Question

The lifetime of a device (in hours) has the Gam(4,0.01) distribution. a) Find the probability that the device will last more than 300 hurs. b) what is the 90th percentile of this distribution?

Answer 1

Here,

X ~ Gamma(4, 0.01)

The probability density function of X is

$f(x) = \frac{0.01^4}{\Gamma (4)}e^{-0.01x}*x^{4-1},x>0$

a) The probability that the device will last more than 300 hours

$=P(X>300) = 1 - 0.352768 = 0.647232$

b) To find the 90th percentile of this distribution, i.e. x such that

$P(X\leq x) = 0.90$

$\Rightarrow x = 668.078$ (using inverse cumulative function of gamma distribution in minitab software)