Problem 30 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

PERSONAL HEALTH Jules decides to go on a diet for 6 weeks, with a goal of losing between 10 and 15 pounds. Based on his body configuration and metabolism, his doctor determines that the amount of weight he will lose can be modeled by a continuous random variable \(X\) with probability density function \(f(x)\) of the form \(f(x)= \begin{cases}k(x-10)^{2} & \text { for } 10 \leq x \leq 15 \\ 0 & \text { otherwise }\end{cases}\) If the doctor's model is valid, how much weight should Jules expect to lose? [Hint: First determine the value of the constant \(k\).]

Step-by-Step Solution

Answer

Jules should expect to lose approximately 22.38 pounds.

Step 1: Formulate the problem

Identify the probability density function (pdf) given as: \[f(x)= \begin{cases} k(x-10)^{2} & \text { for } 10 \leq x \leq 15 \ 0 & \text { otherwise }\end{cases}\]The goal is to determine the expected value of weight Jules should lose.

Step 2: Determine the constant k

To ensure the function is a valid probability density function, its integral over the interval \(10 \leq x \leq 15\) must equal 1. Calculate the integral: \[\int_{10}^{15} k(x-10)^{2} \, dx = 1\]

Step 3: Solve the integral

Calculate the antiderivative:\[k \int_{10}^{15} (x-10)^{2} \, dx = k \left[ \frac{(x-10)^{3}}{3} \right]_{10}^{15}\]Evaluate the definite integral:\[ k \left[ \frac{(15-10)^{3}}{3} - \frac{(10-10)^{3}}{3} \right] = 1 \rightarrow k \frac{125}{3} = 1 \rightarrow k = \frac{3}{125} \]

Step 4: Compute the expected value

The expected value \(E(X)\) for the random variable \(X\) is given by:\[ E(X) = \int_{10}^{15} x f(x) \, dx \]Substitute \(f(x)\) with the determined \(k\): \[ E(X) = \int_{10}^{15} x \cdot \frac{3}{125}(x-10)^{2} \, dx \]

Step 5: Evaluate the new integral

Find the antiderivative of the integral:\[ \frac{3}{125} \int_{10}^{15} x (x-10)^{2} \, dx = \frac{3}{125} \int_{10}^{15} x (x^{2} - 20x + 100) \, dx \]This expands to:\[ \frac{3}{125} \int_{10}^{15} (x^{3} - 20x^{2} + 100x) \, dx \]Evaluate each term separately:\[ \frac{3}{125} \left[ \frac{x^{4}}{4} - \frac{20x^{3}}{3} + 50x^{2} \right]_{10}^{15} \]

Step 6: Calculate the definite integral result

Substitute the limits into the antiderivative formula:\[ \frac{3}{125} \left[ \left( \frac{15^{4}}{4} - \frac{20 \cdot 15^{3}}{3} + 50 \cdot 15^{2} \right) - \left( \frac{10^{4}}{4} - \frac{20 \cdot 10^{3}}{3} + 50 \cdot 10^{2} \right) \right] \]This simplifies to:\[ \frac{3}{125} \left[ \left( \frac{50625}{4} - 22500 + 11250 \right) - \left( \frac{10000}{4} - 6666.67 + 5000 \right) \right] \]

Step 7: Simplify the expressions

Perform the calculations inside the brackets:\[ \frac{3}{125} \left[ 1265.625 - 333.33 \right] = \frac{3}{125} (932.295) \approx 22.38 \]

Key Concepts

Understanding Probability Density Function (PDF)

The probability density function, often abbreviated as PDF, is a crucial concept in probability theory. It describes how the values of a continuous random variable are distributed. For a random variable \(X\) with density function \(f(x)\):

• The function \(f(x)\) must be non-negative for all \(x\)
• The area under the curve of \(f(x)\) over all possible values of \(x\) must equal 1

In Jules' case, the PDF is given as: \[ f(x) = \begin{cases} k(x-10)^2 & \text{for } 10 \leq x \leq 15 \0 & \text{otherwise} \end{cases} \] This PDF is a function that helps us understand the likelihood of Jules losing a specific amount of weight between 10 and 15 pounds. The first step in working with this PDF is to find the constant \(k\) such that the integral of \(f(x)\) over the interval [10, 15] is equal to 1. This ensures that the function \(f(x)\) is a valid PDF.

Integral Calculation for Finding Constant k

Integrals are used to determine the area under the curve of a function, which in the context of probability represents the total probability. For Jules' PDF, we need to calculate: \[ \int_{10}^{15} k(x-10)^2 dx = 1 \] To find \(k\), we first compute the antiderivative of \((x-10)^2\): \[ k \int_{10}^{15} (x-10)^2 dx = 1 \rightarrow k \bigg[ \frac{(x-10)^3}{3} \bigg]_{10}^{15} \] Evaluating the definite integral, we get: \[ k \bigg[ \frac{(15-10)^3}{3} - \frac{(10-10)^3}{3} \bigg] = 1 \rightarrow k \frac{125}{3} = 1 \rightarrow k = \frac{3}{125} \] Now we have the value of \(k\), ensuring that the PDF integrates to 1 over the interval from 10 to 15. This step is crucial as it confirms that the mathematical model for Jules' weight loss is valid.

Expected Value for a Random Variable

The expected value (or mean) of a continuous random variable \(X\) gives us a measure of the central tendency, or the 'average' case scenario. It's calculated using the integral of \(x\) times the PDF \(f(x)\): \[ E(X) = \int_{10}^{15} x f(x) dx \] For Jules' weight loss scenario, substituting the PDF and the calculated constant \(k\), we get: \[ E(X) = \int_{10}^{15} x \frac{3}{125} (x-10)^2 dx \] We expand and evaluate the integrand: \[ E(X) = \frac{3}{125} \int_{10}^{15} x (x^2 - 20x + 100) dx \] Breaking it down further: \[ E(X) = \frac{3}{125} \int_{10}^{15} (x^3 - 20x^2 + 100x) dx \] Calculating each term individually and summing the results gives: \[ \frac{3}{125} \bigg[ \frac{x^4}{4} - \frac{20x^3}{3} + 50x^2 \bigg]_{10}^{15} \] Finally, we evaluate: \[ \frac{3}{125} \bigg[ ( \frac{50625}{4} - 22500 + 11250 ) - ( \frac{10000}{4} - 6666.67 + 5000 ) \bigg] = \frac{3}{125} \bigg( 1265.625 - 333.33 \bigg) = 22.38 \] Therefore, Jules can expect to lose approximately 22.38 pounds over 6 weeks. This value represents the average weight loss Jules should expect according to the given probability model.