| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate Var(aX+b) transformations |
| Difficulty | Moderate -0.3 This is a straightforward probability distribution question testing standard techniques: (a) is simple probability addition, (b) requires calculating E(X), E(X²), Var(X), then applying Var(aX+b)=a²Var(X) with arithmetic, and (c) uses the independence property Var(T+Y)=Var(T)+Var(Y). All are textbook procedures with no novel insight required, though the arithmetic in part (b) provides some computational work. Slightly easier than average due to routine application of formulas. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| \(x\) | 2 | 4 | 6 | 9 |
| P\((X = x)\) | 0.2 | 0.6 | 0.1 | 0.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | Obtains P(X ≤ 6) = 0.9 OE | 1.1b |
| Answer | Marks |
|---|---|
| 5(b) | Applies formula for E(X) |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct Var (X) | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| uses formula for E(3X + 2) | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| E(3X + 2) = 14.9 OE | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| E((3X + 2)2) – (E(3X + 2))2 | 1.1a | M1 |
| Correctly shows that Var (Y) = 32.49 | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| 5(c) | Obtains Var (T + Y) = 37.49 OE | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 7 | |
| Q | Marking Instructions | AO |
Question 5:
--- 5(a) ---
5(a) | Obtains P(X ≤ 6) = 0.9 OE | 1.1b | B1 | P(X ≤ 6) = 0.2 + 0.6 + 0.1 = 0.9
--- 5(b) ---
5(b) | Applies formula for E(X)
or
obtains the values of 3X + 2
PI by correct Var (X) | 1.1a | M1 | E(X) = 2 × 0.2 + 4 × 0.6 + 6 × 0.1 +
9 × 0.1 = 4.3
Var (X) = 22 × 0.2 + 42 × 0.6 + 62 ×
0.1 + 92 × 0.1 – 4.32
= 3.61
Var (Y) = Var (3X + 2) = 32 Var (X)
= 9 × 3.61 = 32.49
Uses the formula for Var (X) PI
or
uses formula for E(3X + 2) | 1.1a | M1
Obtains Var (X) = 3.61 OE
or
E(3X + 2) = 14.9 OE | 1.1b | A1
Applies Var (3X + 2) = 32 Var (X)
or
E((3X + 2)2) – (E(3X + 2))2 | 1.1a | M1
Correctly shows that Var (Y) = 32.49 | 2.1 | R1
--- 5(c) ---
5(c) | Obtains Var (T + Y) = 37.49 OE | 1.1b | B1 | Var (T + Y) = Var (T) + Var (Y)
= 5 + 32.49 = 37.49
Total | 7
Q | Marking Instructions | AO | Marks | Typical SolutionThe discrete random variable $X$ has the following probability distribution.
\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$x$ & 2 & 4 & 6 & 9 \\
\hline
P$(X = x)$ & 0.2 & 0.6 & 0.1 & 0.1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Find P$(X \leq 6)$
[1 mark]
\item Let $Y = 3X + 2$
Show that Var$(Y) = 32.49$
[5 marks]
\item The continuous random variable $T$ is independent of $Y$.
Given that Var$(T) = 5$, find Var$(T + Y)$
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics 2020 Q5 [7]}}