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Chapter 4: Problem 27

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

Short Answer

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

Step by step solution

01

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 \( 0 < x < 2 \).
02

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 \( f(x) = 0.5x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥檚 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.

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Most popular questions from this chapter

An article in Ad Hoc Networks ["Underwater Acoustic Sensor Networks: Target Size Detection and Performance Analysis" \((2009,\) Vol. \(7(4),\) pp. \(803-808)]\) discussed an underwater acoustic sensor network to monitor a given area in an ocean. The network does not use cables and does not interfere with shipping activities. The arrival of clusters of signals generated by the same pulse is taken as a Poisson arrival process with a mean of \(\lambda\) per unit time. Suppose that for a specific underwater acoustic sensor network, this Poisson process has a rate of 2.5 arrivals per unit time. (a) What is the mean time between 2.0 consecutive arrivals? (b) What is the probability that there are no arrivals within 0.3 time units? (c) What is the probability that the time until the first arrival exceeds 1.0 unit of time? (d) Determine the mean arrival rate such that the probability is 0.9 that there are no arrivals in 0.3 time units.

Use the result for the gamma distribution to determine the mean and variance of a chi-square distribution with \(r=7 / 2\).

The lifetime of a mechanical assembly in a vibration test is exponentially distributed with a mean of 400 hours. (a) What is the probability that an assembly on test fails in less than 100 hours? (b) What is the probability that an assembly operates for more than 500 hours before failure? (c) If an assembly has been on test for 400 hours without a failure, what is the probability of a failure in the next 100 hours? (d) If 10 assemblies are tested, what is the probability that at least one fails in less than 100 hours? Assume that the assemblies fail independently. (e) If 10 assemblies are tested, what is the probability that all have failed by 800 hours? Assume that the assemblies fail independently.

A process is said to be of six-sigma quality if the process mean is at least six standard deviations from the nearest specification. Assume a normally distributed measurement. (a) If a process mean is centered between upper and lower specifications at a distance of six standard deviations from each, what is the probability that a product does not meet specifications? Using the result that 0.000001 equals one part per million, express the answer in parts per million. (b) Because it is difficult to maintain a process mean centered between the specifications, the probability of a product not meeting specifications is often calculated after assuming that the process shifts. If the process mean positioned as in part (a) shifts upward by 1.5 standard deviations, what is the probability that a product does not meet specifications? Express the answer in parts per million. (c) Rework part (a). Assume that the process mean is at a distance of three standard deviations. (d) Rework part (b). Assume that the process mean is at a distance of three standard deviations and then shifts upward by 1.5 standard deviations. (e) Compare the results in parts (b) and (d) and comment.

The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. (a) What is the probability that more than three aircraft arrive within an hour? (b) If 30 separate one-hour intervals are chosen, what is the probability that no interval contains more than three arrivals? (c) Determine the length of an interval of time (in hours) such that the probability that no arrivals occur during the interval is \(0.10 .\)
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