| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability of range of values |
| Difficulty | Moderate -0.8 This is a straightforward application of binomial distribution calculations requiring only direct use of cumulative probability tables or calculator functions. All parts involve standard probability computations (P(X≤15), P(X>10), range probabilities) with no conceptual challenges beyond recognizing when to use binomial distribution and basic probability manipulation. Part (b) adds minimal complexity by asking students to state the independence condition. This is easier than average A-level content as it's purely procedural with no problem-solving or 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X \leq 15)\) where \(X \sim B(40, 0.35)\) | M1 | Using correct distribution |
| \(= 0.6rehearsal\) ... reading from tables | A1 | |
| \(= 0.6213\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X > 10) = 1 - P(X \leq 10)\) | M1 | |
| \(= 1 - 0.2376\) | A1 | |
| \(= 0.7624\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(12 < X < 18) = P(X \leq 17) - P(X \leq 12)\) | M1 | Correct structure |
| \(= 0.9005 - 0.4results...\) | A1 | |
| \(= 0.4641\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Mean \(= np = 40 \times 0.35 = 14\) | B1 | Correct mean stated |
| \(P(X = 14) = P(X \leq 14) - P(X \leq 13)\) | M1 | |
| \(= 0.5490 - 0.4results...\) | A1 | |
| \(= 0.1327\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Necessary condition: households are selected independently / each household equally likely to be selected | B1 | Accept equivalent statement about random/independent sampling |
| \(Y \sim B(50, 0.70)\) | M1 | Correct distribution stated or implied |
| \(P(Y < 30) = P(Y \leq 29)\) | M1 | Correct inequality |
| Using \(W \sim B(50, 0.30)\): \(P(W > 20) = 1 - P(W \leq 20)\) | DM1 | Correct complement/complementary probability used |
| \(= 1 - 0.9522\) | A1 | |
| \(= 0.0478\) | A1 | cao |
# Question 6:
## Part (a)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X \leq 15)$ where $X \sim B(40, 0.35)$ | M1 | Using correct distribution |
| $= 0.6rehearsal$ ... reading from tables | A1 | |
| $= 0.6213$ | A1 | cao |
## Part (a)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X > 10) = 1 - P(X \leq 10)$ | M1 | |
| $= 1 - 0.2376$ | A1 | |
| $= 0.7624$ | A1 | cao |
## Part (a)(iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(12 < X < 18) = P(X \leq 17) - P(X \leq 12)$ | M1 | Correct structure |
| $= 0.9005 - 0.4results...$ | A1 | |
| $= 0.4641$ | A1 | cao |
## Part (a)(iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Mean $= np = 40 \times 0.35 = 14$ | B1 | Correct mean stated |
| $P(X = 14) = P(X \leq 14) - P(X \leq 13)$ | M1 | |
| $= 0.5490 - 0.4results...$ | A1 | |
| $= 0.1327$ | A1 | cao |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Necessary condition: households are selected independently / each household equally likely to be selected | B1 | Accept equivalent statement about random/independent sampling |
| $Y \sim B(50, 0.70)$ | M1 | Correct distribution stated or implied |
| $P(Y < 30) = P(Y \leq 29)$ | M1 | Correct inequality |
| Using $W \sim B(50, 0.30)$: $P(W > 20) = 1 - P(W \leq 20)$ | DM1 | Correct complement/complementary probability used |
| $= 1 - 0.9522$ | A1 | |
| $= 0.0478$ | A1 | cao |6
\begin{enumerate}[label=(\alph*)]
\item In a particular country, 35 per cent of the population is estimated to have at least one mobile phone.
A sample of 40 people is selected from the population.\\
Use the distribution $\mathrm { B } ( 40,0.35 )$ to estimate the probability that the number of people in the sample that have at least one mobile phone is:
\begin{enumerate}[label=(\roman*)]
\item at most 15 ;
\item more than 10 ;
\item more than 12 but fewer than 18 ;
\item exactly equal to the mean of the distribution.
\end{enumerate}\item In the same country, 70 per cent of households have a landline telephone connection.
A sample of 50 households is selected from all households in the country.\\
Stating a necessary condition regarding this selection, estimate the probability that fewer than 30 households have a landline telephone connection.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2015 Q6 [13]}}