| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Probability of range of values |
| Difficulty | Moderate -0.3 This is a straightforward binomial distribution question requiring standard cumulative probability calculations and basic mean/variance formulas. Part (a) involves routine use of binomial tables or calculator functions, part (b) is direct application of the binomial formula, and part (c) requires simple recall of E(X)=np and Var(X)=np(1-p) followed by a basic comparison. While multi-part with several calculations, each step is standard S1 material with no problem-solving insight required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(R \leq 7) = 0.924 \text{ to } 0.925\) | M1, A1 | Used somewhere in (a); may be implied. AWFW (0.92466) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(R \geq 11) = 1 – P(R \leq 10) = 1 – (0.9989 \text{ or } 0.9999) = 0.0011\) | M1, A1 | Requires '1 −' and \(\geq 4\) dp accuracy. AWRT (0.001106) |
| Answer | Marks | Guidance |
|---|---|---|
| \(B(14, 0.35)\) expressions stated for at least 3 terms within \(4 \leq R \leq 11\) gives probability \(= 0.353 \text{ to } 0.354\) | M1, M1, A1, (M1), (A2) | Accept 3 dp accuracy. \(p_2 – p_1 \Rightarrow\) M0 M0 A0. \((1 – p_2) – p_1 \Rightarrow\) M0 M0 A0. \(p_1 – (1 – p_2) \Rightarrow\) M1 M0 A0 only providing result > 0. Can be implied by correct answer. AWFW (0.35346) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(R = 4) = \binom{21}{4}(0.35)^7(0.65)^{17} = 0.059 \text{ to } 0.0595\) | M1, A1, A1 | Implied from correct stated formula; do not accept misreads. Can be implied by a correct answer. Ignore any additional terms. AWFW (0.059274) |
| Answer | Marks | Guidance |
|---|---|---|
| Variance = \(np(1 - p) = 7 \times 5/7 \times 2/7 = 10/7 \text{ or } 1.42 \text{ to } 1.43\) | B1, B1 | CAO. Must clearly state variance value if standard deviation (also) stated. CAO / AWFW |
| Answer | Marks | Guidance |
|---|---|---|
| Barry's claim appears/is sound/valid/correct/likely | B1dep, B1dep | Must have scored B1 B1 in (i) or B1 B0 plus 10/7 v 1.5 or \(\sqrt{10/7} \vee \sqrt{1.5}\) stated. Must have scored previous B1dep |
## Part (a)(i)
$R \sim B(14, 0.35)$
$P(R \leq 7) = 0.924 \text{ to } 0.925$ | M1, A1 | Used somewhere in (a); may be implied. AWFW (0.92466) | 2
## Part (a)(ii)
$P(R \geq 11) = 1 – P(R \leq 10) = 1 – (0.9989 \text{ or } 0.9999) = 0.0011$ | M1, A1 | Requires '1 −' and $\geq 4$ dp accuracy. AWRT (0.001106) | 2
## Part (a)(iii)
$P(5 < R < 10) = 0.9940 \text{ or } 0.9989$ ($p_1$)
minus 0.6405 or 0.4227 ($p_2$)
$= 0.353 \text{ to } 0.354$
or
$B(14, 0.35)$ expressions stated for at least 3 terms within $4 \leq R \leq 11$ gives probability $= 0.353 \text{ to } 0.354$ | M1, M1, A1, (M1), (A2) | Accept 3 dp accuracy. $p_2 – p_1 \Rightarrow$ M0 M0 A0. $(1 – p_2) – p_1 \Rightarrow$ M0 M0 A0. $p_1 – (1 – p_2) \Rightarrow$ M1 M0 A0 only providing result > 0. Can be implied by correct answer. AWFW (0.35346) | 3
## Part (b)
$R \sim B(21, 0.35)$
$P(R = 4) = \binom{21}{4}(0.35)^7(0.65)^{17} = 0.059 \text{ to } 0.0595$ | M1, A1, A1 | Implied from correct stated formula; do not accept misreads. Can be implied by a correct answer. Ignore any additional terms. AWFW (0.059274) | 3
## Part (c)(i)
$S \sim B(7, 5/7)$
Mean = $np = 7 \times 5/7 = 5$
If not identified, assume order is $\mu$ then $\sigma^2$
Variance = $np(1 - p) = 7 \times 5/7 \times 2/7 = 10/7 \text{ or } 1.42 \text{ to } 1.43$ | B1, B1 | CAO. Must clearly state variance value if standard deviation (also) stated. CAO / AWFW | 2
## Part (c)(ii)
Means are the same and (both comparisons clearly stated). Variances/standard deviations are similar. Do not accept statements involving correct/incorrect/exact/etc.
Barry's claim appears/is sound/valid/correct/likely | B1dep, B1dep | Must have scored B1 B1 in (i) or B1 B0 plus 10/7 v 1.5 or $\sqrt{10/7} \vee \sqrt{1.5}$ stated. Must have scored previous B1dep | 2
---6 During the winter, the probability that Barry's cat, Sylvester, chooses to stay outside all night is 0.35 , and the cat's choice is independent from night to night.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that, during a period of 2 weeks ( 14 nights) in winter, Sylvester chooses to stay outside:
\begin{enumerate}[label=(\roman*)]
\item on at most 7 nights;
\item on at least 11 nights;
\item on more than 5 nights but fewer than 10 nights.
\end{enumerate}\item Calculate the probability that, during a period of $\mathbf { 3 }$ weeks in winter, Sylvester chooses to stay outside on exactly 4 nights.
\item Barry claims that, during the summer, the number of nights per week, $S$, on which Sylvester chooses to stay outside can be modelled by a binomial distribution with $n = 7$ and $p = \frac { 5 } { 7 }$.
\begin{enumerate}[label=(\roman*)]
\item Assuming that Barry's claim is correct, find the mean and the variance of $S$.
\item For a period of 13 weeks during the summer, the number of nights per week on which Sylvester chose to stay outside had a mean of 5 and a variance of 1.5 .
Comment on Barry's claim.\\
(2 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S1 2010 Q6 [14]}}