/*! 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 1 The current in a certain circuit... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The current in a certain circuit as measured by an ammeter is a continuous random variable \(X\) with the following density function: $$ f(x)=\left\\{\begin{array}{cc} .075 x+.2 & 3 \leq x \leq 5 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Graph the pdf and verify that the total area under the density curve is indeed \(1 .\) b. Calculate \(P(X \leq 4)\). How does this probability compare to \(P(X<4)\) ? c. Calculate \(P(3.5 \leq X \leq 4.5)\) and also \(P(4.5

Short Answer

Expert verified
a. Verify by checking the area under the curve is 1. b. \(P(X \leq 4) = P(X < 4) = 0.4625\). c. \(P(3.5 \leq X \leq 4.5) = 0.5\), \(P(4.5 < X) = 0.5375\).

Step by step solution

01

Identify the Probability Density Function

We have a piecewise function given by:\[ f(x) = \begin{cases} 0.075x + 0.2 & \text{for } 3 \leq x \leq 5 \ 0 & \text{otherwise} \end{cases} \] This function will be used to graph and solve for probabilities.
02

Graph the PDF and Verify Total Area

To graph the function, note that for the relevant interval \([3, 5]\), \( f(x) \) is a linear function. Calculate its values at the boundaries: - \( f(3) = 0.075(3) + 0.2 = 0.425 \) - \( f(5) = 0.075(5) + 0.2 = 0.575 \)Graphing this is a straight line between \((3, 0.425)\) and \((5, 0.575)\). To verify, compute the area under the curve:\[ \int_{3}^{5} (0.075x + 0.2) \, dx = \left[ 0.0375x^2 + 0.2x \right]_3^5 \]\[ = (0.0375(25) + 0.2(5)) - (0.0375(9) + 0.2(3)) = (0.9375 + 1) - (0.3375 + 0.6) = 1 \]Thus, the total area is 1, verifying it's a valid PDF.
03

Calculate P(X ≤ 4) and Compare to P(X < 4)

Calculate \( P(X \leq 4) \):\[ P(X \leq 4) = \int_{3}^{4} (0.075x + 0.2) \, dx = \left[ 0.0375x^2 + 0.2x \right]_3^4 \]\[ = (0.0375(16) + 0.2(4)) - (0.0375(9) + 0.2(3)) = (0.6 + 0.8) - (0.3375 + 0.6) = 0.4625 \]Since \(X\) is continuous, \(P(X \leq 4) = P(X < 4)\).
04

Calculate P(3.5 ≤ X ≤ 4.5)

Calculate \( P(3.5 \leq X \leq 4.5) \):\[ P(3.5 \leq X \leq 4.5) = \int_{3.5}^{4.5} (0.075x + 0.2) \, dx = \left[ 0.0375x^2 + 0.2x \right]_{3.5}^{4.5} \]\[ = (0.0375(20.25) + 0.2(4.5)) - (0.0375(12.25) + 0.2(3.5)) \]\[ = (0.759375 + 0.9) - (0.459375 + 0.7) = 0.5 \]
05

Calculate P(4.5 < X)

Calculate \( P(4.5 < X) \):\[ P(4.5 < X) = \int_{4.5}^{5} (0.075x + 0.2) \, dx = \left[ 0.0375x^2 + 0.2x \right]_{4.5}^{5} \]\[ = (0.9375 + 1) - (0.759375 + 0.9) = 0.5375 \]

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.

Probability Density Function
A probability density function (PDF) is a fundamental concept when dealing with continuous random variables. It describes how the values of a random variable are distributed. In simpler terms, the PDF helps us understand the likelihood of different outcomes in a continuous sample space.

The function itself, which we denote as \(f(x)\), assigns a probability to intervals of values, rather than individual points. This is because, in continuous distributions, the probability of the variable taking an exact value is zero.

One important property of a PDF is that the total area under the curve of this function is equal to 1. This represents the certainty that some value within the existing range of possibilities will occur. In our exercise, this verification is a crucial step in ensuring the PDF is valid. It is achieved by integrating the function over its defined range, checking that the result is indeed 1.
Piecewise Function
A piecewise function is a type of mathematical function defined by multiple sub-equations or "pieces," each applying to a certain interval within the function's domain. This is especially useful when dealing with probability functions that may vary in behavior within different ranges of a random variable.

