| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Derive binomial mean and variance |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard S3 material. Part (a)(i) requires a routine proof of binomial mean using summation, (a)(ii) is algebraic manipulation using Var(X) = E(X²) - [E(X)]², and parts (iii-iv) involve solving simultaneous equations and applying normal approximation with continuity correction. Part (b) is a direct application of normal approximation. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(E(X) = \sum_{x=0}^{n} x \binom{n}{x} p^x(1-p)^{n-x}\) | M1 | Use of \(\sum x \times P(X=x)\) |
| \(= \sum_{x=1}^{n} \frac{n!}{(x-1)!(n-x)!} p^x(1-p)^{n-x}\) | M1 | Expansion of \(^nC_x\); cancelling of \(x\) (Ignore limits) |
| \(= np \times \sum_{x=1}^{n} \frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1}(1-p)^{n-x}\) | M1 | Factors of \(n\) and \(p\) (Ignore limits) |
| \(= np \times \sum P(X=x \mid B(n-1,p)) = np\) | M1 | AG; must be convincing |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\text{Var}(X) = E(X^2) - (E(X))^2\) | M1 | Used |
| \(= [E(X^2) - E(X)] + E(X) - (E(X))^2\) | m1 | Attempted |
| \(= n(n-1)p^2 + np - n^2p^2\) | ||
| \(= np(1-p)\) | A1 | AG; must be convincing |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(np(1-p) = 3(1-p) = 2.97\) | M1 | Substituting \(\mu\) in \(\sigma^2\) |
| \(1-p = \frac{2.97}{3} = 0.99\) | ||
| \(p = 0.01\) | A1 | CAO |
| \(n = 300\) | A1 | CAO |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(B(300, 0.01) \sim \text{Po}(3)\) | B1 | CAO; PI |
| \(P(X>2) = 1 - P(X \leq 2)\) | M1 | Must be applied to Poisson |
| \(= 1 - 0.4232 = 0.577\) | A1 | AWRT |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(Y \sim B(500, 0.45)\) or \(Y \sim \text{Normal}\) with mean \(\mu = 225\) | B1 | PI |
| variance \(\sigma^2 = 123.75\) or standard deviation \(\sigma = 11.124\) | B1 | AWFW 123 to 124 / AWFW 11.05 to 11.15 |
| At least half \(\Rightarrow (\geq) 250\) | B1 | CAO |
| \(P(Y_B \geq 250) = P(Y_N > 249.5)\) | B1 | CAO |
| \(P\left(Z > \frac{249.5 - 225}{\sqrt{123.75}}\right)\) | M1 | Standardising 249.5, 250 or 250.5 with \(\mu\) and \(\sqrt{\sigma^2}\) |
| \(P(Z > 2.20) = 1 - P(Z < 2.20)\) | m1 | Area change |
| \(= 0.0138\) to \(0.014\) | A1 | |
| Total | 7 | |
| Note: Use of \(\frac{0.5-0.45}{\sqrt{0.000495}} \Rightarrow\) max 5 marks | Use of distribution of \(\hat{p}\) | |
| Use of \(\frac{0.499-0.45}{\sqrt{0.000495}} \Rightarrow\) max 7 marks | Use of distribution of \(\hat{p}\) with continuity correction |
# Question 6(a)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $E(X) = \sum_{x=0}^{n} x \binom{n}{x} p^x(1-p)^{n-x}$ | M1 | Use of $\sum x \times P(X=x)$ |
| $= \sum_{x=1}^{n} \frac{n!}{(x-1)!(n-x)!} p^x(1-p)^{n-x}$ | M1 | Expansion of $^nC_x$; cancelling of $x$ (Ignore limits) |
| $= np \times \sum_{x=1}^{n} \frac{(n-1)!}{(x-1)!(n-x)!} p^{x-1}(1-p)^{n-x}$ | M1 | Factors of $n$ and $p$ (Ignore limits) |
| $= np \times \sum P(X=x \mid B(n-1,p)) = np$ | M1 | AG; must be convincing |
| **Total** | **4** | |
---
# Question 6(a)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\text{Var}(X) = E(X^2) - (E(X))^2$ | M1 | Used |
| $= [E(X^2) - E(X)] + E(X) - (E(X))^2$ | m1 | Attempted |
| $= n(n-1)p^2 + np - n^2p^2$ | | |
| $= np(1-p)$ | A1 | AG; must be convincing |
| **Total** | **3** | |
---
# Question 6(a)(iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $np(1-p) = 3(1-p) = 2.97$ | M1 | Substituting $\mu$ in $\sigma^2$ |
| $1-p = \frac{2.97}{3} = 0.99$ | | |
| $p = 0.01$ | A1 | CAO |
| $n = 300$ | A1 | CAO |
| **Total** | **3** | |
---
# Question 6(a)(iv):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $B(300, 0.01) \sim \text{Po}(3)$ | B1 | CAO; PI |
| $P(X>2) = 1 - P(X \leq 2)$ | M1 | Must be applied to Poisson |
| $= 1 - 0.4232 = 0.577$ | A1 | AWRT |
| **Total** | **3** | |
---
# Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $Y \sim B(500, 0.45)$ or $Y \sim \text{Normal}$ with mean $\mu = 225$ | B1 | PI |
| variance $\sigma^2 = 123.75$ or standard deviation $\sigma = 11.124$ | B1 | AWFW 123 to 124 / AWFW 11.05 to 11.15 |
| At least half $\Rightarrow (\geq) 250$ | B1 | CAO |
| $P(Y_B \geq 250) = P(Y_N > 249.5)$ | B1 | CAO |
| $P\left(Z > \frac{249.5 - 225}{\sqrt{123.75}}\right)$ | M1 | Standardising 249.5, 250 or 250.5 with $\mu$ and $\sqrt{\sigma^2}$ |
| $P(Z > 2.20) = 1 - P(Z < 2.20)$ | m1 | Area change |
| $= 0.0138$ to $0.014$ | A1 | |
| **Total** | **7** | |
| Note: Use of $\frac{0.5-0.45}{\sqrt{0.000495}} \Rightarrow$ max 5 marks | | Use of distribution of $\hat{p}$ |
| Use of $\frac{0.499-0.45}{\sqrt{0.000495}} \Rightarrow$ max 7 marks | | Use of distribution of $\hat{p}$ with continuity correction |
---6
\begin{enumerate}[label=(\alph*)]
\item The random variable $X$ has a binomial distribution with parameters $n$ and $p$.
\begin{enumerate}[label=(\roman*)]
\item Prove that $\mathrm { E } ( X ) = n p$.
\item Given that $\mathrm { E } \left( X ^ { 2 } \right) - \mathrm { E } ( X ) = n ( n - 1 ) p ^ { 2 }$, show that $\operatorname { Var } ( X ) = n p ( 1 - p )$.
\item Given that $X$ is found to have a mean of 3 and a variance of 2.97, find values for $n$ and $p$.
\item Hence use a distributional approximation to estimate $\mathrm { P } ( X > 2 )$.
\end{enumerate}\item Dressher is a nationwide chain of stores selling women's clothes. It claims that the probability that a customer who buys clothes from its stores uses a Dressher store card is 0.45 .
Assuming this claim to be correct, use a distributional approximation to estimate the probability that, in a random sample of 500 customers who buy clothes from Dressher stores, at least half of them use a Dressher store card.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2007 Q6 [20]}}