| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Derive binomial mean and variance |
| Difficulty | Moderate -0.3 This is a structured multi-part question requiring standard variance derivation using E(X(X-1)), then solving simultaneous equations, and applying normal approximation with continuity correction. All techniques are routine S3 material with clear signposting and no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02d Binomial: mean np and variance np(1-p) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(X) = E(X^2) - [E(X)]^2\) | M1 | Used; may be implied |
| \(= E[X(X-1)] + E(X) - [E(X)]^2 = n(n-1)p^2 + np - n^2p^2\) | M1 | Rearranging & substitution |
| \(= np - np^2 = np(1-p)\) | A1 | 3 |
| OR: \(E[X(X-1)] = E(X^2) - E(X) = n(n-1)p^2 = n^2p^2 - np^2\) | (M1) | Expansion & substitution |
| \(\text{Var}(X) = E(X^2)-[E(X)]^2 = \{n^2p^2 - np^2 + E(X)\} - n^2p^2\) | (M1) | Used; may be implied |
| \(= np - np^2 = np(1-p)\) | (A1) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mean \(= np = 36\), \(\text{SD} = \sqrt{np(1-p)} = 4.8\) | B1 | Both CAO |
| \(36(1-p) = 4.8^2\) | M1 | Attempt to solve for \(p\) or \(n\) |
| \(n = 100\) & \(p = 0.36\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(30 < X < 40) = P\!\left(Z < \dfrac{39.5-36}{4.8}\right) - P\!\left(Z < \dfrac{30.5-36}{4.8}\right)\) | M1 | Standardising (39.5, 40 or 40.5) or (29.5, 30 or 30.5) with 36 and 4.8 and/or \((36-x)\) |
| B1 | Use of 39.5 & 30.5 | |
| \(P(Z<0.73) - P(Z<-1.15)\) | ||
| \(= P(Z<0.73) - [1-P(Z<1.15)]\) | m1 | Area change |
| \(= 0.76730 - [1-(0.87286 \text{ to } 0.87493)]\) | ||
| \(= 0.64\) to \(0.643\) | A1 | 4 |
# Question 5:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(X) = E(X^2) - [E(X)]^2$ | M1 | Used; may be implied |
| $= E[X(X-1)] + E(X) - [E(X)]^2 = n(n-1)p^2 + np - n^2p^2$ | M1 | Rearranging & substitution |
| $= np - np^2 = np(1-p)$ | A1 | 3 | Or equivalent |
| **OR:** $E[X(X-1)] = E(X^2) - E(X) = n(n-1)p^2 = n^2p^2 - np^2$ | (M1) | Expansion & substitution |
| $\text{Var}(X) = E(X^2)-[E(X)]^2 = \{n^2p^2 - np^2 + E(X)\} - n^2p^2$ | (M1) | Used; may be implied |
| $= np - np^2 = np(1-p)$ | (A1) | 3 | Or equivalent |
**Part (b)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean $= np = 36$, $\text{SD} = \sqrt{np(1-p)} = 4.8$ | B1 | Both CAO |
| $36(1-p) = 4.8^2$ | M1 | Attempt to solve for $p$ or $n$ |
| $n = 100$ & $p = 0.36$ | A1 | 3 | Both CAO |
**Part (b)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(30 < X < 40) = P\!\left(Z < \dfrac{39.5-36}{4.8}\right) - P\!\left(Z < \dfrac{30.5-36}{4.8}\right)$ | M1 | Standardising (39.5, 40 or 40.5) or (29.5, 30 or 30.5) with 36 and 4.8 and/or $(36-x)$ |
| | B1 | Use of 39.5 & 30.5 |
| $P(Z<0.73) - P(Z<-1.15)$ | | |
| $= P(Z<0.73) - [1-P(Z<1.15)]$ | m1 | Area change |
| $= 0.76730 - [1-(0.87286 \text{ to } 0.87493)]$ | | |
| $= 0.64$ to $0.643$ | A1 | 4 | AWFW (0.64112) |
---5 The random variable $X$ has a binomial distribution with parameters $n$ and $p$.
\begin{enumerate}[label=(\alph*)]
\item Given that
$$\mathrm { E } ( X ) = n p \quad \text { and } \quad \mathrm { E } ( X ( X - 1 ) ) = n ( n - 1 ) p ^ { 2 }$$
find an expression for $\operatorname { Var } ( X )$.
\item Given that $X$ has a mean of 36 and a standard deviation of 4.8:
\begin{enumerate}[label=(\roman*)]
\item find values for $n$ and $p$;
\item use a distributional approximation to estimate $\mathrm { P } ( 30 < X < 40 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA S3 2009 Q5 [10]}}