Moderate -0.3 This is a straightforward application of finding the median from a PDF by integrating to find where the cumulative probability equals 0.5. The geometric shape (appears to be a simple linear or piecewise function) makes the integration routine, requiring only basic calculus techniques typical of S2 level with no conceptual challenges beyond the standard definition of median.
2
\includegraphics[max width=\textwidth, alt={}, center]{43b2498f-73e2-4d33-adaf-fc3e460fa36a-2_358_1093_495_520}
A random variable \(X\) takes values between 0 and 4 only and has probability density function as shown in the diagram. Calculate the median of \(X\).
\(\frac{1}{2} \times m \times \left(\frac{m}{4} \times \frac{1}{2}\right) = \frac{1}{2}\)
M1
\(\frac{1}{2} \times m \times (\frac{1}{8}m) = \frac{1}{2}\) or \(\frac{m^2}{16} = \frac{1}{2}\) o.e. N.B. B1 M1 must be consistent; or integrating linear function of form \(y=kx\) with limits 0 and \(m\) or \(m\) and 4 and equated to 0.5
\(m = \sqrt{8}\) or \(2\sqrt{2}\) or 2.83 (3 s.f.)
A1 [3]
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $ht = \frac{1}{2}$ seen | B1 | or $y = \frac{1}{8}x$ |
| $\frac{1}{2} \times m \times \left(\frac{m}{4} \times \frac{1}{2}\right) = \frac{1}{2}$ | M1 | $\frac{1}{2} \times m \times (\frac{1}{8}m) = \frac{1}{2}$ or $\frac{m^2}{16} = \frac{1}{2}$ o.e. N.B. B1 M1 must be consistent; or integrating linear function of form $y=kx$ with limits 0 and $m$ or $m$ and 4 and equated to 0.5 |
| $m = \sqrt{8}$ or $2\sqrt{2}$ or 2.83 (3 s.f.) | A1 [3] | |
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2\\
\includegraphics[max width=\textwidth, alt={}, center]{43b2498f-73e2-4d33-adaf-fc3e460fa36a-2_358_1093_495_520}
A random variable $X$ takes values between 0 and 4 only and has probability density function as shown in the diagram. Calculate the median of $X$.
\hfill \mbox{\textit{CAIE S2 2014 Q2 [3]}}