Let X be a continuous random variable with PDF
[tex]f_x(x) = \begin{cases} kx^2, & 0 \le x \le 1 \\ 0, & \text{otherwise.} \end{cases}[/tex]
Find the median of X. Enter the answer correct to four decimal places.
Hint: The median m is the value such that P(X \le m) = P(X > m) = 0.5
Answer (2)
The median of the continuous random variable X, defined by the PDF f X ( x ) = 3 x 2 for 0 ≤ x ≤ 1 , is approximately 0.7937 . This is found by solving the integral of the PDF to equal 0.5. The solution involves normalizing the PDF first and then calculating the median based on the defined conditions.
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To find the median of a continuous random variable with a given probability density function (PDF), we need to find a value m such that the cumulative probability up to m is 0.5. In other words, we need to solve the equation P ( X ≤ m ) = 0.5.
For the given PDF:
f X ( x ) = { k x 2 , 0 , 0 ≤ x ≤ 1 otherwise.
First, we need to find k by ensuring the total probability is 1.
∫ 0 1 k x 2 d x = 1.
Evaluating the integral,
k ∫ 0 1 x 2 d x = k [ 3 x 3 ] 0 1 = k ( 3 1 ) = 1.
Thus, k = 3.
Now, the PDF is:
f X ( x ) = { 3 x 2 , 0 , 0 ≤ x ≤ 1 otherwise.
To find the median m , we set up the equation:
∫ 0 m 3 x 2 d x = 0.5.
Evaluate the integral,
3 [ 3 x 3 ] 0 m = [ x 3 ] 0 m = m 3 .
We need m 3 = 0.5 .
Solving for m ,
m = 3 0.5 .
Calculating this,
m ≈ 0.7937.
Therefore, the median m of the random variable X is approximately 0.7937 when rounded to four decimal places.
