How Does Gamma Distribution Calculate High Income Probabilities?

[d2j2003]

In a certain country, the distribution of incomes in thousands of dollars is described by a gamma distribution with [itex]\alpha[/itex] = 2 and [itex]\beta[/itex] = 8. What is the probability that a man chosen at random will have the following incomes?

a. More than $14,000
b. At least $12,000

So I know that f(x;[itex]\alpha[/itex],[itex]\beta[/itex]) = [itex]\frac{1}{\Gamma(\alpha)\beta^{\alpha}}[/itex]x[itex]^{\alpha-1}[/itex]e[itex]^{-x/\beta}[/itex] for x>0,[itex]\alpha[/itex],[itex]\beta[/itex]>0

and [itex]\Gamma[/itex]([itex]\alpha[/itex])=[itex]\int[/itex][itex]^{\infty}_{0}[/itex]xe[itex]^{-x}[/itex] = 1

so f(x) = [itex]\frac{1}{64}[/itex] xe[itex]^{-x/8}[/itex]

Not sure where to go from here though.. Hopefully I'm heading in the right direction..

 

[mmwave]

Hello there,

Thank you for your post. It seems like you have a good understanding of the gamma distribution and its parameters. To answer the two questions, we can use the cumulative distribution function (CDF) of the gamma distribution.

a. To find the probability that a man chosen at random will have an income of more than $14,000, we need to find P(X > 14). This can be calculated by using the CDF, which is given by:

F(x) = \int_{0}^{x} f(t) dt = \frac{\gamma(\alpha, x/\beta)}{\Gamma(\alpha)}

where \gamma(\alpha, x/\beta) is the lower incomplete gamma function.

Substituting the values of \alpha = 2, \beta = 8, and x = 14, we get:

F(14) = \frac{\gamma(2, 14/8)}{\Gamma(2)} = \frac{\gamma(2, 1.75)}{1} = 0.578

Therefore, the probability that a man chosen at random will have an income of more than $14,000 is 0.578 or 57.8%.

b. To find the probability that a man chosen at random will have an income of at least $12,000, we need to find P(X \geq 12). This can be calculated by using the CDF, which is given by:

F(x) = \int_{0}^{x} f(t) dt = \frac{\gamma(\alpha, x/\beta)}{\Gamma(\alpha)}

Substituting the values of \alpha = 2, \beta = 8, and x = 12, we get:

F(12) = \frac{\gamma(2, 12/8)}{\Gamma(2)} = \frac{\gamma(2, 1.5)}{1} = 0.483

Therefore, the probability that a man chosen at random will have an income of at least $12,000 is 0.483 or 48.3%.

I hope this helps answer your questions. Let me know if you have any further questions or need clarification. Good luck!