Question

Suppose X has an exponential distribution with mean equal to 23. Determine the following:

Answer 1

Answer:

a) P(X > 10) = 0.6473

b) P(X > 20) = 0.4190

c) P(X < 30) = 0.7288

d) x = 68.87

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean equal to 23.

This means that $m = 23, \mu = \frac{1}{23} = 0.0435$

(a) P(X >10)

$P(X > 10) = e^{-0.0435*10} = 0.6473$

So

P(X > 10) = 0.6473

(b) P(X >20)

$P(X > 20) = e^{-0.0435*20} = 0.4190$

So

P(X > 20) = 0.4190

(c) P(X <30)

$P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288$

So

P(X < 30) = 0.7288

(d) Find the value of x such that P(X > x) = 0.05

So

$P(X > x) = e^{-\mu x}$

$0.05 = e^{-0.0435x}$

$\ln{e^{-0.0435x}} = \ln{0.05}$

$-0.0435x = \ln{0.05}$

$x = -\frac{\ln{0.05}}{0.0435}$

$x = 68.87$