| Exam Board | WJEC |
|---|---|
| Module | Further Unit 2 (Further Unit 2) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Basic sum of two Poissons |
| Difficulty | Standard +0.3 This is a straightforward application of standard Poisson distribution properties with minimal problem-solving required. Part (a) uses the well-known result that sum of independent Poissons is Poisson; part (b)(i) is direct recall of exponential distribution properties; part (b)(ii) uses the memoryless property; part (c) is a routine normal approximation to binomial. All parts are textbook exercises requiring recall rather than insight, though the multi-part structure and Further Maths context place it slightly above average difficulty. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.03a Continuous random variables: pdf and cdf |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | Total number of baskets, \(T\) is \(Po((2.1 + 1.9) \times 4)\) or \(Po(16)\) or \(Po(2.1 \times 4 + 1.9 \times 4)\) | M2 |
| \(P(T = 20) = \frac{16^{20} \times e^{-16}}{20!}\) | m1 | Dependent on M2 Use of formula or calculator |
| \(= 0.0559\) | A1 | cao |
| 3(b)(i) | Exponential distribution Mean time between baskets = standard deviation \(= \frac{1}{2.1} \times 12 = 5.7\) minutes. | B1 M1 A1 |
| 3(b)(ii) | \(P(\text{Klay doesn't score for the rest of the quarter}) = e^{-(1.9 \times 0.75)}\) | M1 |
| \(= 0.2405\) | A1 | |
| Alternative solution \(\lambda = 1.425\) \(P(X = 0) = 0.2405\) | (M1) (A1) | M1 for \(Po(1.9 \times 0.75)\) SC1 for \((e^{-2.1 \times 0.75}) = 0.207\) |
| 3(c) | Let \(F\) be the number of free throws he misses. \(F \sim B(530, 0.04)\) | |
| \(P(F > 25) = 1 - P(F \leq 25) = 0.169(1214 \ldots)\) | M1 A1 | |
| Total [11] |
3(a) | Total number of baskets, $T$ is $Po((2.1 + 1.9) \times 4)$ or $Po(16)$ or $Po(2.1 \times 4 + 1.9 \times 4)$ | M2 | M1 for Poisson and adding. M1 for multiplying by 4.
| $P(T = 20) = \frac{16^{20} \times e^{-16}}{20!}$ | m1 | Dependent on M2 Use of formula or calculator
| $= 0.0559$ | A1 | cao
3(b)(i) | Exponential distribution Mean time between baskets = standard deviation $= \frac{1}{2.1} \times 12 = 5.7$ minutes. | B1 M1 A1 | Must be clear that 5.7... is mean AND standard deviation
3(b)(ii) | $P(\text{Klay doesn't score for the rest of the quarter}) = e^{-(1.9 \times 0.75)}$ | M1 |
| $= 0.2405$ | A1 |
| Alternative solution $\lambda = 1.425$ $P(X = 0) = 0.2405$ | (M1) (A1) | M1 for $Po(1.9 \times 0.75)$ SC1 for $(e^{-2.1 \times 0.75}) = 0.207$
3(c) | Let $F$ be the number of free throws he misses. $F \sim B(530, 0.04)$ | |
| $P(F > 25) = 1 - P(F \leq 25) = 0.169(1214 \ldots)$ | M1 A1 |
| **Total [11]** |
---3. Two basketball players, Steph and Klay, score baskets at random at a rate of $2 \cdot 1$ and $1 \cdot 9$ respectively per quarter of a game. Assume that baskets are scored independently, and that Steph and Klay each play all four quarters of the game.
\begin{enumerate}[label=(\alph*)]
\item Stating the model that you are using, find the probability that they will score a combined total of exactly 20 baskets in a randomly selected game.
\item A quarter of a game lasts 12 minutes.
\begin{enumerate}[label=(\roman*)]
\item State the distribution of the time between baskets for Steph. Give the mean and standard deviation of this distribution.
\item Given that Klay scores at the end of the third minute in a quarter of a game, find the probability that Klay doesn't score for the rest of the quarter.
\end{enumerate}\item When practising, Klay misses $4 \%$ of the free throws he takes. One week he takes 530 free throws. Calculate the probability that he misses more than 25 free throws.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 2 2022 Q3 [11]}}