Problem 5 - 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 median shelf life of a banana using \(p(t) .\) Plot the median on a graph of \(p(t)\). Does it look like half the area is to the right of the median and half the area is to the left?

Step-by-Step Solution

Answer

The median shelf life is approximately 2.65 weeks. When graphed, the median divides the area under \(p(t)\) into two roughly equal parts.

Step 1: Understanding the Median

The median of a continuous probability distribution is the value that divides the probability into two equal parts. For a probability density function \(p(t)\), the median is the value for which the integral of \(p(t)\) from 0 to the median equals 0.5. We need to solve \(\int_0^m p(t) \, dt = 0.5\) to find the median \(m\).

Step 2: Set Up the Integral

The density function is given as \(p(t) = -0.0375 t^2 + 0.225 t\). We need to set up the integral \(\int_0^m (-0.0375 t^2 + 0.225 t) \, dt = 0.5\) and solve for \(m\).

Step 3: Integrate the Function

Calculate the indefinite integral of \(p(t)\):\[\int (-0.0375 t^2 + 0.225 t) \, dt = -0.0125 t^3 + 0.1125 t^2 + C\]We will evaluate this indefinite integral from 0 to \(m\).

Step 4: Evaluate the Definite Integral

The definite integral from 0 to \(m\) is:\[\left[-0.0125 t^3 + 0.1125 t^2 \right]_0^m = -0.0125 m^3 + 0.1125 m^2\]Set this equal to 0.5:\[-0.0125 m^3 + 0.1125 m^2 = 0.5\]

Step 5: Solve for the Median

Rearrange the equation: \[-0.0125 m^3 + 0.1125 m^2 - 0.5 = 0\]Solve this cubic equation for \(m\). This can be done using numerical methods or graphing. The approximate solution is \(m \approx 2.65\).

Step 6: Graph and Interpret the Result

Plot the density function \(p(t)\) over the interval \([0, 4]\). Draw a vertical line at \(t = 2.65\). Visually inspect if the areas on either side of this line appear approximately equal. This confirms whether the line represents the median.

Key Concepts

Median

The concept of the median is crucial in the realm of statistics, specifically in probability distributions. The median of a continuous probability distribution is the point at which the area under the probability density function (PDF) is divided equally. In simpler terms, it's the value that separates the probability of all possible outcomes into two equal halves. For example, if you arranged all possible values of a dataset in order, the median is the middle value. In a continuous probability distribution like the one for banana shelf life \(p(t) = -0.0375 t^2 + 0.225 t\), the median is found by solving the integral equation \(\int_0^m p(t) \, dt = 0.5\),where \(m\) is the median. This means we're looking for the point \(m\) where exactly half of the shelf life values lie to the left, and half lie to the right, when visualized on a graph.

Probability Density Function

A probability density function (PDF) is a function that describes the likelihood of a continuous random variable to take on a particular value. When graphed, the PDF is a curve which shows that different outcomes have different probabilities. The area under the entire curve is equal to one, as this represents the total probability. For a shelf life probability distribution, the function\(p(t) = -0.0375 t^2 + 0.225 t\)is our PDF. It tells us how the probability is distributed over the range of 0 to 4 weeks. The PDF doesn't give us actual probabilities directly. Instead, the probability that the variable falls within a particular range is given by the integral of the PDF over that range. For example, finding the area under the curve from 0 to a particular week gives the probability that a banana lasts up to that number of weeks. This makes understanding the shape and behavior of a PDF essential for analyzing continuous data conveniently.

Integration

Integration is a fundamental concept in calculus, and it plays a significant role in finding probabilities using probability density functions. When dealing with a PDF, integration allows us to sum up (or accumulate) the probability over a continuous range of values.For the function \(p(t) = -0.0375 t^2 + 0.225 t\), integration is used to find the median. You perform an indefinite integration on \(p(t)\):\[\int (-0.0375 t^2 + 0.225 t) \, dt = -0.0125 t^3 + 0.1125 t^2 + C\]The integral evaluates how much of the probability is accumulated up to a particular point. In practice, when you seek the median of the distribution, you set up and solve the definite integral:\[\int_0^m (-0.0375 t^2 + 0.225 t) \, dt = 0.5\]This procedure ensures that you find the exact median where the accumulated probability equals 0.5.

Cubic Equation

A cubic equation is a type of polynomial equation where the highest exponent of the variable is three. These equations can sometimes be tricky to solve because they may have more complex structures, leading to one, two, or three real solutions.In the context of our probability problem, the cubic equation arisen from the integration process looks like this:\[-0.0125 m^3 + 0.1125 m^2 - 0.5 = 0\]To find the median, solving this cubic equation is necessary. This can be done using numerical methods like the Newton-Raphson method, or by graphing the function to estimate the roots. Often, approximation methods are used when a precise solution is challenging to achieve by algebraic means. In our exercise, calculation or an approximation shows that the median value \(m\) is about 2.65 weeks.Understanding how to approach cubic equations is essential because they frequently appear in data modeling, especially when deriving conditions or characteristics in probability and statistics.