Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

The daily summer air quality index \((\mathrm{AQI})\) in \(\mathrm{St}\). Louis is a random variable whose \(\mathrm{PDF}\) is \(f(x)=k x^{2}(180-x)\), \(0 \leq x \leq 180 .\) (a) Find the value of \(k\) that makes this a valid PDF. (b) A day is an "orange alert" day if the AQI is between 100 and 150 . What is the probability that a summer day is an orange alert day? (c) Find the expected value of the summer AQI.

Step-by-Step Solution

Answer

(a) \(k = \frac{1}{29743200}\) (b) Probability ≈ 0.4385 (c) Expected value ≈ 168.93

Step 1: Define the Integral for PDF Validity

For the probability density function (PDF) to be valid, the integral of the PDF over its entire range must equal 1. Thus, we must solve\[\int_0^{180} kx^2 (180-x) \, dx = 1.\]

Step 2: Solve the Integral

First, expand the function, \[kx^2 (180-x) = k(180x^2 - x^3).\]Now integrate term-by-term,\[k\left( \int_0^{180} 180x^2 \, dx - \int_0^{180} x^3 \, dx \right)=1.\]Calculate the integrals:- \( \int 180x^2 \, dx = 180 \cdot \frac{x^3}{3} = 60x^3 \)- \( \int x^3 \, dx = \frac{x^4}{4} \)Thus, the definite integrals are:\[60 \cdot 180^3 - \frac{180^4}{4}.\]

Step 3: Compute the Integral Values

Substituting 180 into the integrals gives:\[\int_0^{180} 180x^2 \, dx = 60 \times 180^3,\]and \[\int_0^{180} x^3 \, dx = \frac{180^4}{4}.\]Simplifying, these become 34992000 and 5248800 respectively. Therefore, \[k(34992000 - 5248800) = 1.\]

Step 4: Solve for k

Simplifying the equation yields \[29743200k = 1.\]Solving for \(k\), we find\[k = \frac{1}{29743200}.\] The value of \(k\) is thus approximately \(3.361 \times 10^{-8}.\)

Step 5: Define the Probability for Orange Alert

The probability of an "orange alert" day, where AQI is between 100 and 150, is given by \[ P(100 \leq x \leq 150) = \int_{100}^{150} f(x) \, dx = \int_{100}^{150} k x^{2}(180-x) \, dx. \]

Step 6: Calculate the Integral for the Alert Probability

Perform similar steps to calculate the integral across the range 100 to 150: Use \[ k \left( \int_{100}^{150} 180x^2 - x^3 \, dx \right).\] Calculate these integrals:- \( \int 180x^2 \, dx \) becomes \[60x^3 \bigg|_{100}^{150},\]resulting in \[60 \times ((150)^3 - (100)^3).\]- \( \int x^3 \, dx \) becomes \[\frac{x^4}{4} \bigg|_{100}^{150},\]resulting in\[\frac{150^4}{4} - \frac{100^4}{4}.\]

Step 7: Solve and Substitute

When computed, these become:- \(60 \times (3375000 - 1000000) = 142500000\), and- \(\frac{50625000 - 2500000}{4} = 12062500.\)Thus, the entire integral equals:\[k \times (142500000 - 12062500) \approx k \times 130437500.\]

Step 8: Compute Probability

The probability is given by multiplying the integral result by \(k\), so\[3.361 \times 10^{-8} \times 130437500 \approx 0.4385.\] Thus, the probability of an orange alert is approximately 0.4385.

Step 9: Calculate Expected Value

The expected value of the random variable is calculated by\[E[X] = \int_{0}^{180} x f(x) \, dx = \int_{0}^{180} x k x^2 (180-x) \, dx.\] Simplifying, we have \(kx^3(180-x)\), which expands to \[k(180x^3 - x^4).\] Solve the definite integral over 0 to 180,

Step 10: Solve the Expected Value Integral

Calculate the integrals separately:- \(k \int 180 x^3 \, dx = 45 x^4,\)- \(k \int x^4 \, dx = \frac{x^5}{5}.\)Evaluate these from 0 to 180:

Step 11: Evaluate and Compute Expected Value

Substitute and evaluate:\[k \times (45 \times 180^4 - \frac{180^5}{5}).\]This evaluates to \[ k \times(52488000000 - 2239488000) = 50248512000k.\]Thus, the expected value is \[50248512000 \times 3.361 \times 10^{-8} \approx 168.93.\] The expected AQI is approximately 168.93.

Key Concepts

Random Variables

In the world of probability and statistics, a random variable is essentially a numerical description of the outcome of a random event. When we talk about a random variable like the Air Quality Index (AQI), we are referring to a mathematical representation that captures the uncertainty and variability of our environment. For example, in St. Louis during summer, the AQI is measured daily and can take on a range of values, making it a random variable. It’s not fixed—it changes according to various environmental factors. Essentially, random variables help us describe the randomness in the phenomena we observe in the real world. This is particularly useful in fields ranging from meteorology to economics. Random variables come in two types:

• Discrete Random Variables: These take on a finite or countably infinite set of values. For example, rolling a die yields integer outcomes between 1 and 6.
• Continuous Random Variables: These can take on an infinite number of values within a given range, such as the exact measurement of AQI on any given day. For the AQI in our exercise, it's best represented by a continuous random variable over the range 0 to 180.

Expected Value

The expected value, often symbolized as \(E[X]\), represents the average value one would anticipate after many trials of an experiment or many occurrences of a random variable. It provides a central measure or mean of the potential outcomes. To find the expected value of a continuous random variable, you need to integrate over its range while factoring in its probability density function (PDF). The formula for this involves: \[E[X] = \int_{a}^{b} x f(x) \, dx \] where \(f(x)\) is the probability density function for the variable.In our exercise, we calculate the expected value to understand the average AQI we might expect on any given summer day in St. Louis. The process involved integrating \(x\) times the PDF, \(f(x) = kx^2(180-x)\), over the entire range from 0 to 180. This calculation revealed that the expected AQI value is approximately 168.93, informing us that on average, summer days tend to have AQI values close to 169, higher than many would expect.

Orange Alert Days

An orange alert day is a classification used in air quality indexing to warn of air conditions that are potentially unhealthy for certain groups. Typically, it means that sensitive groups, like the elderly, children, or those with respiratory problems, may begin to experience health effects. In terms of probability, the problem involves calculating the likelihood that a day falls within this orange alert range, specifically when the AQI is between 100 and 150. We do this by determining the integral of the probability density function over the interval from 100 to 150. This procedure reveals that the probability of experiencing an orange alert day in St. Louis during the summer is approximately 0.4385, or 43.85%. This means that nearly half of the summer days can be expected to reach AQI levels where proactive health measures may be advised for sensitive groups.