/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 The actual tracking weight of a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The actual tracking weight of a stereo cartridge that is set to track at \(3 \mathrm{~g}\) on a particular changer can be regarded as a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cc} k\left[1-(x-3)^{2}\right] & 2 \leq x \leq 4 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Find the value of \(k\). c. What is the probability that the actual tracking weight is greater than the prescribed weight? d. What is the probability that the actual weight is within \(.25 \mathrm{~g}\) of the prescribed weight? e. What is the probability that the actual weight differs from the prescribed weight by more than \(.5 \mathrm{~g}\) ?

Short Answer

Expert verified
k = 0.5; P(X > 3) = 0.25; P(2.75 < X < 3.25) = 0.375; P(|X-3| > 0.5) = 0.5.

Step by step solution

01

Understanding the Function

The function given is a piecewise PDF defined as \[f(x) = k\left[1-(x-3)^{2}\right]\] within the interval \(2 \leq x \leq 4\) and 0 otherwise. This is a quadratic function that is symmetric around \(x = 3\).
02

Determine the Value of k

Since \(f(x)\) is a probability density function, the area under the curve must integrate to 1 over its defined interval: \[\int_2^4 k \left[1-(x-3)^2\right] \, dx = 1.\] Solving this integral, first find:\[\int_2^4 \left(1 - (x - 3)^2\right)\] which becomes:\[\int_2^4 \left(1 - (x^2 - 6x + 9)\right) \, dx = \int_2^4 (-x^2 + 6x - 8) \, dx.\] Evaluating gives 2. Therefore, \[k \cdot 2 = 1\] so \(k = 0.5\).
03

Probability that Actual Weight is Greater than Prescribed

Calculate \(P(X > 3)\):\[P(X > 3) = \int_3^4 0.5 \left[1-(x-3)^2\right] \, dx.\] Evaluating the integral yields \(P(X > 3) = 0.25\).
04

Probability within 0.25 g of Prescribed Weight

Calculate \(P(2.75 < X < 3.25)\):\[P(2.75 < X < 3.25) = \int_{2.75}^{3.25} 0.5 \left[1-(x-3)^2\right] \, dx.\] Evaluating the integral gives \(P(2.75 < X < 3.25) = 0.375\).
05

Probability Weight Differs by More than 0.5 g

Calculate \(P(X < 2.5) + P(X > 3.5)\):\[P(X < 2.5) = \int_2^{2.5} 0.5 \left[1-(x-3)^2\right] \, dx\] and\[P(X > 3.5) = \int_{3.5}^4 0.5 \left[1-(x-3)^2\right] \, dx.\] Evaluating both integrals and summing the results gives \(0.5\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Continuous Random Variable
A continuous random variable is a type of random variable that can take an infinite number of possible values within a given range. In our exercise, the continuous random variable is represented by the weight of a stereo cartridge, denoted as \( X \). This variable can assume any value between 2 and 4 grams. Since it can take a continuum of values, it is particularly useful for modeling quantities that are measured rather than counted.
For continuous random variables, we often describe their behavior with a Probability Density Function (PDF). The PDF assigns a probability to each possible value. However, since a continuous random variable can take countless values, we often look at the probability of the variable falling within a specific range, rather than exact values.
Probability Calculation
Probability calculation for a continuous random variable involves finding the area under its probability density function (PDF) across a particular interval. In the exercise, we're interested in several probability scenarios such as finding the probability that the actual weight is greater than a given weight or falls within a specific weight range.
To calculate these probabilities, we set up integrals of the PDF over the desired interval. For example, to find the probability that the actual weight is greater than 3 grams, we integrate the PDF from 3 to 4. This method allows us to accumulate the total probability across the interval of interest. Remember, the area under the entire PDF curve sums to 1, representing the total probability.
Integration in Probability
Integration is a crucial concept in probability when dealing with continuous random variables. It helps calculate probabilities over intervals for these types of variables by providing the area under the curve of the probability density function (PDF).
The process starts with setting up an integral using the PDF over the specified bounds. For instance, in our exercise, we integrate from 2 to 4 due to the given piecewise function. Performing integration allows us to normalize the probability so that the area under the curve equals 1, and then compute desired probabilities for intervals such as \( P(X > 3) \).
Through integration, we can also confirm that functions are valid PDFs and utilize the fundamental theorem of calculus to evaluate these integrals efficiently.
Piecewise Function Analysis
A piecewise function is a function defined by different expressions over different intervals of the input. In probability, these functions often describe scenarios where conditions change over the range of the random variable. In our task, the PDF \( f(x) = k[1-(x-3)^2] \) is a piecewise function with distinct behavior over the interval from 2 to 4.
Analyzing a piecewise function involves looking at each defined part separately to understand its contribution to the overall function. The PDF goes to zero outside the interval [2, 4], simplifying probability calculations by focusing computation only on relevant bounds. This feature is critical in modeling real-world phenomena where conditions may change, such as the distribution of the stereo cartridge weight in the exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What condition on \(\alpha\) and \(\beta\) is necessary for the standard beta pdf to be symmetric?

