| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Multiple unknowns from expectation and variance |
| Difficulty | Standard +0.8 This is a multi-part Further Maths statistics question requiring: (a) solving a quadratic equation from probability axioms, (b) using the non-trivial relationship E(aX+b) = Var(aX+b) to find two unknowns. Part (b) requires understanding that variance is unaffected by location shifts and careful algebraic manipulation with E(X) and Var(X). More demanding than standard A-level but straightforward for Further Maths students. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | \(p\) | 0.31 | 0.3 | \(p ^ { 2 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p + p^2 = 0.39\) | M1 | Use \(\Sigma p = 1\) |
| \(p \neq -1.3\) as \(0 < p < 1\) | B1 | Explicitly reject \(-1.3\), with reason, allow \(0 \leq p \leq 1\) etc. E.g. "must be positive". Not just "reject" or "invalid" |
| \(p = 0.3\) | A1 [3] | Final answer \(0.3\) oe and nothing else. Can get this even if B1 not gained |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = \Sigma xP(X=x) = 2.18\) | M1 | Use \(\Sigma xP(X=x)\) to get \(E(X)\) |
| \(\text{Var}(X) = 5.68 - 2.18^2\ (= 0.9276)\) | M1 | Use \(\Sigma x^2 P(X=x) - [E(X)]^2\). \(\text{Var}(X) = 3.5\) from \(5.68 - 2.18\): M0 |
| \(0.9276a^2 = 23.19\) | M1 | \(a^2 \times \textit{their}\ \text{Var}(X) = 23.19\) (not \(a^2 E(X)\)). \(2.18a^2 = 23.19\): usually M1, 000, M1A0 |
| \(\Rightarrow a = 5\) | A1 | \(a = 5\) or awrt \(5.00\) *[no fit]*. Ignore negative \(a\) and any value of \(b\) |
| \(2.18a + b = 23.19\) | M1 | Use *their* \(a \times\) *their* \(2.18 + b = 23.19\). Obtained from it e.g. \((-5, 34.09)\) CWO |
| \(\Rightarrow b = 12.29\) | A1 [6] | \(12.3\) or awrt \(12.29\) |
# Question 3:
## Part (a):
$p + p^2 = 0.39$ | **M1** | Use $\Sigma p = 1$
$p \neq -1.3$ as $0 < p < 1$ | **B1** | Explicitly reject $-1.3$, with reason, allow $0 \leq p \leq 1$ etc. E.g. "must be positive". Not just "reject" or "invalid"
$p = 0.3$ | **A1 [3]** | Final answer $0.3$ oe and nothing else. Can get this even if B1 not gained
## Part (b):
$E(X) = \Sigma xP(X=x) = 2.18$ | **M1** | Use $\Sigma xP(X=x)$ to get $E(X)$
$\text{Var}(X) = 5.68 - 2.18^2\ (= 0.9276)$ | **M1** | Use $\Sigma x^2 P(X=x) - [E(X)]^2$. $\text{Var}(X) = 3.5$ from $5.68 - 2.18$: M0
$0.9276a^2 = 23.19$ | **M1** | $a^2 \times \textit{their}\ \text{Var}(X) = 23.19$ (not $a^2 E(X)$). $2.18a^2 = 23.19$: usually M1, 000, M1A0
$\Rightarrow a = 5$ | **A1** | $a = 5$ or awrt $5.00$ *[no fit]*. Ignore negative $a$ and any value of $b$
$2.18a + b = 23.19$ | **M1** | Use *their* $a \times$ *their* $2.18 + b = 23.19$. Obtained from it e.g. $(-5, 34.09)$ CWO
$\Rightarrow b = 12.29$ | **A1 [6]** | $12.3$ or awrt $12.29$
---3 A discrete random variable $X$ has the following probability distribution.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $p$ & 0.31 & 0.3 & $p ^ { 2 }$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $p$.
\item It is given that $\mathrm { E } ( a X + b ) = \operatorname { Var } ( a X + b ) = 23.19$, where $a$ and $b$ are positive constants. Determine the value of $a$ and the value of $b$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2022 Q3 [9]}}