| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| 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 two-unknowns problem using sum of probabilities equals 1 and the given expectation to find a and b, then computing variance with the standard formula Var(aX+b)=a²Var(X). Requires systematic algebraic manipulation but follows a well-established routine with no novel insight needed. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(w\) | 58 | 59 | 60 | 61 | 62 | 63 |
| \(\mathrm { P } ( W = w )\) | \(a\) | \(b\) | 0.2 | 0.2 | 0.1 | 0.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a + b = 0.4\) | B1 | Allow \(a + b + 0.6 = 1\), *or* use both \(2a + b \pm 0.55\) and \(58a + 59b = 23.45\) |
| \(-2a - b + 0.2 + 0.2 + 0.3 = 0.15\) | M1 | Use \(\Sigma(W - 60)P(w) = 0.15\) |
| \(2a + b = 0.55\) | A1 | Correct simplified equation |
| \(a = 0.15,\ b = 0.25\) | A1 | |
| \(4a + b + 1 \times 0.2 + 4 \times 0.1 + 9 \times 0.1\ (= 2.35)\) | M1 | Find \(\Sigma(W-60)^2 P(w)\) |
| \(- 0.15^2\) and \(\times 16\) | M1 | (independent of previous M1) |
| \(= 37.24\) | A1[7] | Or \(\frac{931}{25}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a + b = 0.4\) | B1 | Allow \(a + b + 0.6 = 1\) |
| \(58a + 59b + 60 \times 0.2 + \ldots + 63 \times 0.1 = 60.15\) | M1 | Use \(\Sigma WP(w) = 0.15 + 60\) |
| \(58a + 59b = 23.45\) | A1 | Correct simplified equation |
| \(a = 0.15,\ b = 0.25\) | A1 | |
| \(16 \times (58^2 a + 59^2 b + \ldots + 63^2 \times 0.1)\ (= 57925.6)\) | M1 | \(16 \times \Sigma w^2 P(w)\) |
| \(-(4 \times 60.15)^2\) | M1 | \(-(4 \times \text{their}\ \bar{w})^2\) (independent of previous M1) |
| \(= 37.24\) | A1[7] | Or \(\frac{931}{25}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(58^2 a + 59^2 b \ldots + 63^2 \times 0.1 - 60.15^2\) | M1 | \(= 3620.35 - 3618.0225 = 2.3275\) |
| \(\times 16\) | M1 | Allow for \(16 \times \Sigma w^2 P(w)\) without having subtracted \(60.15^2\) |
| \(= 37.24\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If B0M0, give SC B1 for \(\text{Var}(4W - 60) = 16\text{Var}(W)\) used | B1 |
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a + b = 0.4$ | B1 | Allow $a + b + 0.6 = 1$, *or* use both $2a + b \pm 0.55$ and $58a + 59b = 23.45$ |
| $-2a - b + 0.2 + 0.2 + 0.3 = 0.15$ | M1 | Use $\Sigma(W - 60)P(w) = 0.15$ |
| $2a + b = 0.55$ | A1 | Correct simplified equation |
| $a = 0.15,\ b = 0.25$ | A1 | |
| $4a + b + 1 \times 0.2 + 4 \times 0.1 + 9 \times 0.1\ (= 2.35)$ | M1 | Find $\Sigma(W-60)^2 P(w)$ |
| $- 0.15^2$ and $\times 16$ | M1 | (independent of previous M1) |
| $= 37.24$ | A1[7] | Or $\frac{931}{25}$ |
#### Alternative method 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a + b = 0.4$ | B1 | Allow $a + b + 0.6 = 1$ |
| $58a + 59b + 60 \times 0.2 + \ldots + 63 \times 0.1 = 60.15$ | M1 | Use $\Sigma WP(w) = 0.15 + 60$ |
| $58a + 59b = 23.45$ | A1 | Correct simplified equation |
| $a = 0.15,\ b = 0.25$ | A1 | |
| $16 \times (58^2 a + 59^2 b + \ldots + 63^2 \times 0.1)\ (= 57925.6)$ | M1 | $16 \times \Sigma w^2 P(w)$ |
| $-(4 \times 60.15)^2$ | M1 | $-(4 \times \text{their}\ \bar{w})^2$ (independent of previous M1) |
| $= 37.24$ | A1[7] | Or $\frac{931}{25}$ |
#### Alternative method 2 (last 3 marks):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $58^2 a + 59^2 b \ldots + 63^2 \times 0.1 - 60.15^2$ | M1 | $= 3620.35 - 3618.0225 = 2.3275$ |
| $\times 16$ | M1 | Allow for $16 \times \Sigma w^2 P(w)$ without having subtracted $60.15^2$ |
| $= 37.24$ | A1 | |
#### Special case:
| Answer | Marks | Guidance |
|--------|-------|----------|
| If B0M0, give SC B1 for $\text{Var}(4W - 60) = 16\text{Var}(W)$ used | B1 | |4 A discrete random variable $W$ has the probability distribution shown in the following table, in which $a$ and $b$ are constants.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
$w$ & 58 & 59 & 60 & 61 & 62 & 63 \\
\hline
$\mathrm { P } ( W = w )$ & $a$ & $b$ & 0.2 & 0.2 & 0.1 & 0.1 \\
\hline
\end{tabular}
\end{center}
It is given that $\mathrm { E } ( W - 60 ) = 0.15$.
Determine the value of $\operatorname { Var } ( 4 W - 60 )$.
\hfill \mbox{\textit{OCR Further Statistics AS 2023 Q4 [7]}}