| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Joint distribution with independence testing |
| Difficulty | Standard +0.8 This S4 question requires understanding of independence conditions for joint distributions, applying P(S=s)P(T=t)=P(S=s,T=t) to find unknowns, then computing probabilities and variance of a linear combination. The independence condition adds conceptual depth beyond routine probability calculations, and variance of T-S requires Var(T-S)=Var(T)+Var(S) using independence, making this moderately challenging but still within standard S4 scope. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(S\) | ||||
| \cline { 2 - 5 } | 0 | 1 | 2 | |
| \cline { 2 - 5 } | 1 | \(a\) | 0.18 | \(b\) |
| 2 | 0.08 | 0.12 | 0.20 | |
| \cline { 2 - 5 } | ||||
| \cline { 2 - 5 } | ||||
| Answer | Marks | Guidance |
|---|---|---|
| b=0.3 | M1, M1, A1, A1, A1, A1 | \(P(S=0)=0.08/0.4\) or \(P(S=2)=0.2/0.4\), \(P(S=0)=0.2\), \(P(S=2)=0.5\), \((P(T=1)=0.6)\), a="0.6"x"0.2" or b="0.6"x"0.5" |
| [6] |
| Answer | Marks |
|---|---|
| \(= 0.12 + 0.12 = 0.24\) | M1, A1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(T)= 0.6 + 4 \times 0.4 – (0.6 + 0.8)^2\) | M1 | (\(\text{Var}(T)=0.24\), \(\text{Var}(S)=0.61\)) |
| Allow M1 for sum of vars, even if \(E(T^2)\) and/or \(E(S^2)\) only used.(2.2,2.3) | ||
| \(\text{Var}(S) = 0.3 + 4 \times 0.5 – (0.3 + 1)^2\) | M1 | \(E(T – S) = 0.1 [E([T – S]^2)]\) |
| \(= 0.86\) | Allow \(\text{Var}(T)+\text{Var}(S)-2\text{Cov}(T,S)\) provided Cov obtained from \(E(TS)-E(T)E(S)\) even if incorrectly. | |
| \(\text{Var}(T – S) = 0.85\) | A1 | Var = 0.86 – 0.01 = 0.85 (M1A1) |
| [4] |
**(i)** Use independence to obtain equation in a and/or b
eg 0.4(a + 0.08)=0.08, a=(a+b+0.18)(a+0.08)
0.18+2(b+0.12)+0.8=1.4(0.3+2b+0.4)
Use independence or Σp=1 or P(T=1)=0.6 to obtain 2nd equation. eg a+0.58+b=1 or above
Correct simplified equation eg 0.4a=0.048, a+b=0.42, 0.24=0.8b
2nd correct simplified equation
a=0.12 AG
b=0.3 | M1, M1, A1, A1, A1, A1 | $P(S=0)=0.08/0.4$ or $P(S=2)=0.2/0.4$, $P(S=0)=0.2$, $P(S=2)=0.5$, $(P(T=1)=0.6)$, a="0.6"x"0.2" or b="0.6"x"0.5" | $P(S=0)=0.2$, $a=0.12$ AG A1, b=0.3 A1
| [6] |
**(ii)** $P(T = 2, S = 1) + P(T = 1, S = 0)$
$= 0.12 + 0.12 = 0.24$ | M1, A1 |
| [2] |
## Question 6 (continued):
**(iii)** $\text{Var}(T) + \text{Var}(S)$
$\text{Var}(T)= 0.6 + 4 \times 0.4 – (0.6 + 0.8)^2$ | M1 | ($\text{Var}(T)=0.24$, $\text{Var}(S)=0.61$)
| | | Allow M1 for sum of vars, even if $E(T^2)$ and/or $E(S^2)$ only used.(2.2,2.3)
$\text{Var}(S) = 0.3 + 4 \times 0.5 – (0.3 + 1)^2$ | M1 | $E(T – S) = 0.1 [E([T – S]^2)]$
$= 0.86$ | Allow $\text{Var}(T)+\text{Var}(S)-2\text{Cov}(T,S)$ provided Cov obtained from $E(TS)-E(T)E(S)$ even if incorrectly.
$\text{Var}(T – S) = 0.85$ | A1 | Var = 0.86 – 0.01 = 0.85 (M1A1)
| [4] |6 The random variables $S$ and $T$ are independent and have joint probability distribution given in the table.
\begin{center}
\begin{tabular}{ c | c | c c c | }
& \multicolumn{4}{c}{$S$} \\
\cline { 2 - 5 }
& & 0 & 1 & 2 \\
\cline { 2 - 5 }
& 1 & $a$ & 0.18 & $b$ \\
2 & 0.08 & 0.12 & 0.20 & \\
\cline { 2 - 5 }
& & & & \\
\cline { 2 - 5 }
\end{tabular}
\end{center}
(i) Show that $a = 0.12$ and find the value of $b$.\\
(ii) Find $\mathrm { P } ( T - S = 1 )$.\\
(iii) Find $\operatorname { Var } ( T - S )$.
\hfill \mbox{\textit{OCR S4 2012 Q6 [12]}}