Question

Two programmers started working on their computer programs at 9:00 am. The first programmer will take a Uniform(1hr,4hrs) time to finish the program.The second programmer will need a gamma(5hrs,1hr) time. Which programmer has a better chance to finish before 11:00am?

Answer 1

We need to compute $P(X<2)$ and $P(Y<2)$ for a $Uniform(a=1\ hr, b=4\ hrs)$ variable $X$ and a $Gamma(r=5, \lambda=1\ hrs^{-1})$ variable $Y$ and compare them.

For the Uniform distribution with density $f(x)=\dfrac{1}{4-1}=\dfrac{1}{3}$ for $1<x<4,$

$$P(X<2)=\displaystyle\int_{-\infin}^{2}f(x)dx=\displaystyle\int_{1}^{2}\dfrac{1}{4-1}dx$$

$$=\big[\dfrac{1}{3}x\big]\begin{matrix} 2 \\ 1 \end{matrix}=\dfrac{1}{3}\approx0.33333$$

http://www.math.ntu.edu.tw/~hchen/teaching/StatInference/notes/lecture17.pdf

Gamma-Poisson relationship

There is a relationship between the gamma and Poisson distributions. If $Y$ is a $Gamma(r, \lambda)$ random variable, where $\alpha$ is an integer, then for any $y,$

$$P(Y\leq y)=P(Z\geq r),$$

where $Z\sim Po(y/\lambda).$

For the second programmer, use the Gamma-Poisson formula with a $Gamma(r=5, \lambda=1)$ variable $Y$ and a Poisson $( y/\lambda=2/1=2)$ variable $Z,$

$$P(Y<2)=P(Z\geq r)=P(Z\geq5)\approx 0.05265$$

from the Poisson distribution table with parameter 2.

Thus, the first programmer has a better chance to finish before 11:00 am.