Question

Let \(X\) have a gamma distribution with pdf $$ f(x)=\frac{1}{\beta^{2}} x e^{-x / \beta}, \quad 0

Short answer

The parameter \(\beta\) is 2 and the probability \(P(X<9.49)\) is approximately 0.865.

Step-by-step solution

Find the Parameter \(\beta\)

To find \(\beta\), set the mode equal to \(2\) and solve for \(\beta\):\n\n\(2 = (2-1)\beta \Rightarrow \beta = 2.\)

Substituting the Value of \(\beta\) in pdf function

By using the value of \(\beta\), the pdf function transforms into: \n\n\(f(x) = \frac{1}{2^2} x e^{-x/2}\).

Find \(P(X

The cumulative distribution function is obtained by integrating the pdf from 0 to \(x\). Therefore, to find \(P(X<9.49)\), integrate the provided pdf from 0 to 9.49.\n\nThis can be represented mathematically as:\n\n\(P(X<9.49) = \int_0^{9.49} \frac{1}{4}x e^{-x/2} dx\).\n\nEvaluating this definite integral by using the method of integration by parts, it comes out to be approximately 0.86466.

Key concepts

Probability Density Function

The Probability Density Function, or PDF, is a fundamental concept in probability theory and statistics. For the gamma distribution, the PDF determines how probability is distributed over different values. It is expressed as a function. The PDF of the gamma distribution is given by \(f(x) = \frac{1}{\beta^2} x e^{-x/\beta}\) for \(0 < x < \infty\) and is zero elsewhere. Here, \(x\) represents a possible outcome, and \(\beta\) is a positive parameter that affects the shape of the distribution.

• \(x\) is the random variable associated with the gamma distribution.
• \(\beta\) is known as the scale parameter, controlling how "spread out" the distribution is.
• The exponent \(e^{-x/\beta}\) brings the exponential nature to the distribution, making it tail off as \(x\) increases.

To work with the gamma PDF, you first need to determine all necessary parameters, such as \(\beta\). Without the correct parameters, the PDF won't accurately represent the actual probability distribution.

Mode of Distribution

The mode of a probability distribution is another essential concept that refers to the most commonly occurring value or the "peak" of the distribution. For the gamma distribution, the mode is computed using the relation \(x = (k-1)\beta\), where \(k\) is a known parameter related to the shape of the distribution. In this case, the mode was given as \(x = 2\).

• The mode provides a measure of central tendency, showing where the data points cluster the most.
• In the problem above, setting the mode to \(2\) helped us solve for \(\beta\) with \(k = 2\), yielding \(\beta = 2\).

By finding \(\beta\), you can substitute back into the PDF to get a complete view of how the probability is distributed around this mode. This step is crucial for any further calculations, such as finding probabilities for different ranges.

Cumulative Distribution Function

The Cumulative Distribution Function, or CDF, of a distribution offers insight into the probability that a random variable will be less than or equal to a certain value. For the gamma distribution, the process to find the CDF involves integrating the PDF. This integration aggregates the probability from 0 up to the point of interest, providing us with a cumulated probability.

• The CDF is crucial when you want to know the likelihood of a variable falling beneath a specific threshold.
• In the exercise, finding \(P(X<9.49)\) meant computing the integral \(\int_0^{9.49} \frac{1}{4}x e^{-x/2} dx\), which was worked out to approximately 0.86466.

Integration here can be done by techniques such as integration by parts, especially when handling more complex expressions. Having the CDF makes it easier to calculate various probabilities and is widely used in statistics and real-world applications for making decisions based on probability.