Answer to Question #207029 in Statistics and Probability for Tabraiz

Question

The lifetime T of a device is a Rayleigh random variable. (a) Find the reliability of the device. (b) Find the failure rate function. Does r(t) increase with time? (c) Find the reliability of two devices that are in series. (d) Find the reliability of two devices that are in parallel.

Answer 1

The lifetime of a device T is a random variable that follows Rayleigh distribution.

T ~Ray(α)

The pdf of T is

$f_T(t)= \frac{t}{α^2}e^{-t^2/2a^2}$

here,

t≥0 and α>0

a) Reliability of the device

The reliability is defined as follows:

$R(t) = P[T>t] \\ P[T>t] = \int^{∞}_{t}\frac{t}{α^2}e^{-t^2/2α ^2}dt \\ = \frac{1}{α^2}\int^{∞}_{t}e^{-t^2/2α ^2}dt$

Let,

$-\frac{t}{2α^2}=u \\ -dt=2 α^2du$

Using integration by parts

$P[T>t]=e^{-t^2/2α^2}$

Therefore, the reliability device of the lifetime T is defined as

$R[t]=e^{-t^2/2α^2}$

b) Failure rate function

The failure rate function of a device is defined as

$r(t) = \frac{f_T(t)}{R(t)} \\ = \frac{\frac{t}{α^2}e^{-t^2/2α^2}}{e^{-t^2/2α^2}}\\ = \frac{t}{{α^2}}$

Therefore, the failure rate function of a device is

$r(t) =\frac{t}{{α^2}}$

Yes, as the time increases the reliability increases with it.

If t → ∞, r(t) → ∞ as well.

c) Reliability of two devices that are in series

The devices are in series so the reliability of the two would be multiplied.

R(t) = R_1(t) R_2(T)

Since one are given only one reliability function so, one uses the same function for both R₁(t) and R₂(t)

$R_{series}= e^{-t^2/2α^2} \times e^{-t^2/2α^2} \\ = e^{(-t^2/2α^2)+(-t^2/2α^2)} \\ = e^{-t^2/α^2}$

d) Reliability of two devices that are in parallel

The devices are in parallel so the reliability of the two is

$1-R(t) = (1-R_1(t))(1-R_2(t)) \\ R(t) = 1-(1-R_1(t))(1-R_2(t))$

Since one is given only one reliability function so, one uses the same function for both R₁(t) and R₂(t)

$R_{parallel}=1-(1-e^{-t^2/2α^2}) \times (1-e^{-t^2/2α^2}) \\ = 1-(1-e^{-t^2/2α^2})^2 \\ = 1-[1+(e^{-t^2/2α^2})^2-2e^{-t^2/2α^2}] \\ = 2e^{-t^2/2α^2} -e^{-t^2/2α^2} \\ =e^{-t^2/2α^2}(2e^{t^2/2α^2}-1)$