| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from given distribution |
| Difficulty | Moderate -0.3 This is a standard S1 discrete probability distribution question requiring routine application of expectation and variance formulas. Parts (a)-(e) involve direct formula application with no conceptual challenges. Parts (f)-(g) require basic understanding of independence and conditional probability but follow predictable patterns. The calculations are straightforward with minimal problem-solving required, making it slightly easier than average for A-level. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p) |
| \(s\) | 0 | 1 | 2 | 4 | 5 |
| \(\mathrm { P } ( S = s )\) | 0.2 | 0.2 | 0.1 | 0.3 | 0.2 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(E(S) = 0 + 1 \times 0.2 + 2 \times 0.1 + 4 \times 0.3 + 5 \times 0.2 = [0.2 + 0.2 + 1.2 + 1.0] = \mathbf{2.6}\) | M1, A1 | (2) |
| (b) \(E(S^2) = 0 + 1 \times 0.2 + 2^2 \times 0.1 + 4^2 \times 0.3 + 5^2 \times 0.2\) or \(0.2 + 0.4 + 4.8 + 5 = \mathbf{10.4}\) (*) | M1, A1cso | (2) |
| (c) \(\text{Var}(S) = 10.4 - ("2.6")^2 = \mathbf{3.64}\) or \(\frac{91}{25}\) (o.e.) | M1, A1 | (2) |
| (d)(i) \(5E(S) - 3 = 5 \times "2.6" - 3, = \mathbf{10}\) | M1, A1 | |
| (d)(ii) \(5^2\text{Var}(S) = 25 \times 3.64, = \mathbf{91}\) | M1, A1 | (4) |
| (e) \(5S - 3 > S + 3 \Rightarrow 4S > 6\) or \(S > 1.5\), so need \(P(S \geq 2) = \mathbf{0.6}\) | M1, A1, A1 | (3) |
| (f) \(P(S_1 = 1) \times P(S_2 \leq 4) = 0.2 \times 0.8 = 0.16\) (*) | M1, A1cso | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(S_1 = 0) \times P(S_2 = \text{any value}) = 0.2 \times 1 = 0.20\), Full method – all cases listed, all correct products \(= \mathbf{0.57}\) | M1, A1, A1 | (3) [18] |
**(a)** $E(S) = 0 + 1 \times 0.2 + 2 \times 0.1 + 4 \times 0.3 + 5 \times 0.2 = [0.2 + 0.2 + 1.2 + 1.0] = \mathbf{2.6}$ | M1, A1 | (2) | M1 for attempt at $\sum xP(X = x)$, at least 2 non-zero terms seen. Correct answer 2/2; A1 for 2.6 or exact equivalent
**(b)** $E(S^2) = 0 + 1 \times 0.2 + 2^2 \times 0.1 + 4^2 \times 0.3 + 5^2 \times 0.2$ or $0.2 + 0.4 + 4.8 + 5 = \mathbf{10.4}$ (*) | M1, A1cso | (2) | M1 for correct attempt, at least 3 non-zero terms seen; A1cso for 10.4 provided M1 scored and no incorrect working seen
**(c)** $\text{Var}(S) = 10.4 - ("2.6")^2 = \mathbf{3.64}$ or $\frac{91}{25}$ (o.e.) | M1, A1 | (2) | M1 for 10.4 – $\mu^2$. ft their $\mu$. Must see their value of $\mu$ squared (A1 for 3.64 or any exact equiv.)
**(d)(i)** $5E(S) - 3 = 5 \times "2.6" - 3, = \mathbf{10}$ | M1, A1 | | M1, A1 for correct expression using their 2.6 (A1 for 10)
**(d)(ii)** $5^2\text{Var}(S) = 25 \times 3.64, = \mathbf{91}$ | M1, A1 | (4) | M1 for $25 \times \text{Var}(S)$ – ft their Var(S) (A1 for 91)
**(e)** $5S - 3 > S + 3 \Rightarrow 4S > 6$ or $S > 1.5$, so need $P(S \geq 2) = \mathbf{0.6}$ | M1, A1, A1 | (3) | M1 for solving inequality as far as $pS > q$ where one of $p$ or $q$ are correct; 1st A1 for P(S ≥ 2); 2nd A1 for 0.6 (provided $S > 1.5$ was obtained). Ans only of 0.6 scores 3/3
**(f)** $P(S_1 = 1) \times P(S_2 \leq 4) = 0.2 \times 0.8 = 0.16$ (*) | M1, A1cso | (2) | M1, A1cso for using independence (so multiplying) and attempting P($S_2$ ≤ 4). e.g. $0.2 \times (0.2 + 0.1 + 0.3)$ or $0.04 + 0.04 + 0.02 + 0.06$ score M1 BUT $\frac{1}{4}$ (not from 0.2 × 0.8) is M0A0
**(g)** $P(S_1 = 2) \times P(S_2 \leq 2) = 0.1 \times 0.5 = 0.05$
$P(S_1 = 4) \times P(S_2 \leq 1) = 0.3 \times 0.4 = 0.12$
$P(S_1 = 5) \times P(S_2 = 0) = 0.2 \times 0.2 = 0.04$
$P(S_1 = 0) \times P(S_2 = \text{any value}) = 0.2 \times 1 = 0.20$, **Full method – all cases listed**, all correct products $= \mathbf{0.57}$ | M1, A1, A1 | (3) [18] | M1 for all cases for $S_1$ or all 15 cases for $X$; 1st A1 for all correct probability products for $S_1$ or $X$; 2nd A1 for 0.57 Correct answer scores 3/3. Probabilities out of 25 score A0A0
**A table showing all 25 cases can only score M1 in (g) if the correct cases are indicated.**
---7. The score $S$ when a spinner is spun has the following probability distribution.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$s$ & 0 & 1 & 2 & 4 & 5 \\
\hline
$\mathrm { P } ( S = s )$ & 0.2 & 0.2 & 0.1 & 0.3 & 0.2 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { E } ( S )$.
\item Show that $\mathrm { E } \left( S ^ { 2 } \right) = 10.4$
\item Hence find $\operatorname { Var } ( S )$.
\item Find
\begin{enumerate}[label=(\roman*)]
\item $\mathrm { E } ( 5 S - 3 )$,
\item $\operatorname { Var } ( 5 S - 3 )$.
\end{enumerate}\item Find $\mathrm { P } ( 5 S - 3 > S + 3 )$
The spinner is spun twice.\\
The score from the first spin is $S _ { 1 }$ and the score from the second spin is $S _ { 2 }$\\
The random variables $S _ { 1 }$ and $S _ { 2 }$ are independent and the random variable $X = S _ { 1 } \times S _ { 2 }$
\item Show that $\mathrm { P } \left( \left\{ S _ { 1 } = 1 \right\} \cap X < 5 \right) = 0.16$
\item Find $\mathrm { P } ( X < 5 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2013 Q7 [18]}}