| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | March |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.8 This is a straightforward probability distribution construction requiring a systematic enumeration of outcomes (4×4=16 equally likely cases), followed by standard variance calculation using given E(X). The mechanics are routine for S1 level with no conceptual challenges beyond organizing the sample space. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table with \(X\) values: \(-2, -1, 0, 1, 2\) and \(P(X)\) values: \(\frac{1}{16}, \frac{3}{16}, \frac{5}{16}, \frac{5}{16}, \frac{2}{16}\) (equivalently \(0.0625, 0.1875, 0.3125, 0.3125, 0.125\)) | B1 | Table with correct \(X\) values and at least one probability \(0 < p < 1\). Condone any additional \(X\) values if probability stated as 0. No repeated \(X\) values. |
| 3 correct probabilities linked with correct outcomes | B1 | May not be in table |
| 2 further correct probabilities linked with correct outcomes | B1 | No repeated \(X\) values. SC if less than 3 correct probabilities seen, award SCB1 Sum of *their* probabilities, \(0 < p < 1\), of 4, 5 or 6 \(X\) values \(= 1\) (condone summing to \(1 \pm 0.01\) or better). |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\frac{1}{16}\times(-2)^2 + \frac{3}{16}\times(-1)^2 + \frac{5}{16}\times 0^2 + \frac{5}{16}\times 1^2 + \frac{2}{16}\times 2^2\right] - \left(\frac{1}{4}\right)^2\) | M1 | Appropriate variance formula using \((E(X))^2\) value, accept unsimplified. FT *their* table with at least 3 different \(X\) values even if probabilities not summing to 1, \(0 < p < 1\). Condone 1 error providing all probabilities \(<1\) and \(0.25^2\) used |
| \(\frac{1\times4 + 3\times1 + 5\times0 + 5\times1 + 2\times4}{16} - 0.25^2 = \frac{5}{4} - \frac{1}{16} = \frac{19}{16},\ 1.1875\) | A1 | Condone \(1.188\) or \(1.19\) WWW |
| [2] |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table with $X$ values: $-2, -1, 0, 1, 2$ and $P(X)$ values: $\frac{1}{16}, \frac{3}{16}, \frac{5}{16}, \frac{5}{16}, \frac{2}{16}$ (equivalently $0.0625, 0.1875, 0.3125, 0.3125, 0.125$) | **B1** | Table with correct $X$ values and at least one probability $0 < p < 1$. Condone any additional $X$ values if probability stated as 0. No repeated $X$ values. |
| 3 correct probabilities linked with correct outcomes | **B1** | May not be in table |
| 2 further correct probabilities linked with correct outcomes | **B1** | No repeated $X$ values. **SC** if less than 3 correct probabilities seen, award **SCB1** Sum of *their* probabilities, $0 < p < 1$, of 4, 5 or 6 $X$ values $= 1$ (condone summing to $1 \pm 0.01$ or better). |
| | **[3]** | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\frac{1}{16}\times(-2)^2 + \frac{3}{16}\times(-1)^2 + \frac{5}{16}\times 0^2 + \frac{5}{16}\times 1^2 + \frac{2}{16}\times 2^2\right] - \left(\frac{1}{4}\right)^2$ | **M1** | Appropriate variance formula using $(E(X))^2$ value, accept unsimplified. FT *their* table with at least 3 different $X$ values even if probabilities not summing to 1, $0 < p < 1$. Condone 1 error providing all probabilities $<1$ and $0.25^2$ used |
| $\frac{1\times4 + 3\times1 + 5\times0 + 5\times1 + 2\times4}{16} - 0.25^2 = \frac{5}{4} - \frac{1}{16} = \frac{19}{16},\ 1.1875$ | **A1** | Condone $1.188$ or $1.19$ WWW |
| | **[2]** | |1 A fair red spinner has edges numbered $1,2,2,3$. A fair blue spinner has edges numbered $- 3 , - 2 , - 1 , - 1$. Each spinner is spun once and the number on the edge on which each spinner lands is noted. The random variable $X$ denotes the sum of the resulting two numbers.
\begin{enumerate}[label=(\alph*)]
\item Draw up the probability distribution table for $X$.
\item Given that $\mathrm { E } ( X ) = 0.25$, find the value of $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2022 Q1 [5]}}