| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(aX+b) transformations |
| Difficulty | Moderate -0.5 This is a straightforward two-part question on variance calculations. Part (i) requires finding x from probabilities summing to 1 (solving a quadratic), then computing E(W) and E(W²) from the table—standard bookwork. Part (ii) applies the direct formula Var(aX+b) = a²Var(X), which is immediate recall. While it involves multiple steps, each is routine and the question follows a standard textbook pattern with no problem-solving insight required. |
| Spec | 5.02b Expectation and variance: discrete random variables |
| \(w\) | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( W = w )\) | 0.25 | 0.36 | \(x\) | \(x ^ { 2 }\) |
| Answer | Marks | Guidance |
|---|---|---|
| Subtract \([E(W)]^2\) to get \(\mathbf{0.8571}\) | M1 A1 A1 B1ft B1 M1 A1 AG [7] | M1: Equation using \(\Sigma p = 1\). A1: Correct simplified quadratic (can be implied). A1: Correctly obtain \(x = 0.3\) (method needed). B1ft: Explicitly reject other solution, ft on their quadratic. B1: 2.23 or exact equivalent only; allow \(E(W)^2 = 4.9729\). M1: Use \(\Sigma w^2 p(w)\). A1: Correctly obtain given answer, need 2.23 or 4.9729 and 5.83 or full numerical \(\Sigma w^2 p(w)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(9 \times 0.8571 = \mathbf{7.7139}\) | B1 [1] | Allow 7.71 or 7.714 |
## Question 2:
### Part (i)
$0.25 + 0.36 + x + x^2 = 1$
$x^2 + x - 0.39 = 0$
$x = 0.3$ (or $-1.3$)
$x$ cannot be negative
$E(W) = 2.23$
$E(W^2) = \Sigma w^2 p(w) \quad [= 5.83]$
Subtract $[E(W)]^2$ to get $\mathbf{0.8571}$ | **M1** A1 A1 **B1ft** **B1** **M1** A1 **AG** [7] | M1: Equation using $\Sigma p = 1$. A1: Correct simplified quadratic (can be implied). A1: Correctly obtain $x = 0.3$ (method needed). B1ft: Explicitly reject other solution, ft on their quadratic. B1: 2.23 or exact equivalent only; allow $E(W)^2 = 4.9729$. M1: Use $\Sigma w^2 p(w)$. A1: Correctly obtain given answer, need 2.23 or 4.9729 and 5.83 or full numerical $\Sigma w^2 p(w)$
### Part (ii)
$9 \times 0.8571 = \mathbf{7.7139}$ | **B1** [1] | Allow 7.71 or 7.714
---2 The probability distribution for the discrete random variable $W$ is given in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$w$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( W = w )$ & 0.25 & 0.36 & $x$ & $x ^ { 2 }$ \\
\hline
\end{tabular}
\end{center}
(i) Show that $\operatorname { Var } ( W ) = 0.8571$.\\
(ii) Find $\operatorname { Var } ( 3 W + 6 )$.
\hfill \mbox{\textit{OCR Further Statistics AS 2018 Q2 [8]}}