The article "Second Moment Reliability Evaluation vs. Monte Carlo Simulations for Weld Fatigue Strength" (Quality and Reliability Engr. Intl., 2012: 887-896) considered the use of a uniform distribution with \(A=.20\) and \(B=4.25\) for the diameter \(X\) of a certain type of weld (mm). a. Determine the pdf of \(X\) and graph it. b. What is the probability that diameter exceeds \(3 \mathrm{~mm}\) ? c. What is the probability that diameter is within \(1 \mathrm{~mm}\) of the mean diameter? d. For any value \(a\) satisfying .20

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, 1985: 519-522) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens subjected to \(\mathrm{AC}\) voltage. The values of the parameters depend on the voltage and temperature; suppose \(\alpha=2.5\) and \(\beta=200\) (values suggested by data in the article). a. What is the probability that a specimen's lifetime is at most 250? Less than 250? More than 300 ? b. What is the probability that a specimen's lifetime is between 100 and 250 ? c. What value is such that exactly \(50 \%\) of all specimens have lifetimes exceeding that value?

Let \(X\) denote the lifetime of a component, with \(f(x)\) and \(F(x)\) the pdf and cdf of \(X\). The probability that the component fails in the interval \((x, x+\Delta x)\) is approximately \(f(x) \cdot \Delta x\). The conditional probability that it fails in \((x, x+\Delta x)\) given that it has lasted at least \(x\) is \(f(x) \cdot \Delta x /[1-F(x)]\). Dividing this by \(\Delta x\) produces the failure rate function: $$ r(x)=\frac{f(x)}{1-F(x)} $$ An increasing failure rate function indicates that older components are increasingly likely to wear out, whereas a decreasing failure rate is evidence of increasing reliability with age. In practice, a "bathtub-shaped" failure is often assumed. a. If \(X\) is exponentially distributed, what is \(r(x)\) ? b. If \(X\) has a Weibull distribution with parameters \(\alpha\) and \(\beta\), what is \(r(x)\) ? For what parameter values will \(r(x)\) be increasing? For what parameter values will \(r(x)\) decrease with \(x\) ? c. Since \(r(x)=-(d / d x) \ln [1-F(x)], \ln [1-F(x)]=\) \(-\int r(x) d x\). Suppose $$ r(x)=\left\\{\begin{array}{cc} \alpha\left(1-\frac{x}{\beta}\right) & 0 \leq x \leq \beta \\ 0 & \text { otherwise } \end{array}\right. $$ so that if a component lasts \(\beta\) hours, it will last forever (while seemingly unreasonable, this model can be used to study just "initial wearout"). What are the cdf and pdf of \(X\) ?

Vehicle speed on a particular bridge in China can be modeled as normally distributed ("Fatigue Reliability Assessment for Long-Span Bridges under Combined Dynamic Loads from Winds and Vehicles" \(J\). of Bridge Engr., 2013: 735-747). a. If \(5 \%\) of all vehicles travel less than \(39.12 \mathrm{~m} / \mathrm{h}\) and \(10 \%\) travel more than \(73.24 \mathrm{~m} / \mathrm{h}\), what are the mean and standard deviation of vehicle speed? [Note: The resulting values should agree with those given in the cited article.] b. What is the probability that a randomly selected vehicle's speed is between 50 and \(65 \mathrm{~m} / \mathrm{h}\) ? c. What is the probability that a randomly selected vehicle's speed exceeds the speed limit of \(70 \mathrm{~m} / \mathrm{h}\) ?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.