| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| 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 question requiring systematic enumeration of outcomes (4×3=12 equally likely cases), constructing a table, and applying the standard variance formula. It's below average difficulty as it involves only routine probability calculations with small discrete values and no conceptual challenges beyond basic S1 content. |
| Spec | 5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x\): \(-1, 0, 1, 2, 3\); \(p\): \(\frac{1}{12}=0.0833\), \(\frac{2}{12}=0.167\), \(\frac{4}{12}=0.333\), \(\frac{3}{12}=0.25\), \(\frac{2}{12}=0.167\) | B1 | Table with \(x\) values and at least one probability substituted, \(0 < p < 1\). Condone any additional \(x\) values if probability stated as 0 |
| B1 | 2 correct identified probabilities | |
| B1 | All probabilities correct (accept to 3sf). SC if less than 2 correct probabilities: SC B1 4 or 5 probabilities summing to one |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = -\frac{1}{12} + \frac{4}{12} + \frac{6}{12} + \frac{6}{12} \left[= \frac{15}{12}\right]\) | M1 | May be implied by use in Variance, accept unsimplified expression. Probabilities must sum to \(1 \pm 0.001\) |
| \(\text{Var}(X) = \frac{1}{12} + 0 + \frac{4}{12} + \frac{12}{12} + \frac{18}{12} - \left(\frac{15}{12}\right)^2\) | M1 | Appropriate variance formula using *their* \((E(X))^2\). FT accept probabilities not summing to 1. Condone \(\frac{35}{12} - \left(\frac{15}{12}\right)^2\) or \(\frac{35}{12} - \frac{25}{9}\) from correct table |
| \(\left[\frac{35}{12} - \frac{25}{16} =\right] \frac{65}{48}, 1.35\) | A1 | WWW |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x$: $-1, 0, 1, 2, 3$; $p$: $\frac{1}{12}=0.0833$, $\frac{2}{12}=0.167$, $\frac{4}{12}=0.333$, $\frac{3}{12}=0.25$, $\frac{2}{12}=0.167$ | B1 | Table with $x$ values and at least one probability substituted, $0 < p < 1$. Condone any additional $x$ values if probability stated as 0 |
| | B1 | 2 correct identified probabilities |
| | B1 | All probabilities correct (accept to 3sf). **SC** if less than 2 correct probabilities: **SC B1** 4 or 5 probabilities summing to one |
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## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = -\frac{1}{12} + \frac{4}{12} + \frac{6}{12} + \frac{6}{12} \left[= \frac{15}{12}\right]$ | M1 | May be implied by use in Variance, accept unsimplified expression. Probabilities must sum to $1 \pm 0.001$ |
| $\text{Var}(X) = \frac{1}{12} + 0 + \frac{4}{12} + \frac{12}{12} + \frac{18}{12} - \left(\frac{15}{12}\right)^2$ | M1 | Appropriate variance formula using *their* $(E(X))^2$. **FT** accept probabilities not summing to 1. Condone $\frac{35}{12} - \left(\frac{15}{12}\right)^2$ or $\frac{35}{12} - \frac{25}{9}$ from correct table |
| $\left[\frac{35}{12} - \frac{25}{16} =\right] \frac{65}{48}, 1.35$ | A1 | WWW |
---4 A fair spinner has edges numbered $0,1,2,2$. Another fair spinner has edges numbered $- 1,0,1$. Each spinner is spun. The number on the edge on which a spinner comes to rest is noted. The random variable $X$ is the sum of the numbers for the two spinners.
\begin{enumerate}[label=(\alph*)]
\item Draw up the probability distribution table for $X$.
\item Find $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q4 [6]}}