/*! 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 26 The weekly demand for propane ga... [FREE SOLUTION] | 91Ó°ÊÓ

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The weekly demand for propane gas (in 1000 ' 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 \(\mu\) ? 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
CDF is \( F(x) = 2x - \frac{2}{x} \). Mean \( \mu = 3 - 2\ln 2 \). Expected amount left is calculated from integral.

Step by step solution

01

Determine the Cumulative Distribution Function (CDF)

Given the pdf, the cdf is found by integrating the pdf over its range. The cdf, denoted as \( F(x) \), is defined as \( F(x) = \int_{1}^{x} f(t) \, dt \) for \( 1 \leq x \leq 2 \). Calculating this we have:\[ F(x) = \int_{1}^{x} 2\left(1-\frac{1}{t^{2}}\right) \, dt = \left[2t - \frac{2}{t}\right]_{1}^{x} = (2x - \frac{2}{x}) - (2 \times 1 - \frac{2}{1}) \]\[ = 2x - \frac{2}{x} - 0 \]Thus, \( F(x) = 2x - \frac{2}{x} \) for \( 1 \leq x \leq 2 \).
02

Find the Percentile Function

The \((100p)\)th percentile, \( \ \ x_p \), is found by setting \( F(x_p) = p \). From Step 1, \( F(x) = 2x - \frac{2}{x} \), setting this equal to \( p \) gives:\[ 2x - \frac{2}{x} = p \]Solving for \( x \), multiply through by \( x \) and rearrange:\[ 2x^2 - px - 2 = 0 \]Solve this quadratic equation to find an expression for \( x_p \) in terms of \( p \).
03

Calculate the Mean \( E(X) \)

The expected value \( E(X) \) is calculated by integrating \( x \cdot f(x) \) over the range:\[ E(X) = \int_{1}^{2} x \, \cdot \, 2\left(1-\frac{1}{x^{2}}\right) \, dx = \int_{1}^{2} (2x - 2 \, \frac{1}{x}) \, dx \]Simplify and integrate:\[ E(X) = \left[ x^2 - 2\ln x \right]_{1}^{2} = (4 - 2\ln 2) - (1 - 0) \]\[ = 3 - 2\ln 2 \].
04

Calculate the Variance \( V(X) \)

The variance \( V(X) = E(X^2) - [E(X)]^2 \). First, calculate \( E(X^2) \) by integrating:\[ E(X^2) = \int_{1}^{2} x^2 \, \cdot \, 2\left(1-\frac{1}{x^{2}}\right) \, dx = \int_{1}^{2} (2x^2 - 2) \, dx \]\[ E(X^2) = \left[ \frac{2x^3}{3} - 2x \right]_{1}^{2} = \left(\frac{16}{3} - 4\right) - \left(\frac{2}{3} - 2\right) \]\[ = \frac{12}{3} - \frac{6}{3} = 2 \]So, \( V(X) = 2 - (3 - 2\ln 2)^2 \).
05

Compute the Expected Amount Left

The amount left when demand \( = x \) is given by \( h(x) = 1.5 - x \) if \( x < 1.5 \) and zero otherwise. Thus, the expected amount left is:\[ E(h(X)) = \int_{1}^{1.5} (1.5 - x) \, \cdot \, 2\left(1-\frac{1}{x^{2}}\right) \, dx \]Calculate this integral:\[ = 2\int_{1}^{1.5} (1.5-x)(1-\frac{1}{x^2}) \, dx \]Simplify and integrate to find the expected amount of propane left.

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Key Concepts

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

Probability Density Function
The probability density function (pdf) is a crucial concept in statistics and probability. It represents the likelihood of a continuous random variable taking on a specific value. Understanding pdfs helps us model real-world scenarios where outcomes are not discrete but rather form a continuous spectrum.

For example, the pdf given in the exercise for the demand of propane gas describes how probable it is for the usage to be within a certain range. The formula is:
  • \( f(x) = 2 \left( 1 - \frac{1}{x^2} \right) \) for \( 1 \leq x \leq 2 \)
  • \( f(x) = 0 \) otherwise

This function tells us that amounts of gas between 1,000 and 2,000 gallons have variable probabilities, influenced by the shape of the curve (graph) defined by the function. The area under the curve of the pdf over its domain sums to 1, illustrating the total probability.
Expected Value
The expected value or mean of a random variable provides a measure of its central tendency. In simpler terms, it's the average outcome if an experiment is repeated many times.

We calculate the expected value of a continuous random variable by integrating the product of the variable's value and its pdf over the entire range:
\[E(X) = \int_{a}^{b} x \cdot f(x) \, dx\]
In our exercise, the expected demand for propane is calculated using:
  • The integral \( \int_{1}^{2} x \cdot 2 \left( 1 - \frac{1}{x^2} \right) \, dx \)
  • Simplifying the function within the integral before performing the integration
The outcome, after calculation, gives the expected amount of propane needed in a typical week.
Percentile Function
Percentiles are statistical measures indicating the value below which a given percentage of observations fall. For continuous random variables, percentiles help us understand the distribution better by describing different points along the range.

For the \(100p\)th percentile, denoted as \ x_p \, we solve for \ x \ from the cumulative distribution function (CDF), where \( F(x) = p \).

In the exercise, this involves solving the equation:
  • \( 2x - \frac{2}{x} = p \)
  • Multiplying through by \( x \) and rearranging terms to form a quadratic equation: \( 2x^2 - px - 2 = 0 \)
The roots of this equation will provide the desired percentiles for different probabilities (\( p \)) in the context of propane demand.
Variance Calculation
Variance is a statistical measure indicating how much the values of a random variable deviate from the mean. It's a measure of dispersion or variability in the data.

To calculate the variance \( V(X) \), we first find \( E(X^2) \), the expected value of the square of the random variable:
  • Integrate \( x^2 \cdot f(x) \) over the given range
Then, apply the formula:
\[V(X) = E(X^2) - [E(X)]^2\]
From our exercise, after determining \( E(X) \) and \( E(X^2) \), the variance informs us about the variability in weekly propane gas demand. This insight helps in planning and stocking the facility efficiently.
Integration in Statistics
Integration is a fundamental tool in statistics, especially when dealing with continuous random variables. It helps us find aggregate quantities like totals and averages from continuous functions.

The integration process transforms the pdf into the cumulative distribution function (CDF) and calculates expected values, variances, and more.
The CDF is derived by integrating the pdf over its defined range:
  • \( F(x) = \int_{lower}^{x} f(t) \), provides the probability that the random variable is less than or equal to \( x \)
In statistical analysis, this lets us compute probabilities and other essential metrics about an entire data set or distribution.
Understanding how to integrate functions is a key skill for interpreting statistical models, as evidenced by the various calculations performed in the propane exercise.

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