Mathematics 9709/71 — October/November 2012

Walkthrough

The median of a continuous random variable $X$ with PDF $f(x)$ is the value $m$ such that the probability $P(X \leq m) = 0.5$. This means the area under the PDF from the lower bound of the support up to $m$ must equal $\frac{1}{2}$.

Step 1: Determine the equation of $f(x)$ from the graph.

The graph shows that $f(x) = 0$ outside the interval $[0, 2]$. Inside $[0, 2]$, the PDF is a straight line from $(0, 0)$ to $(2, 1)$. The gradient is $\frac{1-0}{2-0} = \frac{1}{2}$, and since the line passes through the origin, the equation is $f(x) = \frac{1}{2}x$.

We can verify this is a valid PDF by checking the total area: $\int_0^2 \frac{1}{2}x \, dx = \left[\frac{x^2}{4}\right]_0^2 = \frac{4}{4} = 1$. ✓

Step 2: Set up the median equation.

Since $f(x) = 0$ for $x < 0$, the median equation simplifies to:

Step 3: Evaluate the integral.

Step 4: Solve for $m$.

Setting $\frac{m^2}{4} = \frac{1}{2}$ gives $m^2 = 2$, so $m = \sqrt{2} \approx 1.41$ (taking the positive root since $m \geq 0$).

Key Takeaways

• The median of a continuous random variable is found by solving $\int_{-\infty}^m f(x)\,dx = \frac{1}{2}$.
• When the PDF is zero over part of the range, the integral limits can be adjusted accordingly.
• Reading a linear PDF from a graph requires finding the equation of the line (gradient and intercept).
• Always verify the solution lies within the support of the PDF.

Common Mistakes

• Forgetting to set the integral equal to $\frac{1}{2}$: Some students integrate and forget the right-hand side should be $0.5$, not $1$.
• Using $\pm\sqrt{2}$ as the final answer: The median must be positive (and within $[0, 2]$), so only the positive root is valid. The mark scheme explicitly awards A0 for $\pm\sqrt{2}$.
• Incorrect PDF equation: Misreading the gradient or intercept of the line from the graph.
• Not checking that $m$ is in the valid range: The median must satisfy $0 \leq m \leq 2$.

Things to Be Careful About

• The PDF is zero for $x < 0$, so the lower limit of integration is $0$, not $-\infty$.
• The answer $m = -\sqrt{2}$ is mathematically valid for $m^2 = 2$ but physically meaningless here since $m$ must be in $[0, 2]$. Always state the valid root.
• When writing the final answer, give $\sqrt{2}$ to 3 significant figures as $1.41$ if required by the mark scheme.
• Verify that the total area under the PDF equals 1 as a sanity check before proceeding.