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)=\left\\{\begin{array}{ll}5(1-x)^{4} & 0
Short Answer
Expert verified
The tank capacity must be 6.23 gallons to have a 1% chance of being exhausted in a week.
Step by step solution
01
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 \( \int_{0}^{1} 5(1-x)^4 dx = 1 \).
02
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].
03
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 \(x\).
04
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.
05
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.
06
Calculate CDF
Now, we calculate the CDF:
\[F(x) = \int_{0}^{x} 5(1-t)^4 dt = 5 \frac{-(1-t)^5}{5} \Big|_{0}^{x} = -(1-x)^5 + 1\]
07
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.
08
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.
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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) 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:
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
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:
\[F(x) = \int_{0}^{x} 5(1-t)^4 dt = -(1-x)^5 + 1\]
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.
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:
\(- (1-x)^5 + 1 = 0.99\)
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.
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:
- \(F(x) = 0.99\)
\(- (1-x)^5 + 1 = 0.99\)
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.