| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Piecewise or conditional probability function |
| Difficulty | Moderate -0.8 This is a routine S1 probability distribution question requiring standard techniques: summing probabilities to find k, calculating E(X) and E(X²) by definition, and applying the variance formula Var(aX+b) = a²Var(X). All steps are mechanical with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| Table with \(x\): \(-1, 0, 1, 2\) and \(P(X=x)\): \(4k, k, 0, k\) | M1 | For attempt at \(P(X=x)\) with at least 2 correct. Do not give for 4, 1, etc but \(\frac{4}{6}, \frac{1}{6}\) are OK |
| \(4k + k + 0 + k = 1\) | A1 | For at least \(4k + k + k = 1\) seen. Allow \(\frac{4}{6}+\frac{1}{6}+\frac{1}{6}=1\) [Must see = 1] |
| \(6k = 1 \Rightarrow k = \frac{1}{6}\) | A1cso (3) | Provided previous 2 marks scored and no incorrect working seen. To score final A1cso there must be a comment such as "therefore \(k=\frac{1}{6}\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(E(X) = -4k + (0+0) + 2k\) or \(-2k\) or \(-1\times\frac{4}{6}+2\times\frac{1}{6}\) | M1 | For a full correct expression for \(E(X)\), ft their probabilities. Allow in terms of \(k\). Division by 4 (or any other \(n\)) is M0. Do not apply ISW |
| \(= -\frac{1}{3}\) (or \(-0.\overline{3}\)) | A1 (2) | For \(-\frac{1}{3}\) or exact equivalent only. Just \(-\frac{1}{3}\) scores M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(E(X^2) = (-1)^2\times 4k + (0+0) + 2^2k\) or \(4k+4k\) or \((-1)^2\times\frac{4}{6}+2^2\times\frac{1}{6}\) | M1 | For evidence of both non-zero terms seen. May be simplified but 2 terms needed |
| \(= \frac{4}{3}\) | A1cso (2) | For M1 seen leading to \(\frac{4}{3}\) or any exact equivalent. Condone \(-1^2\times 4k\) but not \(-4k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(\text{Var}(X) = \frac{4}{3} - \left(-\frac{1}{3}\right)^2\) or \(8k - 4k^2 = \frac{11}{9}\) | M1 | For correct attempt at \(\text{Var}(X)\) - follow through their \(E(X)\) and allow in terms of \(k\). Award if correct formula seen and some correct substitution made |
| \(\text{Var}(1-3X) = (-3)^2\text{Var}(X)\) or \(9\text{Var}(X)\) | M1 | For correct use of \(\text{Var}(aX+b)\). Condone \(-3^2\text{Var}(X)\) if it eventually yields \(9\text{Var}(X)\) |
| \(= 11\) | A1 cao (3) [10] | For 11 only |
# Question 1:
## Part (a)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| Table with $x$: $-1, 0, 1, 2$ and $P(X=x)$: $4k, k, 0, k$ | M1 | For attempt at $P(X=x)$ with at least 2 correct. Do not give for 4, 1, etc but $\frac{4}{6}, \frac{1}{6}$ are OK |
| $4k + k + 0 + k = 1$ | A1 | For at least $4k + k + k = 1$ seen. Allow $\frac{4}{6}+\frac{1}{6}+\frac{1}{6}=1$ [Must see = 1] |
| $6k = 1 \Rightarrow k = \frac{1}{6}$ | A1cso (3) | Provided previous 2 marks scored and no incorrect working seen. To score final A1cso there must be a comment such as "therefore $k=\frac{1}{6}$" |
## Part (b)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $E(X) = -4k + (0+0) + 2k$ or $-2k$ or $-1\times\frac{4}{6}+2\times\frac{1}{6}$ | M1 | For a full correct expression for $E(X)$, ft their probabilities. Allow in terms of $k$. **Division by 4 (or any other $n$) is M0. Do not apply ISW** |
| $= -\frac{1}{3}$ (or $-0.\overline{3}$) | A1 (2) | For $-\frac{1}{3}$ or exact equivalent only. Just $-\frac{1}{3}$ scores M1A1 |
## Part (c)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $E(X^2) = (-1)^2\times 4k + (0+0) + 2^2k$ or $4k+4k$ or $(-1)^2\times\frac{4}{6}+2^2\times\frac{1}{6}$ | M1 | For evidence of both non-zero terms seen. May be simplified but 2 terms needed |
| $= \frac{4}{3}$ | A1cso (2) | For M1 seen leading to $\frac{4}{3}$ or any exact equivalent. Condone $-1^2\times 4k$ but not $-4k$ |
## Part (d)
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\text{Var}(X) = \frac{4}{3} - \left(-\frac{1}{3}\right)^2$ or $8k - 4k^2 = \frac{11}{9}$ | M1 | For correct attempt at $\text{Var}(X)$ - follow through their $E(X)$ and allow in terms of $k$. Award if correct formula seen and some correct substitution made |
| $\text{Var}(1-3X) = (-3)^2\text{Var}(X)$ or $9\text{Var}(X)$ | M1 | For correct use of $\text{Var}(aX+b)$. Condone $-3^2\text{Var}(X)$ if it eventually yields $9\text{Var}(X)$ |
| $= 11$ | A1 cao (3) **[10]** | For 11 only |\begin{enumerate}
\item A discrete random variable $X$ has the probability function
\end{enumerate}
$$\mathrm { P } ( X = x ) = \begin{cases} k ( 1 - x ) ^ { 2 } & x = - 1,0,1 \text { and } 2 \\ 0 & \text { otherwise } \end{cases}$$
(a) Show that $k = \frac { 1 } { 6 }$\\
(b) Find $\mathrm { E } ( X )$\\
(c) Show that $\mathrm { E } \left( X ^ { 2 } \right) = \frac { 4 } { 3 }$\\
(d) Find $\operatorname { Var } ( 1 - 3 X )$\\
\hfill \mbox{\textit{Edexcel S1 2012 Q1 [10]}}