Problem 27 - Applied Statistics and Probability for Engineers

Question

The gap width is an important property of a magnetic recording head. In coded units, if the width is a continuous random variable over the range from \(0

Step-by-Step Solution

Answer

CDF: \( F(x) = 0.25x^2 \); PDF: \( f(x) = 0.5x \).

Step 1: Determine the Cumulative Distribution Function (CDF)

To determine the CDF from the given probability density function (PDF), we need to integrate the PDF. The PDF is given as \( f(x) = 0.5x \) for \( 0 < x < 2 \). The CDF \( F(x) \) is the integral of \( f(x) \) with respect to \( x \): \[ F(x) = \int_{0}^{x} 0.5t \, dt \]Calculating the integral, we find:\[ F(x) = \left[ 0.5 \cdot \frac{t^2}{2} \right]_{0}^{x} = 0.25x^2 \]Therefore, the CDF is \( F(x) = 0.25x^2 \) for \( 0 < x < 2 \).

Step 2: Determine the Probability Density Function from the CDF

To find the probability density function (PDF) from a given cumulative distribution function (CDF), we need to differentiate the CDF. For the CDF \( F(x) = 0.25x^2 \), the PDF \( f(x) \) is given by the derivative \( \frac{d}{dx} F(x) \):\[ f(x) = \frac{d}{dx}(0.25x^2) = 0.5x \]Thus, the PDF derived from this CDF is \( f(x) = 0.5x \).

Key Concepts

Probability Density Function

A Probability Density Function, or PDF, is a crucial concept in statistics. It describes the probability of a continuous random variable falling within a particular range of values. Unlike discrete probabilities, a PDF indicates a relative likelihood instead of assigning a probability to each possible value.
The PDF has several important properties:

• The PDF must be non-negative for all possible values.
• The area under the PDF over the entire range must equal 1. This represents the total probability.
• The PDF can be used to find probabilities over intervals. These probabilities are calculated as the area under the curve of the PDF over the interval of interest.

In the given exercise, the PDF is expressed as \( f(x) = 0.5x \) which is defined over the interval \(0 < x < 2\). This function indicates how densely packed probability is over the interval. Since it's continuous, rather than depicting probability for an exact \(x\) value, it allows us to compute probability over a specified range.

Integration in Statistics

Integration is a fundamental tool in statistics, used for finding cumulative probabilities, areas under curves, and in transforming probability density functions into cumulative distribution functions (CDFs). The CDF gives us the probability that a random variable is less than or equal to a particular value.
The process to calculate the CDF \( F(x) \) from a PDF \( f(x) \) involves integrating the PDF from its lower limit up to \(x\):

• Start with the PDF. In the exercise, this is \( f(x) = 0.5x \).
• Integrate the function from the lower bound up to \(x\), i.e., \( F(x) = \int_{0}^{x} 0.5t \, dt \).
• The result gives \( F(x) = 0.25x^2 \), which represents the CDF for the range \(0 < x < 2\).

This integration shows how much probability accumulates up to a particular point, giving insight into the overall distribution of the random variable.

Differentiation in Statistics

Differentiation is another key concept in statistics which is often used to reverse the process of integration in the context of probability. It helps us derive a PDF from a given CDF. The derivative of the CDF provides the density function, showing how probability is distributed locally.
In the exercise, we start with the CDF \( F(x) = 0.25x^2 \). Here’s how differentiation helps:

• Find the derivative of the CDF with respect to \(x\): \( f(x) = \frac{d}{dx}(0.25x^2) \).
• The differentiation yields \( f(x) = 0.5x \), returning us to the original PDF used to find the CDF initially.
• This connection highlights the relationship between cumulative probabilities and density functions.

Differentiating shows the rate at which probability is accumulating and helps verify that the process of integration was completed correctly.