Problem 49
Question
Find \(c\) so that \(\int_{0}^{c} \frac{1}{3 \sqrt{2 \pi}} x^{3 / 2} e^{-x / 2} d x=0.90\).
Step-by-Step Solution
Verified Answer
Use Gamma CDF to find \(c\) such that \(F(c) = 0.90\), where \(\alpha = 2.5\) and \(\theta = 2\).
1Step 1: Understand the Problem
We need to determine the value of \(c\) such that the integral of the function \(\frac{1}{3 \sqrt{2\pi}} x^{3/2} e^{-x/2}\) from 0 to \(c\) equals 0.90. The function given is related to a probability distribution.
2Step 2: Recognize the Distribution
The function \(\frac{1}{3 \sqrt{2\pi}} x^{3/2} e^{-x/2}\) is the probability density function of a Gamma distribution with shape parameter \(\alpha = 2.5\) and scale parameter \(\theta = 2\). This is a type of distribution commonly used in statistics.
3Step 3: Use the CDF of the Gamma Distribution
We need the cumulative distribution function (CDF) of a Gamma distribution to solve this. For a given Gamma distribution with parameters \(\alpha\) and \(\theta\), the CDF is \(F(x)\). We want \(F(c) = 0.90\).
4Step 4: Calculate \(c\) Using Statistical Software or Tables
Using statistical software or a Gamma distribution table, find \(c\) such that the cumulative probability is 0.90. Given \(\alpha = 2.5\) and \(\theta = 2\), look up or calculate numerically to find the value of \(c\) matching this critera.
5Step 5: Verify the Calculation
After determining the value of \(c\), verify by checking \(F(c)\). Ensure that the Gamma CDF for the chosen \(c\) closely approximates 0.90 to ensure accuracy.
Key Concepts
Probability Density FunctionCumulative Distribution FunctionGamma Distribution Parameters
Probability Density Function
The probability density function (PDF) is central to understanding probability distributions. In the context of the Gamma distribution, it describes how the probability of a random variable is distributed across different values.
For a Gamma distribution with shape parameter \(\alpha\) and scale parameter \(\theta\), the PDF can be expressed as follows:
For a Gamma distribution with shape parameter \(\alpha\) and scale parameter \(\theta\), the PDF can be expressed as follows:
- \( f(x; \alpha, \theta) = \frac{x^{\alpha - 1} e^{-x/\theta}}{\theta^\alpha \Gamma(\alpha)} \) for \( x > 0 \).
- Here, \(\Gamma(\alpha)\) is the gamma function, which generalizes the factorial function to non-integer values.
- The expression \(x^{3/2}e^{-x/2}\) from the original problem indicates it's a gamma distribution PDF, specifically with \(\alpha = 2.5\) and \(\theta = 2\).
Cumulative Distribution Function
The cumulative distribution function (CDF) is a vital tool in statistics. It represents the probability that a random variable will take a value less than or equal to a specific value.
For a Gamma distribution, the CDF is denoted as \(F(x; \alpha, \theta)\), and it is the integral of the PDF from zero to \(x\):
For a Gamma distribution, the CDF is denoted as \(F(x; \alpha, \theta)\), and it is the integral of the PDF from zero to \(x\):
- \( F(x; \alpha, \theta) = \int_0^x \frac{t^{\alpha - 1} e^{-t/\theta}}{\theta^\alpha \Gamma(\alpha)} \, dt \)
- In simpler terms, it is the accumulated probability up to point \(x\).
- In the exercise, finding \(c\) such that \(F(c) = 0.90\) confirms that 90% of the distribution's probability is captured up to that value.
Gamma Distribution Parameters
The Gamma distribution is defined by two key parameters: the shape parameter \(\alpha\), and the scale parameter \(\theta\). These parameters determine the distribution's characteristics:
- Shape Parameter (\(\alpha\)): This controls the skewness and the form of the distribution. In simple terms, it affects how 'pointed' or 'spread out' the distribution looks.
- Scale Parameter (\(\theta\)): This affects the spread or the scale on the x-axis of the distribution curve. A larger \(\theta\) results in a wider distribution.
- The given problem specifies \(\alpha = 2.5\) and \(\theta = 2\), meaning the distribution will be moderately skewed and spread, suitable for modeling quantities that naturally accumulate over time like waiting times.
Other exercises in this chapter
Problem 48
Consider the logistic equation with initial condition \(y(0)=y_{0}\). Assuming \(y_{0}
View solution Problem 48
Perform the indicated integrations. $$ \int \frac{1}{x^{2}-4 x+9} d x $$
View solution Problem 49
Perform the indicated integrations. $$ \int \frac{d x}{9 x^{2}+18 x+10} $$
View solution Problem 50
Find \(c\) so that \(\int_{-c}^{c} \frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} d x=0.95 .\) Hint: Use symmetry.
View solution