Final answer:
To find the shape and scale parameters of a Gamma distribution from its mean and variance, use the alpha (shape) and beta (scale) formulas. For the given mean of 6 and variance of 18, alpha is 2 and beta is 3. The pdf of X is f(x) = (x e^{-x/3})/9 for x > 0.
Step-by-step explanation:
To find the shape (\(\alpha\)) and scale (\(\beta\)) parameters for the Gamma distribution given the mean (\(\mu\)) and variance (\(\sigma^2\)), we use the following relationships:
- \(\alpha = \frac{\mu^2}{\sigma^2}\)
- \(\beta = \frac{\sigma^2}{\mu}\)
Given \(\mu = 6\) and \(\sigma^2 = 18\), we calculate:
- \(\alpha = \frac{6^2}{18} = \frac{36}{18} = 2\)
- \(\beta = \frac{18}{6} = 3\)
Therefore, the shape parameter \(\alpha\) is 2 and the scale parameter \(\beta\) is 3. The probability density function (pdf) of X, f(x), for the Gamma distribution is given by:
\(f(x) = \frac{x^{\alpha - 1}e^{-x/\beta}}{\beta^{\alpha}\Gamma(\alpha)}\), for \(x > 0\)
Substituting the calculated values of \(\alpha\) and \(\beta\), the pdf becomes:
\(f(x) = \frac{x^{2 - 1}e^{-x/3}}{3^{2}\Gamma(2)}\), for \(x > 0\)
Since \(\Gamma(2) = 1! = 1\), the final pdf expression simplifies to:
\(f(x) = \frac{x e^{-x/3}}{9}\), for \(x > 0\)