| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2019 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Verify conditions in context |
| Difficulty | Moderate -0.3 This is a straightforward S2 binomial distribution question requiring standard applications: stating an assumption (independence/constant probability), calculating basic probabilities using the binomial formula, and computing an expected value. Part (c) requires careful setup of the profit function but involves only routine calculation. Slightly easier than average due to the direct nature of all parts with no novel problem-solving required. |
| Spec | 5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.03c Calculate mean/variance: by integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Each person turns up independently/randomly of others | B1 | Must be in context containing the words in bold. Allow "turning up rather than not turning up" and vice versa |
| Probability/proportion/percentage of not turning up remains constant (5%) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim B(50, 0.05)\) | ||
| \(P(X=0) = 0.076944\ldots\) awrt 0.0769 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X=1) = 0.2794 - 0.0769 = 0.20248\ldots\) awrt 0.202/0.203 | M1A1 | M1 for correct use of tables or correct use of formula \(50(0.05)(0.95)^{49}\). Allow \(P(X \leq 1) - P(X=0)\) written or used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifying 3 possible total earnings: \(T = 880, 940, 1000\) with \(P(T=t) =\) "0.0769", "0.2025", \(1-(0.0769+0.2025)=[0.7206]\) | M1 | 1st M1 for identifying the 3 possible total earnings 880, 940, 1000 oe |
| \(E(T) = 880 \times \text{"0.0769"} + 940 \times \text{"0.2025"} + 1000 \times (1-(\text{"0.0769"}+\text{"0.2025"}))\) | dM1 | 2nd dM1 for a correct ft expression for \(E(T)\), e.g. \(880 \times \text{b(i)} + 940 \times \text{b(ii)} + 1000 \times (1-(\text{b(i)}+\text{b(ii)}))\) |
| \(= \pounds978.62\) awrt (£) 979 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(S\) = possible pay out; \(S = 60, 120\) with \(P(S=s) =\) "0.2025", "0.0769" | M1 | 1st M1 for identifying the 2 possible pay outs 120 and 60 oe |
| \(E(T) = 1000 - (120 \times \text{"0.0769"} + 60 \times \text{"0.2025"})\) | dM1 | 2nd dM1 for a correct ft expression for \(E(T)\) |
| \(= \pounds978.62\) awrt (£) 979 | A1 |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Each person **turns up** independently/randomly of others | B1 | Must be in context containing the words in bold. Allow "turning up rather than not turning up" and vice versa |
| Probability/proportion/percentage of not **turning up** remains constant (5%) | | |
### Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(50, 0.05)$ | | |
| $P(X=0) = 0.076944\ldots$ awrt **0.0769** | B1 | |
### Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X=1) = 0.2794 - 0.0769 = 0.20248\ldots$ awrt **0.202/0.203** | M1A1 | M1 for correct use of tables or correct use of formula $50(0.05)(0.95)^{49}$. Allow $P(X \leq 1) - P(X=0)$ written or used |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifying 3 possible total earnings: $T = 880, 940, 1000$ with $P(T=t) =$ "0.0769", "0.2025", $1-(0.0769+0.2025)=[0.7206]$ | M1 | 1st M1 for identifying the 3 possible total earnings 880, 940, 1000 oe |
| $E(T) = 880 \times \text{"0.0769"} + 940 \times \text{"0.2025"} + 1000 \times (1-(\text{"0.0769"}+\text{"0.2025"}))$ | dM1 | 2nd dM1 for a correct ft expression for $E(T)$, e.g. $880 \times \text{b(i)} + 940 \times \text{b(ii)} + 1000 \times (1-(\text{b(i)}+\text{b(ii)}))$ |
| $= \pounds978.62$ awrt (£) **979** | A1 | |
**Alternative method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $S$ = possible pay out; $S = 60, 120$ with $P(S=s) =$ "0.2025", "0.0769" | M1 | 1st M1 for identifying the 2 possible pay outs 120 and 60 oe |
| $E(T) = 1000 - (120 \times \text{"0.0769"} + 60 \times \text{"0.2025"})$ | dM1 | 2nd dM1 for a correct ft expression for $E(T)$ |
| $= \pounds978.62$ awrt (£) **979** | A1 | |
**Total: [7 marks]**\begin{enumerate}
\item A bus company sells tickets for a journey from London to Oxford every Saturday. Past records show that $5 \%$ of people who buy a ticket do not turn up for the journey.
\end{enumerate}
The bus has seats for 48 people.\\
Each week the bus company sells tickets to exactly 50 people for the journey.\\
The random variable $X$ represents the number of these people who do not turn up for the journey.\\
(a) State one assumption required to model $X$ as a binomial distribution.
For this week's journey find,\\
(b) (i) the probability that all 50 people turn up for the journey,\\
(ii) $\mathrm { P } ( X = 1 )$
The bus company receives $\pounds 20$ for each ticket sold and all 50 tickets are sold. It must pay out $\pounds 60$ to each person who buys a ticket and turns up for the journey but does not have a seat.\\
(c) Find the bus company's expected total earnings per journey.\\
\hfill \mbox{\textit{Edexcel S2 2019 Q1 [7]}}