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: Write the median condition.
Since \(m\) is the median of \(Z=|X-1|\), we have
\[ P(Z\leq m)=\frac12 \] That is,
\[ P(|X-1|\leq m)=\frac12 \] \[ P(1-m\leq X\leq 1+m)=\frac12 \]

Step 2: Transform the density around \(x=1\).
Let
\[ u=x-1 \] Then
\[ x=1+u \] and
\[ x(2-x)=(1+u)(1-u)=1-u^2 \] So,
\[ f(x)=\frac34(1-u^2) \]

Step 3: Compute \(P(|X-1|\leq m)\).
\[ P(|X-1|\leq m) = \int_{1-m}^{1+m}\frac34 x(2-x)\,dx \] Using \(u=x-1\), the limits become \(-m\) to \(m\). Hence,
\[ P(|X-1|\leq m) = \int_{-m}^{m}\frac34(1-u^2)\,du \] \[ = \frac34\left[u-\frac{u^3}{3}\right]_{-m}^{m} \] \[ = \frac34\left(2m-\frac{2m^3}{3}\right) \] \[ = \frac32m-\frac12m^3 \]

Step 4: Use the median condition.
\[ \frac32m-\frac12m^3=\frac12 \] Multiplying by \(2\),
\[ 3m-m^3=1 \]

Step 5: Find the required value.
We need
\[ 12m-4m^3 \] Taking \(4\) common,
\[ 12m-4m^3=4(3m-m^3) \] Since
\[ 3m-m^3=1, \] we get
\[ 12m-4m^3=4 \]

Step 6: Final conclusion.
Hence,
\[ \boxed{4} \]