| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Statistics (Further Paper 3 Statistics) |
| Year | 2021 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Calculate E(X) from constructed distribution |
| Difficulty | Moderate -0.5 This is a straightforward probability distribution question requiring students to set up equations from verbal constraints (P(20)=2P(30) and P(20)=0.5P(10)), use ΣP=1 to solve for probabilities, then calculate E(X) and apply a linear transformation E(aX+b). All steps are routine with no novel insight required, making it slightly easier than average for Further Maths statistics. |
| Spec | 5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets up probability distribution table with \(P(X=x)\) as \(4a\), \(2a\), \(a\) for \(x = 10, 20, 30\) respectively | M1 (AO3.3) | Condone slips in representing relationships between probabilities |
| \(4a + 2a + a = 1\), \(7a = 1\), \(a = \frac{1}{7}\) | ||
| \(E(X) = 10 \times 4 \times \frac{1}{7} + 20 \times 2 \times \frac{1}{7} + 30 \times \frac{1}{7}\) | M1 (AO3.4) | Uses formula for mean of discrete random variable |
| Mean \(= 15.71\) points | A1 (AO3.2a) | AWRT 15.71 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(5X - 10) = 5E(X) - 10\) | M1 (AO3.1b) | Condone sign error; or adjusts discrete random variable and attempts to calculate mean |
| \(= 5 \times 15.71 - 10 = 69\) pence | A1F (AO3.2a) | AWRT 69; follow through their mean from part (a) |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets up probability distribution table with $P(X=x)$ as $4a$, $2a$, $a$ for $x = 10, 20, 30$ respectively | M1 (AO3.3) | Condone slips in representing relationships between probabilities |
| $4a + 2a + a = 1$, $7a = 1$, $a = \frac{1}{7}$ | | |
| $E(X) = 10 \times 4 \times \frac{1}{7} + 20 \times 2 \times \frac{1}{7} + 30 \times \frac{1}{7}$ | M1 (AO3.4) | Uses formula for mean of discrete random variable |
| Mean $= 15.71$ points | A1 (AO3.2a) | AWRT 15.71 |
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# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(5X - 10) = 5E(X) - 10$ | M1 (AO3.1b) | Condone sign error; or adjusts discrete random variable and attempts to calculate mean |
| $= 5 \times 15.71 - 10 = 69$ pence | A1F (AO3.2a) | AWRT 69; follow through their mean from part (a) |
---3 In a game, it is only possible to score 10, 20 or 30 points.
The probability of scoring 20 points is twice the probability of scoring 30 points.\\
The probability of scoring 20 points is half the probability of scoring 10 points.\\
3
\begin{enumerate}[label=(\alph*)]
\item Find the mean points scored when the game is played once, giving your answer to two decimal places.\\
3
\item Mina plays the game.\\
Her father, Michael, tells her that he will multiply her score by 5 and then subtract 10 He will then give her the value he has calculated in pence rounded to the nearest penny.
Calculate the expected value in pence that Mina receives.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Statistics 2021 Q3 [5]}}