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Chapter 5: Problem 5

A filling station is supplied with gasoline once a week. If its weekly volume of sales in thousands of gallons is a random variable with probability density function $$ f(x)= \begin{cases}5(1-x)^{4} & 0

Short Answer

Expert verified
The required tank capacity that would give a 1% chance of the supply being exhausted in a given week is approximately 0.985 thousand gallons or 985 gallons.

Step by step solution

01

Validate the probability density function

In order to make sure that the given function is a valid probability density function, we need to confirm that its integral over its domain is equal to 1. The given pdf is: \(f(x) = \begin{cases} 5(1-x)^4 & 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}\) We need to compute the integral of the given pdf: \(\int_{-\infty}^{+\infty}f(x)dx =\int_{0}^{1}5(1-x)^4 dx\)
02

Calculate the integral of the pdf

Compute the definite integral between 0 and 1: \(\int_{0}^{1}5(1-x)^4 dx = 5\int_{0}^{1}(1-x)^4 dx = 5[-\frac{(1-x)^5}{5}]\Big|_0^1\)
03

Verify the pdf property

Evaluate the integral's value: \(5[-\frac{(1-x)^5}{5}]\Big|_0^1 = [0 - (-1)] = 1\) Since the integral of the pdf equals 1, the provided function is a valid probability density function.
04

Calculate the cumulative distribution function (CDF)

In order to find the tank capacity where the probability of the supply being exhausted is 0.01 (1% chance), we need to find x such that the cumulative distribution function (CDF) equals 0.99. The CDF is the integral of the pdf: \(F(x) = \int_{0}^{x} 5(1-t)^4 dt = 5[-\frac{(1-t)^5}{5}]\Big|_0^x\)
05

Solve for the required CDF

Since we want the chance of the supply being exhausted to be 0.01, we need to find the x that gives us a CDF of 0.99: \(0.99 = 5[-\frac{(1-x)^5}{5}]\) Solve for x: \(-\frac{(1-x)^5}{5} = 0.198\) \((1-x)^5 = 0.99\)
06

Find the value of x

Solve for x: \(1-x = 0.99^{1/5}\) \(x = 1 - 0.99^{1/5}\) \(x \approx 0.985\) The required tank capacity that would give a 1% chance of the supply being exhausted in a given week is approximately 0.985 thousand gallons or 985 gallons.

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Key Concepts

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

Cumulative Distribution Function
Understanding the Cumulative Distribution Function (CDF) is key to solving problems involving probabilistic events. The CDF is essentially a tool that tells us the probability that a random variable takes on a value less than or equal to a certain point. It provides a comprehensive picture of how probabilities are distributed across different possible values of a random variable.

For instance, if we say that the CDF of a variable X at a particular value x equals 0.99, it means there is a 99% probability that X will assume a value equal to or less than x. In terms of integral calculus, the CDF is defined as the integral of the probability density function (PDF) from the minimum possible value of the random variable up to x. It accumulates the probabilities, thus forming the CDF.

In our exercise, the task was to figure out the maximum sales volume (or tank capacity) for gasoline where the chance of not running out of supply is 0.99, represented by the CDF. Through integration of the PDF, the CDF was calculated and used to find the requisite capacity.
Integral Calculus
Integral calculus is the mathematical process of finding the integral of a function. In the context of probability and statistics, it is used primarily to derive the Cumulative Distribution Function (CDF) from the Probability Density Function (PDF).

The fundamental idea behind integration in this context is that it allows us to sum an infinite number of infinitesimally small probabilities to find a cumulative probability. Specifically, the CDF is found by integrating the PDF from the lower bound of the random variable's domain up to a particular value x.

In our exercise, we computed the integral of the PDF to ensure that it summed to one over its entire domain, validating it as a probability density function. Following this validation, integration was again used to find the CDF so that we could solve for the tank capacity providing a 1% chance of depletion.
Random Variables
Random variables are fundamental concepts in probability and statistics. They are variables whose possible values are numerical outcomes of a random phenomenon. Random variables can either be discrete, with finite or countable outcomes, or continuous, with infinite outcomes within a range.

In our gasoline-selling scenario, the weekly volume of sales represents a continuous random variable. This is because it can take on any value within a specified range (specifically between 0 and 1 in the exercise). To analyze the behavior of such a variable, we use probability functions, such as the Probability Density Function (PDF) and the Cumulative Distribution Function (CDF).

These functions help us describe how probabilities are distributed over the range of the random variable, providing a mathematical framework to solve problems like the determination of tank capacity in our exercise.

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Most popular questions from this chapter

Let \(X\) be a normal random variable with mean 12 and variance \(4 .\) Find the value of \(c\) such that \(P[X>c\\}=.10\).

A point is chosen at random on a line segment of length \(L\). Interpret this statement and find the probability that the ratio of the shorter to the longer segment is less than \(\frac{1}{4}\).

The life of a certain type of automobile tire is normally distributed with mean 34,000 miles and standard deviation 4000 miles. (a) What is the probability that such a tire lasts over 40,000 miles? (b) What is the probability that it lasts between 30,000 and 35,000 miles? (c) Given that it has survived 30,000 miles, what is the conditional probability that it survives another 10,000 miles?

Let \(X\) be a random variable that takes on values between 0 and \(c\). That is, \(P\\{0 \leq X \leq c\\}=1\). Show that $$ \operatorname{Var}(X) \leq \frac{c^{2}}{4} $$ HINT: One approach is to first argue that $$ E\left[X^{2}\right] \leq c E[X] $$ Then use this to show that $$ \operatorname{Var}(X) \leq c^{2}[\alpha(1-\alpha)] \quad \text { where } \quad \alpha=\frac{E[X]}{c} $$

Find the distribution of \(R=A \sin \theta\), where \(A\) is a fixed constant and \(\theta\) is uniformly distributed on \((-\pi / 2, \pi / 2)\). Such a random variable \(R\) arises in the theory of ballistics. If a projectile is fired from the origin at an angle \(\alpha\) from the earth with a speed \(v\), then the point \(R\) at which it returns to the earth can be expressed as \(R=\left(v^{2} / g\right) \sin 2 \alpha\), where \(g\) is the gravitational constant, equal to 980 centimeters per second squared.
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