| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Standard +0.3 This is a standard S1 probability distribution question requiring systematic application of ΣP=1 and E(X)=Σxp formulas to find unknowns, then routine calculations of variance and E(1/X). Part (f) requires careful comparison of values but no novel insight. Slightly above average due to multiple parts and the need to compare S vs R systematically, but all techniques are textbook exercises. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | - 2 | - 1 | \(\frac { 1 } { 2 }\) | \(\frac { 3 } { 2 }\) | 2 |
| \(\mathrm { P } ( X = x )\) | \(p\) | \(q\) | 0.2 | 0.3 | \(p\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p + q + 0.2 + 0.3 + p = 1\) or \(2p + q = 0.5\) | B1 | Any correct equation based on sum of probabilities = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{E}(X) = -2p - q + \frac{1}{2}(0.2) + \frac{3}{2}(0.3) + 2p = 0.4\) or \(-q + 0.1 + 0.45 = 0.4\) | M1 | Attempt at expression based on E(X), at most 2 errors or omissions |
| Correct equation | A1 | May be implied by correct answer |
| \(q = 0.15\) | A1 | For \(q = 0.15\) or exact equivalent e.g. \(\frac{6}{40}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(2p + \text{"0.15"} = 0.5\) | M1 | Correct equation using their \(q\), provided \(q \in [0,1]\) |
| \(p = 0.175\) | A1 | For \(p = 0.175\) or exact equivalent e.g. \(\frac{7}{40}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(X) = 2.275 - (0.4)^2 = \mathbf{2.115}\) | M1 | Correct numerical expression (M0 if followed by division by \(k\)) |
| Accept 2.12 | A1 | For 2.115 or awrt 2.12 (also accept \(\frac{423}{200}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Table with \(r\) values \(-\frac{1}{2}, -1, 2, \frac{2}{3}, \frac{1}{2}\) and probabilities \(p, q, 0.2, 0.3, p\) | M1 | Correct values for \(R\), allow 1 error, allow unsimplified |
| \(\text{E}(R) = -\frac{1}{2}p - q + 0.4 + 0.2 + \frac{1}{2}p\) | dM1 | Dependent on 1st M1; attempt at E(\(R\)), ft \(p\) and \(q\); at least 3 correct products |
| \(= 0.6 - q = \mathbf{0.45}\) (or \(\frac{9}{20}\)) | A1ft | For 0.45 or \((0.6 - \text{their } q)\) provided \(q\) is a probability |
| Answer | Marks | Guidance |
|---|---|---|
| \(S > R\) when \(x = 1.5\) and \(2\) | M1 | For identifying correct values of \(X\) |
| \(\text{P(Sarah wins)} = 0.3 + p = \mathbf{0.475}\) (or \(\frac{19}{40}\)) | A1ft | For 0.475 or \(0.3 + \text{their } p\), provided answer is a probability |
| Answer | Marks | Guidance |
|---|---|---|
| \(R > S\) when \(x = -2\) and \(\frac{1}{2}\), or \(r = -\frac{1}{2}\) and \(2\) | M1 | For identifying correct values of \(X\) or \(R\) |
| \(\text{P(Rebecca wins)} = 0.2 + p = \mathbf{0.375}\) (or \(\frac{15}{40}\)) | A1ft | For 0.375 or \(0.2 + \text{their } p\) or \(1 - \text{their } 0.475 - \text{their } q\), provided answer is a probability |
# Question 2:
## Part (a)
| $p + q + 0.2 + 0.3 + p = 1$ or $2p + q = 0.5$ | B1 | Any correct equation based on sum of probabilities = 1 |
## Part (b)
| $\text{E}(X) = -2p - q + \frac{1}{2}(0.2) + \frac{3}{2}(0.3) + 2p = 0.4$ or $-q + 0.1 + 0.45 = 0.4$ | M1 | Attempt at expression based on E(X), at most 2 errors or omissions |
| Correct equation | A1 | May be implied by correct answer |
| $q = 0.15$ | A1 | For $q = 0.15$ or exact equivalent e.g. $\frac{6}{40}$ |
## Part (c)
| $2p + \text{"0.15"} = 0.5$ | M1 | Correct equation using their $q$, provided $q \in [0,1]$ |
| $p = 0.175$ | A1 | For $p = 0.175$ or exact equivalent e.g. $\frac{7}{40}$ |
## Part (d)
| $\text{Var}(X) = 2.275 - (0.4)^2 = \mathbf{2.115}$ | M1 | Correct numerical expression (M0 if followed by division by $k$) |
| Accept 2.12 | A1 | For 2.115 or awrt 2.12 (also accept $\frac{423}{200}$) |
## Part (e)
| Table with $r$ values $-\frac{1}{2}, -1, 2, \frac{2}{3}, \frac{1}{2}$ and probabilities $p, q, 0.2, 0.3, p$ | M1 | Correct values for $R$, allow 1 error, allow unsimplified |
| $\text{E}(R) = -\frac{1}{2}p - q + 0.4 + 0.2 + \frac{1}{2}p$ | dM1 | Dependent on 1st M1; attempt at E($R$), ft $p$ and $q$; at least 3 correct products |
| $= 0.6 - q = \mathbf{0.45}$ (or $\frac{9}{20}$) | A1ft | For 0.45 or $(0.6 - \text{their } q)$ provided $q$ is a probability |
## Part (f)(i)
| $S > R$ when $x = 1.5$ and $2$ | M1 | For identifying correct values of $X$ |
| $\text{P(Sarah wins)} = 0.3 + p = \mathbf{0.475}$ (or $\frac{19}{40}$) | A1ft | For 0.475 or $0.3 + \text{their } p$, provided answer is a probability |
## Part (f)(ii)
| $R > S$ when $x = -2$ and $\frac{1}{2}$, or $r = -\frac{1}{2}$ and $2$ | M1 | For identifying correct values of $X$ or $R$ |
| $\text{P(Rebecca wins)} = 0.2 + p = \mathbf{0.375}$ (or $\frac{15}{40}$) | A1ft | For 0.375 or $0.2 + \text{their } p$ or $1 - \text{their } 0.475 - \text{their } q$, provided answer is a probability |
---2. The discrete random variable $X$ has the following probability distribution, where $p$ and $q$ are constants.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 2 & - 1 & $\frac { 1 } { 2 }$ & $\frac { 3 } { 2 }$ & 2 \\
\hline
$\mathrm { P } ( X = x )$ & $p$ & $q$ & 0.2 & 0.3 & $p$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Write down an equation in $p$ and $q$
Given that $\mathrm { E } ( X ) = 0.4$
\item find the value of $q$
\item hence find the value of $p$
Given also that $\mathrm { E } \left( X ^ { 2 } \right) = 2.275$
\item find $\operatorname { Var } ( X )$
Sarah and Rebecca play a game.\\
A computer selects a single value of $X$ using the probability distribution above.\\
Sarah's score is given by the random variable $S = X$ and Rebecca's score is given by the random variable $R = \frac { 1 } { X }$
\item Find $\mathrm { E } ( R )$
Sarah and Rebecca work out their scores and the person with the higher score is the winner. If the scores are the same, the game is a draw.
\item Find the probability that
\begin{enumerate}[label=(\roman*)]
\item Sarah is the winner,
\item Rebecca is the winner.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2016 Q2 [15]}}