| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2006 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | One unknown from sum constraint only |
| Difficulty | Easy -1.3 This is a straightforward application of basic probability distribution properties: part (i) uses the fact that probabilities sum to 1 (simple algebra), and part (ii) requires standard formula application for expectation and variance. Both are routine textbook exercises with no problem-solving or conceptual challenge beyond recall and calculation. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | 0.26 | \(q\) | \(3 q\) | 0.05 | 0.09 |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Equation with \(q\) in summing probs to 1 must be probs | |
| A1 | 2 marks | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| B1 ft, M1 | Correct final answer, ft on wrong \(q\). Subst in \(\sum x^2 - \text{mean}^2\) formula | |
| A1 | 3 marks | Correct final answer |
**(i)** $q + 3q + 0.26 + 0.05 + 0.09 = 1$
$q = 0.15$
| M1 | Equation with $q$ in summing probs to 1 must be probs |
| A1 | 2 marks | Correct answer |
**(ii)** $E(X) = 1.56$
$\text{Var}(X) = 0.15 + 1.8 + 0.45 + 1.44 - \text{mean}^2 = 1.41$
| B1 ft, M1 | Correct final answer, ft on wrong $q$. Subst in $\sum x^2 - \text{mean}^2$ formula |
| A1 | 3 marks | Correct final answer |
---2 The discrete random variable $X$ has the following probability distribution.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & 0.26 & $q$ & $3 q$ & 0.05 & 0.09 \\
\hline
\end{tabular}
\end{center}
(i) Find the value of $q$.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\hfill \mbox{\textit{CAIE S1 2006 Q2 [5]}}