| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Discrete CDF to PMF |
| Difficulty | Easy -1.2 This is a straightforward S1 question requiring only direct recall of the relationship between CDF and PMF. Students simply subtract consecutive F(x) values to find P(X=x), which is a mechanical procedure with no problem-solving or conceptual challenge beyond basic definition knowledge. |
| Spec | 2.04a Discrete probability distributions |
| \(x\) | 1 | 2 | 3 |
| \(\mathrm {~F} ( x )\) | 0.4 | 0.65 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(x\) | 1 |
| P(X = x) | 0.4 | 0.25 |
| \(P(X = 2) = F(2) - F(1)\) (o.e.), \(P(X = 2) = \mathbf{0.25}\), \(P(X = 3) = \mathbf{0.35}\) | M1, A1, A1 | (3) |
| (b) \([F(1.8) = P(X < 1.8) = P(X \leq 1)] = \mathbf{0.4}\) | B1 | (1) [4] |
**(a)** | $x$ | 1 | 2 | 3 |
| P(X = x) | 0.4 | 0.25 | 0.35 |
$P(X = 2) = F(2) - F(1)$ (o.e.), $P(X = 2) = \mathbf{0.25}$, $P(X = 3) = \mathbf{0.35}$ | M1, A1, A1 | (3) | M1 for P(X = 1) = 0.4 and evidence of correct method for finding P(X = 2) or P(X = 3). Implied by correct ans.; 1st A1 for P(X = 2) = 0.25; 2nd A1 for P(X = 3) = 0.35
**(b)** $[F(1.8) = P(X < 1.8) = P(X \leq 1)] = \mathbf{0.4}$ | B1 | (1) [4] | B1 for 0.4
---2.The discrete random variable $X$ takes the values 1,2 and 3 and has cum\\
function $\mathrm { F } ( x )$ given by
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 \\
\hline
$\mathrm {~F} ( x )$ & 0.4 & 0.65 & 1 \\
\hline
\end{tabular}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{4cf4f2d7-d912-4b65-a666-caa37009661a-04_24_37_182_2010}\\
\hfill \mbox{\textit{Edexcel S1 2013 Q2 [4]}}