SOLUTION: The time X in months until failure of a certain product has a probability density function of https://i.gyazo.com/8c054188451da18fcdc1c3919fea6bf2.png
for x > 0 and 0 otherwise. F
Algebra.Com
Question 1136622: The time X in months until failure of a certain product has a probability density function of https://i.gyazo.com/8c054188451da18fcdc1c3919fea6bf2.png
for x > 0 and 0 otherwise. Find the cumulative distribution and the median
value of X.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'll break this into two parts. Part A will show how to find the cumulative distribution function (CDF) while part B shows how to find the median. Part B will use the result from part A.
========================================================================================================
Part A
To find the CDF, we apply the integral to the probability distribution function (PDF) like so
 = \frac{3x^2}{64}e^{-\left(\frac{x}{4}\right)^3})
dx = \int \frac{3x^2}{64}e^{-\left(\frac{x}{4}\right)^3}dx)
dx = \frac{1}{64}\int 3x^2*e^{-\left(\frac{x}{4}\right)^3}dx)
dx = \frac{1}{64}\int 3x^2dx*e^{-\left(\frac{x}{4}\right)^3})
From here, we apply u-substitution.
Let u = (-x/4)^3, so
du = -1*3(1/4)*(-x/4)^2dx
du = -(3/4)*(-x/4)^2dx
du = -(3/4)*((x^2)/16)dx
du = (-3x^2dx)/64
64du = -3x^2*dx
3x^2*dx = -64du
Replace the (-x/4)^3 exponent with 'u'; replace the 3x^2dx with -64du
dx = \frac{1}{64}\int -64du*e^{u})
dx = \frac{-64}{64}\int e^{u}du)
dx = -\int e^{u}du)
dx = -e^{u}+C)
dx = -e^{-\left(\frac{x}{4}\right)^3}+C)
------------
Let g(x) = -e^(-(x/4)^3)+C
The area under the f(x) curve is exactly equal to g(k) - g(0) where k approaches positive infinity
In other words,
dx = g(k) - g(0))
dx = \left(-e^{-\left(\frac{k}{4}\right)^3}+C\right)-\left(-e^{-\left(\frac{0}{4}\right)^3}+C\right))
dx = -e^{-\left(\frac{k}{4}\right)^3}+C+e^{-\left(\frac{0}{4}\right)^3}-C)
dx = -e^{-\left(\frac{k}{4}\right)^3}+e^{-\left(\frac{0}{4}\right)^3})
dx = -e^{-\left(\frac{k}{4}\right)^3}+e^{0^3})
dx = -e^{-\left(\frac{k}{4}\right)^3}+e^{0})
dx = -e^{-\left(\frac{k}{4}\right)^3}+1)
------------
Let's see what happens when k approaches positive infinity
dx\right] = \lim_{k\to\infty}\left[-e^{-\left(\frac{k}{4}\right)^3}+1\right])
dx\right] = \lim_{k\to\infty}\left[-e^{\frac{-k^3}{64}}+1\right])
dx\right] = \lim_{k\to\infty}\left[-\frac{1}{e^{\frac{k^3}{64}}}+1\right])
dx\right] = 0+1)
dx\right] = 1)
So we get an area of 1 as expected, indicating that  = -e^{-\left(\frac{x}{4}\right)^3}+1)
is the proper CDF for the PDF
note: g(k) = P(x < k) meaning that the CDF returns the area under the f(x) curve to the left of the x = k value.
========================================================================================================
Part B
Set the CDF function equal to 0.5 and solve for x. This will yield the median value (as the median is at the halfway point)
Put another way, the area to the left of the median value is exactly 0.5
^3}+1 = 0.5)
^3} = 0.5-1)
^3} = -0.5)
^3} = 0.5)
^3 = \ln(0.5))
^3 = -\ln(0.5))
})
})
})
\right)^{1/3})
The median is approximately x = 3.53998817800208
========================================================================================================
In summary,
The CDF is which is equivalent to  = -e^{-\frac{x^3}{64}}+1)
.
The median is approximately x = 3.53998817800208
Round this however you need to.
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