its PDF

In the exercise context, evaluating these integrals accurately provides crucial insights into the behavior of impurity levels across batches of chemicals. Mastery of these techniques offers a strong foundation for analyzing more complex probabilistic models.