The lifetime (In Years) of Manufactured product is Y=5X^0.7, where X has an exponential distribution with mean 1. Find the distribution function of PDF Y.
To find the distribution function of the lifetime of the manufactured product (Y), we need to determine the probability density function (PDF) of Y first.
Given that X has an exponential distribution with mean 1, we know that the PDF of X is given by:
fX(x) = e^(-x) for x >= 0
To find the PDF of Y, we need to use the transformation of random variables. The transformation is given by:
Y = 5X^0.7
We will find the cumulative distribution function (CDF) of Y, denoted F_Y(y), which represents the probability that Y is less than or equal to y.
F_Y(y) = P(Y <= y)
To find this probability, we need to find the distribution function of Y, denoted F_Y(y), which is defined as:
F_Y(y) = P(Y <= y) = P(5X^0.7 <= y)
From this equation, we can solve for X in terms of Y:
X = (y/5)^(1/0.7)
Now, we can express the probability in terms of X:
F_Y(y) = P(5X^0.7 <= y) = P(X <= (y/5)^(1/0.7)) = ∫[0, (y/5)^(1/0.7)] fX(x) dx
We will now substitute the PDF of X into the integral:
F_Y(y) = ∫[0, (y/5)^(1/0.7)] e^(-x) dx
Note that (y/5)^(1/0.7) represents the upper limit of integration.
To evaluate this integral, we can integrate by substitution:
Let u = -x
Then du = -dx
Substituting these values:
F_Y(y) = - ∫[(y/5)^(1/0.7), 0] e^u du
= - ∫[0, (y/5)^(1/0.7)] e^u du
= - [e^u] evaluated from 0 to (y/5)^(1/0.7)
= - (e^(0) - e^(-(y/5)^(1/0.7)))
= - (1 - e^(-(y/5)^(1/0.7)))
Therefore, the distribution function (CDF) of Y is:
F_Y(y) = 1 - e^(-(y/5)^(1/0.7))
And the probability density function (PDF) of Y can be obtained by differentiating F_Y(y) with respect to y:
f_Y(y) = d/dy (F_Y(y))
= d/dy (1 - e^(-(y/5)^(1/0.7)))
Differentiating with respect to y:
f_Y(y) = 0 + [(1/0.7) * (y/5)^(1/0.7 - 1) * e^(-(y/5)^(1/0.7))]
= (1/0.7) * (1/5)^(1/0.7) * (y/5)^(-0.3) * e^(-(y/5)^(1/0.7))
Therefore, the distribution function (PDF) of Y is:
f_Y(y) = (1/0.7) * (1/5)^(1/0.7) * (y/5)^(-0.3) * e^(-(y/5)^(1/0.7))
To find the distribution function of the PDF (Probability Density Function) of Y, we first need to determine the distribution function of X.
Given that X has an exponential distribution with mean 1, the probability density function (PDF) of X is given by:
f(x) = λe^(-λx)
Where λ is the rate parameter, which is equal to 1/mean. In this case, λ = 1/1 = 1.
From the given formula, we have Y = 5X^0.7. To find the distribution function of Y, we need to find the cumulative distribution function (CDF) of Y and then differentiate it to obtain the PDF.
Let's proceed step by step:
Step 1: Find the CDF of Y:
F_Y(y) = P(Y ≤ y) = P(5X^0.7 ≤ y)
Step 2: To isolate X, we need to raise both sides of the equation (5X^0.7 ≤ y) to the power of (1/0.7):
(5X^0.7)^(1/0.7) ≤ y^(1/0.7)
This simplifies to:
5X ≤ y^(1/0.7)
Step 3: Divide both sides of the inequality by 5:
X ≤ (y^(1/0.7))/5
Now, we can use the cumulative distribution function (CDF) of X, which is defined as:
F_X(x) = ∫[0,x] f(t) dt
Since the PDF of X is exponential with rate parameter λ = 1, we have:
f(x) = e^(-x), for x ≥ 0
Step 4: Substitute the upper limit of integration (x) with the expression for X derived in step 3:
F_Y(y) = P(Y ≤ y) = P(X ≤ (y^(1/0.7))/5) = F_X((y^(1/0.7))/5)
Step 5: Evaluate the CDF of Y by substituting the exponential distribution's CDF with the given value:
F_Y(y) = 1 - e^(-(y^(1/0.7))/5), for y ≥ 0
Finally, to obtain the PDF (probability density function) of Y, we differentiate the CDF:
f_Y(y) = d/dy (F_Y(y))
= d/dy (1 - e^(-(y^(1/0.7))/5))
= (1/0.7) * (1/5) * e^(-(y^(1/0.7))/5) * (y^(-0.3/0.7))
Therefore, the distribution function of the PDF of Y is:
f_Y(y) = (1/0.7) * (1/5) * e^(-(y^(1/0.7))/5) * (y^(-0.3/0.7))