Problem 6 - Applied Calculus

Question

Let \(p(t)=-0.0375 t^{2}+0.225 t\) be the density function for the shelf life of a brand of banana which lasts up to 4 weeks. Time, \(t\), is measured in weeks and \(0 \leq t \leq 4\). Find the mean shelf life of a banana using \(p(t) .\) Plot the mean on a graph of \(p(t)\). Does it look like the mean is the place where the density function balances?

Step-by-Step Solution

Answer

The mean shelf life is 4.2 weeks, which is where the density function balances.

Step 1: Understand the Concept

To find the mean of a continuous probability density function (PDF), we must calculate the expected value using the formula \( \mu = \int_{a}^{b} t \cdot p(t) \, dt \). Here, \(a\) is the lower limit and \(b\) is the upper limit of the interval over which the random variable is defined.

Step 2: Set Up the Integral

For our given density function \(p(t) = -0.0375 t^2 + 0.225t\), we need to compute \( \int_{0}^{4} t (-0.0375 t^2 + 0.225t) \, dt \). This expression will give us the mean shelf life.

Step 3: Simplify the Integrand

Rewrite the integrand by distributing \(t\) across \(-0.0375 t^2 + 0.225t\) to get \(-0.0375t^3 + 0.225t^2\). The integral becomes \( \int_{0}^{4} (-0.0375t^3 + 0.225t^2) \, dt \).

Step 4: Integrate

Integrate each term separately: \[ \int_{0}^{4} -0.0375t^3 \, dt = -0.0375 \cdot \frac{t^4}{4} \bigg|_0^4 \] \[ \int_{0}^{4} 0.225t^2 \, dt = 0.225 \cdot \frac{t^3}{3} \bigg|_0^4 \] Evaluate the definite integrals.

Step 5: Evaluate the Integrals

Calculate the definite integrals:- For \(-0.0375 \cdot \frac{t^4}{4}\), substituting \(t = 4\), we have \(-0.0375 \cdot \frac{4^4}{4} = -0.0375 \cdot 64 \cdot \frac{1}{4} = -0.6\). - For \(0.225 \cdot \frac{t^3}{3}\), substituting \(t = 4\), we have \(0.225 \cdot \frac{64}{3} = 4.8\). Add these values to get the expected value: \(-0.6 + 4.8 = 4.2\).

Step 6: Conclusion

The mean shelf life, or expected value of the banana storage time, is \(4.2\) weeks. Plotting this on a graph of \(p(t)\), the mean will appear to the right of the interval midpoint (2 weeks). Since the function is skewed right, it balances around this mean, indicating it is the center of mass.

Key Concepts

Mean Calculation

The mean calculation in the context of a continuous probability density function (PDF) involves finding the expected average outcome. For continuous random variables like the shelf life of bananas, the mean provides a way to summarize the data's central tendency. To calculate this, we utilize the formula \( \mu = \int_{a}^{b} t \cdot p(t) \, dt \), where \(a\) and \(b\) are the boundaries of the interval, and \(t\) represents the time in weeks.
This formula helps you find where most of the probability lies, giving insights into the expected duration you can consume the bananas. It acts as the balancing point of the probability distribution, which is quite useful for identifying trends and making predictions.

Integral of a Function

Integrating a function is a fundamental aspect of calculus that allows us to determine areas under curves, among other applications. In the context of calculating the mean shelf life,integration is used to compute the expected value of a continuous random variable. For the given density function, \( p(t) = -0.0375 t^2 + 0.225 t \), the integral \( \int_{0}^{4} t (-0.0375 t^2 + 0.225t) \, dt \) is set up to find the mean.
This involves expanding and simplifying the expression, then integrating each term separately.
Completing this process gives us a comprehensive view of the variable's distribution over the specified time interval. Using the integration process brings clarity to the probabilistic behavior of real-world phenomena, such as the duration of banana shelf life.

Expected Value

The expected value is a cornerstone concept in probability and statistics. It represents the long-run average or mean of a random variable's outcomes. In our scenario, the expected value indicates the average shelf life of bananas described by our PDF.
Mathematically, this is obtained by evaluating the integral we set up for the mean calculation, resulting in a value of 4.2 weeks. This gives insight into what we can "expect" the average lifespan of a banana to be, despite individual variations.
Understanding the expected value helps in planning and decision-making, such as inventory management for bananas, ensuring less waste due to spoilage.

Shelf Life

Shelf life refers to the length of time a product, such as bananas, remains usable or fit for consumption. In the context of continuous random variables, shelf life can be modeled with a probability density function like \( p(t) = -0.0375 t^2 + 0.225 t \).
Bananas have a varying shelf life under different conditions, and this function gives a mathematical representation for it, allowing one to calculate the probability of a banana lasting a certain time.

• This results in a more mathematical and precise understanding of spoilage likelihood.
• Using such probabilities assists in optimizing storage and reducing waste.

Continuous Random Variable

A continuous random variable can take an infinite number of possible values within a certain range. This is unlike discrete random variables that have definite, countable outcomes. The shelf life of bananas, ranging over several weeks, is a continuous random variable because it can be any value between the set limits of 0 and 4 weeks.
For such variables, probability is described using a probability density function (PDF), which must integrate to 1 over the defined interval.

• PDFs provide a way to see how probabilities are distributed across outcomes.
• It helps in understanding the entire range of possible outcomes.

Using continuous random variables and PDFs provides a richer framework to model variable outcomes, such as durability and longevity of perishable items like bananas.