| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Proof and derivation of E(X) and Var(X) |
| Difficulty | Challenging +1.2 This is a multi-part S3 question combining standard proof of E(X) (routine algebraic manipulation of binomial series), straightforward variance derivation using given result, and mode inequalities requiring algebraic manipulation of binomial probability ratios. Part (c) involves standard Poisson approximation. While lengthy and requiring careful algebra, all techniques are standard S3 material with no novel insights needed—moderately above average difficulty due to the proof component and multi-step nature. |
| Spec | 2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(X) = \sum_{x=0}^{n} x \binom{n}{x} p^x (1-p)^{n-x}\) | M1 | Writing out sum, x=0 term vanishes |
| \(= \sum_{x=1}^{n} x \cdot \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x}\) | M1 | Cancelling \(x\) with \(x!\) to get \((x-1)!\), factoring out \(np\) |
| \(= np \sum_{x=1}^{n} \binom{n-1}{x-1} p^{x-1}(1-p)^{n-x} = np(p+(1-p))^{n-1} = np\) | A1 | Recognising remaining sum = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(X) = E(X(X-1)) + E(X) - [E(X)]^2\) | M1 | Using \(\text{Var}(X) = E(X^2)-[E(X)]^2\) and \(E(X^2)=E(X(X-1))+E(X)\) |
| \(= n(n-1)p^2 + np - n^2p^2 = np(1-p)\) | A1 | Correct simplified expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=m) \geqslant P(X=m-1)\) so \(\binom{n}{m}p^m(1-p)^{n-m} \geqslant \binom{n}{m-1}p^{m-1}(1-p)^{n-m+1}\) | M1 | Writing out both probabilities |
| Dividing: \(\frac{(n-m+1)p}{m(1-p)} \geqslant 1\) | M1 | Forming ratio |
| \((n-m+1)p \geqslant m(1-p)\) | M1 | Rearranging |
| \(np - mp + p \geqslant m - mp\), so \(m \leqslant (n+1)p\) | A1 | Completing proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| From inequalities: \((n+1)p - 1 \leqslant m \leqslant (n+1)p\) | M1 | Combining both results |
| For B(10, 0.65): \((n+1)p = 11 \times 0.65 = 7.15\), so \(6.15 \leqslant m \leqslant 7.15\), unique mode \(m=7\) | A1 | Correct deduction of single mode |
| For B(35, 0.5): \((n+1)p = 18\), so \(17 \leqslant m \leqslant 18\), modes are \(m = 17\) and \(m = 18\) | A1 | Both values stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y \sim B(4000, 0.00095)\), approximate by Poisson with \(\lambda = 4000 \times 0.00095 = 3.8\) | B1 | Correct Poisson approximation stated |
| Mode \(k\): \((n+1)p = 4001 \times 0.00095 = 3.80095\), so \(k = 3\) or use Poisson mode where \(\lambda\) is non-integer, \(k=3\) | M1 | Finding mode of approximating distribution |
| \(P(Y \leqslant 3) \approx e^{-3.8}\left(1 + 3.8 + \frac{3.8^2}{2!} + \frac{3.8^3}{3!}\right) \approx 0.4735\) | A1 | Correct probability |
# Question 7:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = \sum_{x=0}^{n} x \binom{n}{x} p^x (1-p)^{n-x}$ | M1 | Writing out sum, x=0 term vanishes |
| $= \sum_{x=1}^{n} x \cdot \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x}$ | M1 | Cancelling $x$ with $x!$ to get $(x-1)!$, factoring out $np$ |
| $= np \sum_{x=1}^{n} \binom{n-1}{x-1} p^{x-1}(1-p)^{n-x} = np(p+(1-p))^{n-1} = np$ | A1 | Recognising remaining sum = 1 |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X) = E(X(X-1)) + E(X) - [E(X)]^2$ | M1 | Using $\text{Var}(X) = E(X^2)-[E(X)]^2$ and $E(X^2)=E(X(X-1))+E(X)$ |
| $= n(n-1)p^2 + np - n^2p^2 = np(1-p)$ | A1 | Correct simplified expression |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=m) \geqslant P(X=m-1)$ so $\binom{n}{m}p^m(1-p)^{n-m} \geqslant \binom{n}{m-1}p^{m-1}(1-p)^{n-m+1}$ | M1 | Writing out both probabilities |
| Dividing: $\frac{(n-m+1)p}{m(1-p)} \geqslant 1$ | M1 | Forming ratio |
| $(n-m+1)p \geqslant m(1-p)$ | M1 | Rearranging |
| $np - mp + p \geqslant m - mp$, so $m \leqslant (n+1)p$ | A1 | Completing proof |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| From inequalities: $(n+1)p - 1 \leqslant m \leqslant (n+1)p$ | M1 | Combining both results |
| For B(10, 0.65): $(n+1)p = 11 \times 0.65 = 7.15$, so $6.15 \leqslant m \leqslant 7.15$, unique mode $m=7$ | A1 | Correct deduction of single mode |
| For B(35, 0.5): $(n+1)p = 18$, so $17 \leqslant m \leqslant 18$, modes are $m = 17$ and $m = 18$ | A1 | Both values stated |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y \sim B(4000, 0.00095)$, approximate by Poisson with $\lambda = 4000 \times 0.00095 = 3.8$ | B1 | Correct Poisson approximation stated |
| Mode $k$: $(n+1)p = 4001 \times 0.00095 = 3.80095$, so $k = 3$ or use Poisson mode where $\lambda$ is non-integer, $k=3$ | M1 | Finding mode of approximating distribution |
| $P(Y \leqslant 3) \approx e^{-3.8}\left(1 + 3.8 + \frac{3.8^2}{2!} + \frac{3.8^3}{3!}\right) \approx 0.4735$ | A1 | Correct probability |7
\begin{enumerate}[label=(\alph*)]
\item The random variable $X$ has a binomial distribution with parameters $n$ and $p$.
\begin{enumerate}[label=(\roman*)]
\item Prove, from first principles, that $\mathrm { E } ( X ) = n p$.
\item Hence, given that $\mathrm { E } ( X ( X - 1 ) ) = n ( n - 1 ) p ^ { 2 }$, find, in terms of $n$ and $p$, an expression for $\operatorname { Var } ( X )$.
\end{enumerate}\item The mode, $m$, of $X$ is such that
$$\mathrm { P } ( X = m ) \geqslant \mathrm { P } ( X = m - 1 ) \quad \text { and } \quad \mathrm { P } ( X = m ) \geqslant \mathrm { P } ( X = m + 1 )$$
\begin{enumerate}[label=(\roman*)]
\item Use the first inequality to show that
$$m \leqslant ( n + 1 ) p$$
\item Given that the second inequality results in
$$m \geqslant ( n + 1 ) p - 1$$
deduce that the distribution $\mathrm { B } ( 10,0.65 )$ has one mode, and find the two values for the mode of the distribution $B ( 35,0.5 )$.
\end{enumerate}\item The random variable $Y$ has a binomial distribution with parameters 4000 and 0.00095 .
Use a distributional approximation to estimate $\mathrm { P } ( Y \leqslant k )$, where $k$ denotes the mode of $Y$.\\
(3 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2012 Q7 [15]}}