In our given problem, the probability density function \(f(x)\) is piecewise, which means it takes different forms depending on the value of \(x\). Specifically, it is defined as a linear function \(0.075x + 0.2\) within the interval from 3 to 5, and equals zero outside of this interval.

Piecewise functions are common in continuous probability scenarios because they allow the PDF to effectively model scenarios where the distribution of values can change significantly over different ranges.
Area Under Curve
The concept of the area under the curve (AUC) is vital when working with probability density functions. It represents the total probability of a random variable falling within a certain range.

In terms of calculus, AUC involves integrating the PDF over the desired interval of the random variable. This mathematical process allows us to calculate the probability for continuous distributions. For instance, in our exercise, to verify the valid PDF and calculate probabilities like \(P(X \leq 4)\), we integrate the function over the specified range.

By evaluating these areas, we can effectively determine how likely an event is to occur within certain confines, reflecting the nuanced variations of probability in continuous spaces.
Probability Calculation
Calculating probabilities in the context of continuous random variables often involves integrating the probability density function over a given interval. This calculated area corresponds to the likelihood that the random variable falls within that interval.

For the exercise in question, we perform these calculations by setting the lower and upper bounds for integration according to the desired interval. For example:
  • To compute \(P(X \leq 4)\), we integrate from 3 to 4.
  • For \(P(3.5 \leq X \leq 4.5)\), we set boundaries at 3.5 and 4.5.
  • Finally, to find \(P(4.5 < X)\), integration is done from 4.5 to 5.
Continuity of the random variable allows us to equate \(P(X \leq a)\) with \(P(X < a)\), simplifying calculations unless specifically stated otherwise. This methodology ensures all probabilities are accurately determined, respecting the continuous nature of the variable.

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

The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean \(\mu\), the actual temperature of the medium, and standard deviation \(\sigma\). What would the value of \(\sigma\) have to be to ensure that \(95 \%\) of all readings are within \(1^{\circ}\) of \(\mu\) ?

The article " Determination of the MTF of Positive Photoresists Using the Monte Carlo Method" (Photographic Sct. and Engr., 1983: 254-260) proposes the exponential distribution with parameter \(\lambda=.93\) as a model for the distribution of a photon's free path length \((\mu \mathrm{m})\) under certain circumstances. Suppose this is the correct model. a. What is the expected path length, and what is the standard deviation of path length? b. What is the probability that path length exceeds \(3.0\) ? What is the probability that path length is between \(1.0\) and \(3.0\) ? c. What value is exceeded by only \(10 \%\) of all path lengths?

Suppose the diameter at breast height (in.) of trees of a certain type is normally distributed with \(\mu=8.8\) and \(\sigma=2.8\), as suggested in the article "Simulating a Harvester-Forwarder Softwood Thinning" (Forest Products J., May 1997: 36-41). a. What is the probability that the diameter of a randomly selected tree will be at least 10 in.? Will exceed 10 in.? b. What is the probability that the diameter of a randomly selected tree will exceed 20 in.? c. What is the probability that the diameter of a randomly selected tree will be between 5 and 10 in.? d. What value \(c\) is such that the interval \((8.8-c, 8.8+c)\) includes \(98 \%\) of all diameter values? e. If four trees are independently selected, what is the probability that at least one has a diameter exceeding 10 in.?

The error involved in making a certain measurement is a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cc} .09375\left(4-x^{2}\right) & -2 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Compute \(P(X>0)\). c. Compute \(P(-1.5)\).

Let \(X\) be the total medical expenses (in 1000 s of dollars) incurred by a particular individual during a given year. Although \(X\) is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf \(f(x)=k(1+x / 2.5)^{-7}\) for \(x \geq 0\). a. What is the value of \(k\) ? b. Graph the pdf of \(X\). c. What are the expected value and standard deviation of total medical expenses? d. This individual is covered by an insurance plan that entails a \(\$ 500\) deductible provision (so the first \(\$ 500\) worth of expenses are paid by the individual). Then the plan will pay \(80 \%\) of any additional expenses exceeding \(\$ 500\), and the maximum payment by the individual (including the deductible amount) is \(\$ 2500\). Let \(Y\) denote the amount of this individual's medical expenses paid by the insurance company. What is the expected value of \(Y\) ? [Hint: First figure out what value of \(X\) corresponds to the maximum out-of- pocket expense of \(\$ 2500\). Then write an expression for \(Y\) as a function of \(X\) (which involves several different pieces) and calculate the expected value of this function.]

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.