Question

Suppose that two electronic components in the guidance system for a missile operate independently and that each has a length of life governed by the exponential distribution with mean 3 (with measurements in hundreds of hours). Find the probability density function for the average length of life of the two components.

Answer 1

Answer:

attached below

Step-by-step explanation:

Assuming that Y1 and Y2 are independent exponential random variable with mean 3

where : U = Y1 + Y2 applying the property of exponential distribution sum of exponentials will follow a gamma distribution

i.e. ∝ = 2 , and β = 1

note : U = 2X

Hence probability distribution function X =

Fx (x) = p( X ≤ x )

= p( U ≤ 2x )

= Fu (2x)

probability density function  X

f (x) = $\frac{dF(2x)}{dx}$  = 2 Fu (2x)

attached below is the remaining part of the solution

Answer 2

In this exercise we have to use probability knowledge to calculate the density of the function, so we have to:

$f(u) \left \{ {{ue^u: u>0} \atop {0}} \right.$

$f(x) \left \{ {{4xe^{-2x}: x>0} \atop {0}} \right.$

Assuming that $Y_1$ and $Y_2$ are independent exponential random variable where:

$U = Y_1 + Y_2$

Applying the property of exponential distribution sum of exponentials will follow a gamma distribution, we have:

• $\alpha = 2$

• $\beta = 1$

• $U = 2X$

Hence probability distribution function X , will be:

$F_x (x) = p( X \leq x )\\= p( U \leq 2x )\\= F_u (2x)$

So we can say that:  

$f (x) = \frac{dF}{du} = 2 F_u (2x)$

See more about probability at brainly.com/question/795909