| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2008 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Moderate -0.3 This is a standard S1 question requiring systematic application of probability axioms (sum=1) and expectation formula to solve simultaneous equations for p and q, followed by routine variance and linear transformation calculations. While it involves multiple parts and algebraic manipulation, it follows a well-practiced template with no conceptual surprises, making it slightly easier than average. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| \(x\) | - 1 | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | \(p\) | \(q\) | 0.2 | 0.15 | 0.15 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-1 \times p + 1 \times 0.2 + 2 \times 0.15 + 3 \times 0.15 = 0.55\) | M1dM1 | M1 for at least 2 correct terms on LHS; dM1 dependent on first M1, equate to 0.55 and solve |
| \(p = 0.4\) | A1 | Award M1M1A1 for \(p=0.4\) with no working |
| \(p + q + 0.2 + 0.15 + 0.15 = 1\) | M1 | M1 for adding probabilities and equating to 1 |
| \(q = 0.1\) | A1 | Award M1A1 for \(q=0.1\) with no working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(X) = (-1)^2 \times p + 1^2 \times 0.2 + 2^2 \times 0.15 + 3^2 \times 0.15 - 0.55^2\) | M1A1, M1 | M1 attempting \(E(X^2)\) with at least 2 correct terms; A1 fully correct expression or 2.55; M1 for subtracting mean squared |
| \(= 2.55 - 0.3025 = 2.2475\) | A1 | awrt 2.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(2X-4) = 2E(X) - 4\) | M1 | M1 for \(2\times(\text{their mean}) - 4\) |
| \(= -2.9\) | A1 | Award 2 marks for \(-2.9\) with no working |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-1 \times p + 1 \times 0.2 + 2 \times 0.15 + 3 \times 0.15 = 0.55$ | M1dM1 | M1 for at least 2 correct terms on LHS; dM1 dependent on first M1, equate to 0.55 and solve |
| $p = 0.4$ | A1 | Award M1M1A1 for $p=0.4$ with no working |
| $p + q + 0.2 + 0.15 + 0.15 = 1$ | M1 | M1 for adding probabilities and equating to 1 |
| $q = 0.1$ | A1 | Award M1A1 for $q=0.1$ with no working |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X) = (-1)^2 \times p + 1^2 \times 0.2 + 2^2 \times 0.15 + 3^2 \times 0.15 - 0.55^2$ | M1A1, M1 | M1 attempting $E(X^2)$ with at least 2 correct terms; A1 fully correct expression or 2.55; M1 for subtracting mean squared |
| $= 2.55 - 0.3025 = 2.2475$ | A1 | awrt 2.25 |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(2X-4) = 2E(X) - 4$ | M1 | M1 for $2\times(\text{their mean}) - 4$ |
| $= -2.9$ | A1 | Award 2 marks for $-2.9$ with no working |
---3. The random variable $X$ has probability distribution given in the table below.
\begin{center}
\begin{tabular}{ l l l l l l }
$x$ & - 1 & 0 & 1 & 2 & 3 \\
$\mathrm { P } ( X = x )$ & $p$ & $q$ & 0.2 & 0.15 & 0.15 \\
\end{tabular}
\end{center}
Given that $\mathrm { E } ( X ) = 0.55$, find
\begin{enumerate}[label=(\alph*)]
\item the value of $p$ and the value of $q$,
\item $\operatorname { Var } ( X )$,
\item $\mathrm { E } ( 2 X - 4 )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2008 Q3 [11]}}