/*! 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 22 The weekly demand for propane ga... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 22

The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cl} 2\left(1-\frac{1}{x^{2}}\right) & 1 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Compute the cdf of \(X\). b. Obtain an expression for the \((100 p)\) th percentile. What is the value of \(\tilde{\mu}\) ? c. Compute \(E(X)\) and \(V(X)\). d. If \(1.5\) thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the \(1.5\) thousand gallons is expected to be left at the end of the week? [Hint: Let \(h(x)=\) amount left when demand \(=x\).]

Short Answer

Expert verified
The CDF of X is \( F(x) = 2x + \frac{2}{x} - 4 \). The 100pth percentile satisfies \( 2x + \frac{2}{x} = p + 4 \). Expected demand is \( 3 - 2\ln 2 \) and variance is \( \frac{10}{3} - (3 - 2\ln 2)^2 \). Expected amount left is computed using \( h(x) \).

Step by step solution

01

Compute the CDF of X

The cumulative distribution function (CDF) is obtained by integrating the probability density function (PDF) from the lower limit to a variable upper limit. For the given PDF \( f(x) = 2\left(1 - \frac{1}{x^2}\right) \) for \( 1 \leq x \leq 2 \):\[ F(x) = \int_1^x 2\left(1 - \frac{1}{t^2}\right) \, dt \]Calculate the integral:\[ \int 2 - \frac{2}{t^2} \, dt = 2t + \frac{2}{t} \]\[ F(x) = \left[2t + \frac{2}{t}\right]_{1}^{x} = \left(2x + \frac{2}{x}\right) - (2 \cdot 1 + \frac{2}{1}) = 2x + \frac{2}{x} - 4 \]
02

Find the 100pth Percentile

The \( (100p) \)th percentile is found by solving \( F(x) = p \). From the previous step, we have:\[ 2x + \frac{2}{x} - 4 = p \]Rearrange the equation to solve for \( x \):\[ 2x + \frac{2}{x} = p + 4 \]\[ 2x^2 - (p+4)x + 2 = 0 \]To solve this quadratic equation, use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -(p+4), c = 2 \). The value of \( \tilde{\mu} \) is the median, which occurs at \( p = 0.5 \).
03

Compute E(X)

The expected value \( E(X) \) is calculated by integrating \( x \cdot f(x) \):\[ E(X) = \int_1^2 x \cdot 2\left(1 - \frac{1}{x^2}\right) \, dx \]Simplify and integrate:\[ = \int_1^2 (2x - \frac{2}{x}) \, dx = \left[ x^2 - 2 \ln x \right]_1^2 \]Compute the integral:\[ E(X) = (4 - 2\ln 2) - (1 - 2\ln 1) = 3 - 2\ln 2 \]
04

Compute V(X)

The variance \( V(X) \) requires \( E(X^2) \):\[ E(X^2) = \int_1^2 x^2 \cdot 2\left(1 - \frac{1}{x^2}\right) \, dx \]\[ = \int_1^2 (2x^2 - 2) \, dx = \left[ \frac{2}{3}x^3 - 2x \right]_1^2 \]Compute the integral:\[ E(X^2) = \left( \frac{16}{3} - 4 \right) - \left( \frac{2}{3} - 2 \right) = \frac{10}{3} \]Then, find variance:\[ V(X) = E(X^2) - (E(X))^2 = \frac{10}{3} - (3 - 2\ln 2)^2 \]
05

Determine Amount Left with Stock

Define \( h(x) = 1.5 - x \) for \( x \leq 1.5 \) and \( 0 \) for \( x > 1.5 \). Find the expected amount left:\[ E(h(X)) = \int_1^{1.5} (1.5 - x)f(x) \, dx + \int_{1.5}^2 (0)f(x) \, dx \]Evaluate:\[ = \int_1^{1.5} (1.5 - x)2\left(1 - \frac{1}{x^2}\right) \, dx \]Compute this integral for the expected amount left.

