Problem 101 - An Introduction to Mathematical Statistics and Its Applications

Question

A box is to be constructed so that its height is five inches and its base is \(Y\) inches by \(Y\) inches, where \(Y\) is a random variable described by the pdf, \(f_{Y}(y)=6 y(1-y)\), \(0

Step-by-Step Solution

Answer

The expected volume of the box is 1.5 cubic inches.

Step 1: Define the volume function

Given that the base of the box is Y inches by Y inches and its height is 5 inches, the volume V of the box can be written as \(V=Y^2*5=5Y^2\). This is the variable whose expected value needs to be computed.

Step 2: Write down the integral for the expected value

The expected value E[V] of a continuous random variable is given by the integral of the product of the random variable and its pdf integrated over the range of the variable. So, in this case, E[V] can be written as \(E[V]=\int_0^1 v*f_{V}(v) dv\)

Step 3: Substitute for V and f_{V}(v)

Substitute for V and f_{V}(v) in the above integral. This gives \(E[V]=\int_0^1 5y^2*6y(1-y) dy\)

Step 4: Compute the integral

Compute the integral to get the expected volume. This results in \(E[V] = \int_0^1 30y^3-30y^4 dy = [7.5y^4 - 6y^5]_0^1 = 7.5 - 6 = 1.5 \) cubic inches

Key Concepts

Continuous Random Variable

A continuous random variable is a type of random variable that can take on an infinite set of values within a given range. Unlike discrete random variables, which have specific separate outcomes, continuous random variables can assume any value within a specified interval. This means there's no gaps or jumps between the values it can take. For example, in our exercise, the variable \( Y \), which represents the base dimension of the box, can be any value between 0 and 1.
This makes continuous random variables particularly useful for modeling real-world situations, like measuring the dimensions of physical objects or natural phenomena like temperature, where values can vary fluidly across a spectrum.
To analyze continuous random variables effectively, we utilize mathematical tools like probability density functions and integrals. These tools help us find probabilities, expected values, and other properties related to the variable.

Probability Density Function

A Probability Density Function, or simply PDF, is a function used in statistics to describe the likelihood of a continuous random variable taking on a particular value. While PDFs themselves don't give probabilities directly (since probabilities must be between 0 and 1), they do provide the relative likelihood of values.
In the context of the exercise, the PDF given is \( f_Y(y) = 6y(1-y) \). This function describes how likely the variable \( Y \) (the base of the box) is to assume a specific value within the interval from 0 to 1.

• The PDF is crucial for calculations involving continuous random variables because it enables us to compute quantities like expected values through integration.
• Important properties of PDFs include that they are non-negative and the integral over their entire range is equal to 1.

For example, integrating the PDF over the specified interval gives us the expected value of a function of \( Y \), like the volume of a box.

Expected Value

The expected value is a fundamental concept in probability and statistics that provides a measure of the "average" or "mean" outcome one can expect from a random experiment, particularly when repeated numerous times. For continuous random variables, it's computed by the integral of the product of the random variable and its probability density function.
In the exercise, the expected volume \( E[V] \) of the box is found using:\[E[V] = \int_0^1 5y^2 \cdot 6y(1-y) \, dy\]This integral evaluates the average value of the volume based on the various possible sizes of the base \( Y \), as informed by its PDF. The expected value acts as a central measure that summarizes the characteristic outcome of the random variable over the interval.

Integral Calculus

Integral calculus is a branch of mathematics that deals with the study of integrals and their properties. It's essential for finding areas under curves, which in probability is used to compute quantities like cumulative probabilities and expected values.
In our exercise, we apply integral calculus to calculate the expected value of the volume of the box. The integral for \( E[V] \) is set up as:\[E[V] = \int_0^1 (30y^3 - 30y^4) \, dy\]Step-by-step, this involved:

• Substituting the volume function and PDF into the integral.
• Applying basic integration techniques to solve it, which often includes techniques like substitution or recognizing polynomial forms.
• Evaluating the integral over the given range, translating calculus into practical, statistical results like the expected volume, which was calculated to be 1.5 cubic inches.

The final result illustrates the power of integral calculus in translating continuous probability expressions into meaningful expected outcomes.