Problem 5 - A First Course in Probability

Question

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)=\left\\{\begin{array}{ll}5(1-x)^{4} & 0

Step-by-Step Solution

Answer

The tank capacity must be 6.23 gallons to have a 1% chance of being exhausted in a week.

Step 1: Validate the pdf

First, let's validate that the function given is a legitimate pdf. A pdf must integrate to 1 over its domain. For the function \(f(x)\), the domain is [0,1]. So, we need to solve \( \int_{0}^{1} 5(1-x)^4 dx = 1 \).

Step 2: Calculate CDF

The cumulative distribution function (CDF) is obtained by integrating the pdf. Calculate the CDF \(F(x) = \int_{0}^{x} 5(1-t)^4 dt\), for \(x\) in the domain [0, 1].

Step 3: Find the 99% percentile

We need to find the x-value such that \(F(x) = 0.99\). This is the same as solving the equation \( \int_{0}^{x} 5(1-t)^4 dt = 0.99 \) for \(x\).

Step 4: Solve for tank capacity

The tank capacity, in thousands of gallons, will be equal to the x-value found in step 3. Since we want the capacity such that there is only a 1% chance of running out, we can set the capacity equal to the 99% percentile value. Now let's proceed with the calculations.

Step 1: Validate the pdf

We need to solve the integral \( \int_{0}^{1} 5(1-x)^4 dx \). Using the power rule for integration, we get: \[ 5 \int_{0}^{1} (1-x)^4 dx = 5[\frac{- (1-x)^5}{5}]_{0}^{1} = 5[-(1 - 2x + x^2 - x^3 + x^4)]_{0}^{1} = 5(1-0) =1\] The integral equals 1, which validates the pdf.

Step 2: Calculate CDF

Now, we calculate the CDF: \[F(x) = \int_{0}^{x} 5(1-t)^4 dt = 5 \frac{-(1-t)^5}{5} \Big|_{0}^{x} = -(1-x)^5 + 1\]

Step 3: Find the 99% percentile

Solve \(F(x) = 0.99\): \(0.99 = -(1-x)^5 + 1\) \((1-x)^5 = 0.01\) Now, take the fifth root: \(1 - x = 0.01^{\frac{1}{5}} = 0.99377\) \(x = 1 - 0.99377 = 0.00623\) The x-value corresponding to the 99% percentile is approximately 0.00623.

Step 4: Solve for tank capacity

Since our x-value is in thousands of gallons, we can convert it back to gallons by simply multiplying by 1000: \(Tank\ capacity = 0.00623 * 1000 = 6.23\ gallons\) So, the tank capacity must be 6.23 gallons to have a 1% chance of being exhausted in a week.

Key Concepts

Probability Density Function

The Probability Density Function (PDF) provides the likelihood of a random variable falling within a particular range of values. The PDF is applicable only for continuous random variables and is an integral part of determining the probability of events within a specified interval.

For a function to be recognized as a legitimate PDF, it must integrate to 1 over the domain of the variable. In simpler terms, if you imagine the PDF as a curve over a graph, the total area under this curve must be equal to 1. This condition ensures that the sum of all probabilities across all possible outcomes equals certainty, which is 100% or 1.

In our specific example, the function is defined as:

• \(f(x) = 5(1-x)^4\) for \(0 < x < 1\)
• \(f(x) = 0\) otherwise

Here, \(f(x)\) describes a range of probabilities over the interval \(x\) from 0 to 1. To confirm this is a valid PDF, we compute the integral of \(5(1-x)^4\) from 0 to 1. Successfully computing this as 1 confirms its legitimacy, ensuring calculations based on this PDF will yield valid probabilities.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a critical concept when working with probability, especially in the context of continuous distributions. It represents the probability that a random variable \(X\) will take a value less than or equal to \(x\).

To find the CDF, we integrate the PDF over the range from the lowest bound up to a specified point \(x\). In our explained solution, the PDF \(f(x) = 5(1-x)^4\) is integrated as follows:

\[F(x) = \int_{0}^{x} 5(1-t)^4 dt = -(1-x)^5 + 1\]

• This function continuously increases as \(x\) increases from 0 to 1.
• At \(x=0\), the CDF is 0, and at \(x=1\), the CDF becomes 1, reflecting that the entire probability distribution is captured.

In any probability question, understanding the CDF helps assess the likelihood of different outcomes which can include the probability of a variable falling short of, or exceeding specific values.

Percentiles

Percentiles are a useful statistical measure that indicate the value below which a given percentage of data in a data set falls. When dealing with probability distributions, specifically continuous ones like ours, percentiles reflect key probabilities and potential cutoff points across the dataset.

To solve for a certain percentile, say the 99th percentile as in our example, we equate the CDF to this desired probability. So we solve for \(x\) in:

• \(F(x) = 0.99\)

This involves manipulating the CDF equation:

\(- (1-x)^5 + 1 = 0.99\)
Solving for \(x\) gives us the specific value (here \(x = 0.00623\)), which represents the maximum threshold under which 99% of data, or in this case, gas consumption, lies.

Understanding percentiles is valuable in real-life scenarios like determining tank capacity or inventory needs where you wish to avoid running out of resources with a certain confidence level. Thus, for this example, a tank capacity where only 1% risk of exhaustion exists corresponds to the value at the calculated 99th percentile.