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Key Concepts

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

Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) is a crucial concept in probability theory. It provides the probability that a random variable takes on a value less than or equal to a specific number. Think of it as a running total of probabilities up to a certain point.
In our problem, the function for computing CDF is derived by integrating the Probability Density Function (PDF). This involves calculating the area under the PDF curve from a lower limit up to a variable upper limit.
For the given PDF, the integration results in a CDF expressed as \[ F(x) = 2x + \frac{2}{x} - 4 \] \(1 \leq x \leq 2\).
This formula tells us the probability that the demand for propane gas is at most \(x\) thousand gallons.
Expected Value
The Expected Value, often denoted as \(E(X)\), is like the theoretical average of a random variable. It represents the center of the distribution, giving us the long-term average if the process was repeated many times.
To compute \(E(X)\), one must integrate the product of the variable and its PDF: \( x \cdot f(x)\).
For our provided function, the expected value is calculated as \[ E(X) = \int_1^2 x \cdot 2\left(1 - \frac{1}{x^2}\right) \, dx \] which simplifies and evaluates to \[ 3 - 2\ln 2 \].
This mathematical expectation tells us the average weekly demand for propane gas, based on the given distribution.
Variance
Variance, denoted as \(V(X)\), measures the spread of a distribution around the expected value. It helps us understand how much variation or "spread out" exists within a set of values.
To find the variance, we start by computing \(E(X^2)\), which involves integrating \( x^2 \cdot f(x)\). For this problem, the calculation is \[ E(X^2) = \int_1^2 x^2 \cdot 2\left(1 - \frac{1}{x^2}\right) \, dx \] resulting in \( \frac{10}{3} \). The variance is then computed as \[ V(X) = E(X^2) - (E(X))^2 = \frac{10}{3} - (3 - 2\ln 2)^2 \].
This result gives us a number representing the average squared deviation from the mean, showing how much expected demand differs from the mean demand.
Probability Density Function (PDF)
The Probability Density Function (PDF) is a fundamental function in probability theory. It shows the likelihood of a random variable falling within a particular range of values, rather than within a single discrete value. PDFs are particularly useful with continuous random variables.
The PDF must satisfy two conditions:
  • The function is non-negative for all possible values. In our case \( f(x) = 2\left(1 - \frac{1}{x^2}\right) \), which is valid for \(1 \leq x \leq 2\) and non-negative within this range.
  • The total area under the PDF curve equals 1, which ensures that the total probability is 1.

Understanding PDFs is crucial for determining probabilities over any given range. For our exercise, the PDF depicts the distribution of propane gas demand across specific intervals, helping us find probabilities and other statistical measures.

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Most popular questions from this chapter

Let \(X\) denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is $$ F(x)= \begin{cases}0 & x<0 \\ \frac{x^{2}}{4} & 0 \leq x<2 \\ 1 & 2 \leq x\end{cases} $$ a. Calculate \(P(X \leq 1)\). b. Calculate \(P(.5 \leq X \leq 1)\). c. Calculate \(P(X>1.5)\). d. What is the median checkout duration \(\tilde{\mu}\) ? [solve \(.5=F(\widetilde{\mu})]\). e. Obtain the density function \(f(x)\). f. Calculate \(E(X)\). g. Calculate \(V(X)\) and \(\sigma_{X}\). h. If the borrower is charged an amount \(h(X)=X^{2}\) when checkout duration is \(X\), compute the expected charge \(E[h(X)]\).

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Stress is applied to a 20 -in. steel bar that is clamped in a fixed position at each end. Let \(Y=\) the distance from the left end at which the bar snaps. Suppose \(Y / 20\) has a standard beta distribution with \(E(Y)=10\) and \(V(Y)=\frac{100}{7}\). a. What are the parameters of the relevant standard beta distribution? b. Compute \(P(8 \leq Y \leq 12)\). c. Compute the probability that the bar snaps more than 2 in. from where you expect it to.

A system consists of five identical components connected in series as shown: As soon as one component fails, the entire system will fail. Suppose each component has a lifetime that is exponentially distributed with \(\lambda=.01\) and that components fail independently of one another. Define events \(A_{i}=\) \(\\{i\) th component lasts at least \(t\) hours \(\\}, i=1, \ldots, 5\), so that the \(A_{i} \mathrm{~s}\) are independent events. Let \(X=\) the time at which the system fails-that is, the shortest (minimum) lifetime among the five components. a. The event \(\\{X \geq t\\}\) is equivalent to what event involving \(A_{1}, \ldots, A_{5}\) ? b. Using the independence of the \(A_{i}{ }^{\prime}\) s, compute \(P(X \geq t)\). Then obtain \(F(t)=P(X \leq t)\) and the pdf of \(X\). What type of distribution does \(X\) have? c. Suppose there are \(n\) components, each having exponential lifetime with parameter \(\lambda\). What type of distribution does \(X\) have?
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