| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Moderate -0.3 This is a straightforward application of binomial distribution formulas with clearly given probabilities. Students must identify the correct p-values from the table (including combining categories for part (iii)) and apply standard binomial probability calculations. While it requires careful reading and multiple calculations, all steps are routine S1 techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.01a Permutations and combinations: evaluate probabilities |
| Number of vehicles registered | 0 | 1 | 2 | \(\geqslant 3\) |
| Percentage of households | 18 | 47 | 25 | 10 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim B(30, 0.18)\) | M1 | Correct binomial stated or implied |
| \(P(X = 3) = \binom{30}{3}(0.18)^3(0.82)^{27}\) | M1 | Correct expression |
| \(= 0.1847\) (awrt \(0.185\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Y \sim B(30, 0.10)\) | M1 | Correct \(p = 0.10\) used |
| \(P(Y \leq 5) = 0.9268\) (awrt \(0.927\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(W \sim B(30, 0.35)\), \(P(W > 10) = 1 - P(W \leq 10)\) | M1 | Correct \(p = 0.35\) |
| \(= 1 - 0.7304\) | M1 | Correct method |
| \(= 0.2696\) (awrt \(0.270\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(5 < X_2 < 10)\) where \(X_2 \sim B(30, 0.25)\) | M1 | Correct \(p = 0.25\) |
| \(= P(X_2 \leq 9) - P(X_2 \leq 5)\) | M1 | Correct differencing |
| \(= 0.8791 - 0.1935 = 0.6856\) (awrt \(0.686\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(p(\text{1 or 2 vehicles}) = 0.47 + 0.25 = 0.72\) | B1 | Correct probability identified |
| Mean \(= 150 \times 0.72 = 108\) | B1 | Correct mean and variance: Variance \(= 150 \times 0.72 \times 0.28 = 30.24\) |
# Question 5:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(30, 0.18)$ | M1 | Correct binomial stated or implied |
| $P(X = 3) = \binom{30}{3}(0.18)^3(0.82)^{27}$ | M1 | Correct expression |
| $= 0.1847$ (awrt $0.185$) | A1 | |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y \sim B(30, 0.10)$ | M1 | Correct $p = 0.10$ used |
| $P(Y \leq 5) = 0.9268$ (awrt $0.927$) | A1 | |
## Part (a)(iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $W \sim B(30, 0.35)$, $P(W > 10) = 1 - P(W \leq 10)$ | M1 | Correct $p = 0.35$ |
| $= 1 - 0.7304$ | M1 | Correct method |
| $= 0.2696$ (awrt $0.270$) | A1 | |
## Part (a)(iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(5 < X_2 < 10)$ where $X_2 \sim B(30, 0.25)$ | M1 | Correct $p = 0.25$ |
| $= P(X_2 \leq 9) - P(X_2 \leq 5)$ | M1 | Correct differencing |
| $= 0.8791 - 0.1935 = 0.6856$ (awrt $0.686$) | A1 | |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $p(\text{1 or 2 vehicles}) = 0.47 + 0.25 = 0.72$ | B1 | Correct probability identified |
| Mean $= 150 \times 0.72 = 108$ | B1 | Correct mean and variance: Variance $= 150 \times 0.72 \times 0.28 = 30.24$ |5 An analysis of the number of vehicles registered by each household within a city resulted in the following information.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Number of vehicles registered & 0 & 1 & 2 & $\geqslant 3$ \\
\hline
Percentage of households & 18 & 47 & 25 & 10 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item A random sample of 30 households within the city is selected.
Use a binomial distribution with $n = 30$, together with relevant information from the table in each case, to find the probability that the sample contains:
\begin{enumerate}[label=(\roman*)]
\item exactly 3 households with no registered vehicles;
\item at most 5 households with three or more registered vehicles;
\item more than 10 households with at least two registered vehicles;
\item more than 5 households but fewer than 10 households with exactly two registered vehicles.
\end{enumerate}\item If a random sample of $\mathbf { 1 5 0 }$ households within the city were to be selected, estimate the mean and the variance for the number of households in the sample that would have either one or two registered vehicles.\\[0pt]
[2 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{8aeacd54-a5a1-4f2d-b936-2faf635ffce7-16_1075_1707_1628_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2014 Q5 [13]}}