| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | June |
| 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 construction requiring systematic enumeration of outcomes (3×3=9 cases), basic probability calculations (each outcome has probability 1/9), then standard application of E(X) and Var(X) formulas. It's below average difficulty as it involves only routine procedures with no conceptual challenges or problem-solving insight required. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Table with \(X\) values: \(-1, 0, 1, 2, 3\) | B1 | Table with correct \(X\) values and at least one probability. Condone any additional \(X\) values if probability stated as 0. |
| \(P(X)\): \(\frac{1}{9}, \frac{2}{9}, \frac{1}{9}, \frac{3}{9}, \frac{2}{9}\) | B1 | 2 correct probabilities linked with correct outcomes, may not be in table. |
| B1 | 3 further correct probabilities linked with correct outcomes, may not be in table. SC if less than 2 correct probabilities seen, award SCB1 for sum of *their* 4 or 5 probabilities in table \(= 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[E(X) = \frac{-1 \times 1 + (0 \times 2) + 1 \times 1 + 2 \times 3 + 3 \times 2}{9}\right] = \frac{-1+1+6+6}{9}\) | M1 | May be implied by use in variance, accept unsimplified expression. FT *their* table if *their* 3 or more probabilities sum to 1 or 0.999 |
| \(\text{Var}(X) = \left[\frac{-1^2 \times 1 + (0^2 \times 2) + 1^2 \times 1 + 2^2 \times 3 + 3^2 \times 2}{9} - (\textit{their } E(X))^2\right]\) \(= \frac{1+0+1+12+18}{9} - (\textit{their } E(X))^2\) | M1 | Appropriate variance formula using *their* \((E(X))^2\) value. FT *their* table even if *their* 3 or more probabilities not summing to 1. |
| \(E(X) = \frac{4}{3}\) or \(1.33\) and \(\text{Var}(X) = \frac{16}{9}\) or \(1.78\) | A1 | Answers for \(E(X)\) and \(\text{Var}(X)\) must be identified. N.B. If method FT for M marks from *their* incorrect table, expressions for \(E(X)\) and \(\text{Var}(X)\) must be seen unsimplified with all probabilities \(< 1\) |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Table with $X$ values: $-1, 0, 1, 2, 3$ | B1 | Table with correct $X$ values and at least one probability. Condone any additional $X$ values if probability stated as 0. |
| $P(X)$: $\frac{1}{9}, \frac{2}{9}, \frac{1}{9}, \frac{3}{9}, \frac{2}{9}$ | B1 | 2 correct probabilities linked with correct outcomes, may not be in table. |
| | B1 | 3 further correct probabilities linked with correct outcomes, may not be in table. **SC** if less than 2 correct probabilities seen, award **SCB1** for sum of *their* 4 or 5 probabilities in table $= 1$ |
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## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[E(X) = \frac{-1 \times 1 + (0 \times 2) + 1 \times 1 + 2 \times 3 + 3 \times 2}{9}\right] = \frac{-1+1+6+6}{9}$ | M1 | May be implied by use in variance, accept unsimplified expression. FT *their* table if *their* 3 or more probabilities sum to 1 or 0.999 |
| $\text{Var}(X) = \left[\frac{-1^2 \times 1 + (0^2 \times 2) + 1^2 \times 1 + 2^2 \times 3 + 3^2 \times 2}{9} - (\textit{their } E(X))^2\right]$ $= \frac{1+0+1+12+18}{9} - (\textit{their } E(X))^2$ | M1 | Appropriate variance formula using *their* $(E(X))^2$ value. FT *their* table even if *their* 3 or more probabilities not summing to 1. |
| $E(X) = \frac{4}{3}$ or $1.33$ and $\text{Var}(X) = \frac{16}{9}$ or $1.78$ | A1 | Answers for $E(X)$ and $\text{Var}(X)$ must be identified. **N.B.** If method FT for M marks from *their* incorrect table, expressions for $E(X)$ and $\text{Var}(X)$ must be seen unsimplified with all probabilities $< 1$ |
---4 A fair spinner has sides numbered 1, 2, 2. Another fair spinner has sides numbered $- 2,0,1$. Each spinner is spun. The number on the side 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 $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2021 Q4 [6]}}