Let *X* be a continuous random variable with PDF
f_X(x)= \begin{cases} kx^2, & 0\leq x \leq 1\\ 0, & \text{otherwise.} \end{cases}*fX*​(*x*)={*kx*2,0,​0≤*x*≤1otherwise.​
Find the median of *X*. Enter the answer correct to four decimal places.

**Hint**: The median *m* is the value such that P(X \leq m) = P(X > m) = 0.5*P*(*X*≤*m*)=*P*(*X*>*m*)=0.5

We are given a probability density function (PDF) for a continuous random variable \( X \) as:

\[
f_X(x) = 
\begin{cases} 
kx^2, & 0 \leq x \leq 1, \\
0, & \text{otherwise.} 
\end{cases}
\]

### Step 1: Determine the constant \( k \)
To find \( k \), we use the fact that the total area under the PDF must equal 1, since the total probability must sum to 1. This means we need to solve the equation:

\[
\int_0^1 f_X(x) \, dx = 1.
\]

Substituting \( f_X(x) = kx^2 \):

\[
\int_0^1 kx^2 \, dx = 1.
\]

Now, calculate the integral:

\[
k \int_0^1 x^2 \, dx = k \left[ \frac{x^3}{3} \right]_0^1 = k \times \frac{1^3}{3} = \frac{k}{3}.
\]

So we have:

\[
\frac{k}{3} = 1.
\]

Solving for \( k \):

\[
k = 3.
\]

Thus, the PDF becomes:

\[
f_X(x) = 
\begin{cases} 
3x^2, & 0 \leq x \leq 1, \\
0, & \text{otherwise.} 
\end{cases}
\]

### Step 2: Find the median \( m \)
The median \( m \) is defined as the value such that:

\[
P(X \leq m) = 0.5.
\]

This is equivalent to finding \( m \) such that the cumulative distribution function (CDF) up to \( m \) equals 0.5:

\[
\int_0^m 3x^2 \, dx = 0.5.
\]

First, calculate the integral:

\[
3 \int_0^m x^2 \, dx = 3 \left[ \frac{x^3}{3} \right]_0^m = m^3.
\]

Now, set this equal to 0.5:

\[
m^3 = 0.5.
\]

Solving for \( m \):

\[
m = \sqrt[3]{0.5} \approx 0.7937.
\]

### Final Answer:
The median of \( X \) is approximately \( \boxed{0.7937} \).

Would you like further details or have any questions?

Here are 5 related questions:
1. How do you compute the expectation \( E[X] \) for this distribution?
2. What is the variance of \( X \) for the given PDF?
3. How do you find the mode of this distribution?
4. Can you derive the CDF \( F_X(x) \) from the PDF?
5. What is the probability that \( X \) is greater than the median?

**Tip:** Always ensure that the area under a probability density function equals 1, as this is a key property of PDFs.

Let X be a continuous random variable with PDF f_X(x) = {kx^2, 0≤x≤1; 0 otherwise}. Find the median of X. Enter the answer correct to four decimal places.

Learn how to find the median of a continuous random variable with the given probability density function (PDF) f_X(x) = 3x^2 for 0 ≤ x ≤ 1. This solution includes step-by-step calculations using the CDF and properties of continuous random variables. Suitable for undergraduate-level probability and statistics students.

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