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. The \(P(X<9.49)\) can be found by calculating the cumulative distribution function (CDF) for a Gamma distribution at \(x=9.49\) with \(k=2\) and \(\beta=2\), denoted as \(CDF_{\Gamma}(9.49, k=2, \beta=2)\).

Step-by-step solution

Find the parameter \(\beta\)

To find the \(\beta\) parameter, we can use the mode formula for the Gamma distribution. The formula is \(mode = (k - 1) * \beta\). The mode is given as \(x=2\). Since we know that the shape parameter \(k\) is always positive and greater than 1 in a Gamma distribution, we assume \(k=2\) here. That simplifies the mode formula to \(2=1*\beta\), therefore, the parameter \(\beta = 2\).

Find the cumulative distribution function (CDF)

In order to find \(P(X<9.49)\), it's necessary to calculate the cumulative distribution function (CDF) at \(x=9.49\). The CDF of a Gamma distribution for \(x > 0\) is given by the lower incomplete gamma function divided by the gamma function of the shape parameter \(k\). These calculations are typically done by specialized statistical software or a scientific calculator with gamma functions. For this exercise, we will denote this value by \(CDF_{\Gamma}(9.49, k=2, \beta =2)\).

Calculate \(P(X

Using the CDF calculated in step 2, we get that \(P(X<9.49) = CDF_{\Gamma}(9.49, k=2, \beta=2)\).