Question

Let \(X\) be a random variable with the probability density function

Hint:

For \(Z=|X-a|\), convert the event \(Z\leq m\) into \(a-m\leq X\leq a+m\).

Answer 1

Correct Answer: 4

Step 1: Shift to centre at $1$.
With $u=x-1$, the density $\frac34 x(2-x)$ becomes $\frac34(1-u^2)$, symmetric about $u=0$.

Step 2: Median condition.
$P(|X-1|\le m)=\frac12$ turns into $\int_{-m}^m\frac34(1-u^2)du=\frac12$.

Step 3: Integrate.
This equals $\frac32 m-\frac12 m^3=\frac12$, so $3m-m^3=1$.

Step 4: Build the wanted value.
$12m-4m^3=4(3m-m^3)=4\cdot1=4$.

Step 5: Conclude.
\[ \boxed{4} \]