| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Identify distribution and parameters |
| Difficulty | Moderate -0.8 This is a straightforward S1 binomial distribution question requiring only direct application of standard formulas and calculator use. Parts (a)(i-iv) involve routine probability calculations and mean/variance formulas (np, np(1-p)), while part (b) tests basic understanding of binomial assumptions—all standard textbook material with no problem-solving or novel insight required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| \(B(16\) or \(25\) or \(40, 0.45)\) \(P(S = 3) = \binom{16}{3}(0.45)^3(0.55)^{13} = 0.021\) to \(0.022\) | M1 (A1) (A1) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(S < 10) = 0.3843\) or \(0.2424 = 0.242\) to \(0.243\) | B1 (B1) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(15 \leq S \leq 20) = 0.7870\) or \(0.6844\) (\(p_1\)); minus \(0.1326\) or \(0.2142\) (\(p_2\)) \(= 0.654\) to \(0.655\) | M1 (M1) (A1) | 3 |
| OR \(B(40, 0.45)\) expressions stated for at least 3 terms within \(14 \leq S \leq 20\) gives probability \(= 0.654\) to \(0.655\) | M1 (A2) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Mean, \(\mu = np = 50 \times 0.45 = 22.5\) or \(22\frac{1}{2}\); Variance, \(\sigma^2 = np(1 - p) = 50 \times 0.45 \times 0.55 = 12.3\) to \(12.4\); Accept \(12\frac{3}{8}\) or \(\frac{99}{8}\) | B1 (B1) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Non-independence of senior citizens travel; Senior citizens tend to travel in pairs/groups | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7.15 am is outside 9.30 am to 11.30 am; Cannot use SCPs before 9.30 am; Cannot use SCPs @ 7.15 am; Cannot use SCPs during morning 'rush hour'; Value of \(p\) likely to be smaller/different/zero; Data not available; Senior citizens not out at this time; Passengers likely to be workers/school children | B1 | 1 |
**7(a)(i)**
| $B(16$ or $25$ or $40, 0.45)$ $P(S = 3) = \binom{16}{3}(0.45)^3(0.55)^{13} = 0.021$ to $0.022$ | M1 (A1) (A1) | 3 | Used at least once in (d)(i) to (iii); May be implied by correct answer; Ignore any additional terms; AWFW (0.0215) |
**7(a)(ii)**
| $P(S < 10) = 0.3843$ or $0.2424 = 0.242$ to $0.243$ | B1 (B1) | 2 | Accept 3 dp accuracy from tables or calculation; AWFW (0.2424) |
**7(a)(iii)**
| $P(15 \leq S \leq 20) = 0.7870$ or $0.6844$ ($p_1$); minus $0.1326$ or $0.2142$ ($p_2$) $= 0.654$ to $0.655$ | M1 (M1) (A1) | 3 | Accept 3 dp accuracy; $p_2 - p_1 ⟹ M0$ M0 A0; $p_1 - (1 - p_2) ⟹ M1$ M0 A0; Accept 3 dp accuracy / truncation; AWFW (0.6544) |
| OR $B(40, 0.45)$ expressions stated for at least 3 terms within $14 \leq S \leq 20$ gives probability $= 0.654$ to $0.655$ | M1 (A2) | 3 | Or implied by a correct answer; AWFW |
**7(a)(iv)**
| Mean, $\mu = np = 50 \times 0.45 = 22.5$ or $22\frac{1}{2}$; Variance, $\sigma^2 = np(1 - p) = 50 \times 0.45 \times 0.55 = 12.3$ to $12.4$; Accept $12\frac{3}{8}$ or $\frac{99}{8}$ | B1 (B1) | 2 | CAO (22.5 = 22 or 23 ⟹ ISW); AWFW (12.375) |
**7(b)(i)**
| Non-independence of senior citizens travel; Senior citizens tend to **travel in pairs/groups** | B1 | 1 | Or equivalent; but must be a clear indication of non-independent events |
**7(b)(ii)**
| 7.15 am is outside 9.30 am to 11.30 am; Cannot use SCPs before 9.30 am; Cannot use SCPs @ 7.15 am; Cannot use SCPs during morning 'rush hour'; Value of $p$ likely to be smaller/different/zero; Data not available; Senior citizens not out at this time; Passengers likely to be workers/school children | B1 | 1 | Or equivalent; Accept other sensible reasons; Distribution of **types of passenger** different |
**Total for Q7: 12 marks**
**Total Paper: 75 marks**7 The proportion of passengers who use senior citizen bus passes to travel into a particular town on 'Park \& Ride' buses between 9.30 am and 11.30 am on weekdays is 0.45 .
It is proposed that, when there are $n$ passengers on a bus, a suitable model for the number of passengers using senior citizen bus passes is the distribution $\mathrm { B } ( n , 0.45 )$.
\begin{enumerate}[label=(\alph*)]
\item Assuming that this model applies to the 10.30 am weekday 'Park \& Ride' bus into the town:
\begin{enumerate}[label=(\roman*)]
\item calculate the probability that, when there are $\mathbf { 1 6 }$ passengers, exactly 3 of them are using senior citizen bus passes;
\item determine the probability that, when there are $\mathbf { 2 5 }$ passengers, fewer than 10 of them are using senior citizen bus passes;
\item determine the probability that, when there are $\mathbf { 4 0 }$ passengers, at least 15 but at most 20 of them are using senior citizen bus passes;
\item calculate the mean and the variance for the number of passengers using senior citizen bus passes when there are $\mathbf { 5 0 }$ passengers.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Give a reason why the proposed model may not be suitable.
\item Give a different reason why the proposed model would not be suitable for the number of passengers using senior citizen bus passes to travel into the town on the 7.15 am weekday 'Park \& Ride' bus.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2009 Q7 [12]}}