| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2017 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Find parameter from normal approximation |
| Difficulty | Standard +0.8 This S2 question requires working backwards from a normal approximation probability to find the original binomial parameter n. Students must apply continuity correction, use inverse normal tables with P(Z < -2.4) = 0.0082, set up the equation (49.5 - n/6)/√(5n/36) = -2.4, and solve the resulting quadratic. This reverse-engineering with multiple interconnected steps and algebraic manipulation is notably harder than standard forward normal approximation problems. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| END |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(X \sim N\left(\frac{1}{6}n, \frac{5}{36}n\right)\) | M1A1 | M1 Using Normal with mean \(\frac{1}{6}n\); A1 mean and variance both correct |
| \(P(X < 50) = P\left(Z < \frac{49.5 - \frac{1}{6}n}{\sqrt{\frac{5}{36}n}}\right)\) | M1 dM1 | M1 for \(\pm\left(\frac{(48.5 \text{ or } 49 \text{ or } 49.5 \text{ or } 50 \text{ or } 50.5) - \text{their mean}}{\text{their sd}}\right)\); dM1 dep on previous M1 for continuity correction \(49 \pm 0.5\) or \(50 \pm 0.5\) |
| \(\frac{49.5 - \frac{1}{6}n}{\sqrt{\frac{5}{36}n}} = -2.4\) | M1 A1 | M1 setting expression \(= z\) value \(\ |
| \(49.5 - \frac{1}{6}n = -2.4\frac{\sqrt{5n}}{6}\) | Rearranging step | |
| \(n - 2.4\sqrt{5}\sqrt{n} - 297 = 0\) | M1 A1 | M1 rearranging to get a 3TQ in \(\sqrt{n}\) or \(n\); A1 correct 3TQ equation e.g. \(n - 2.4\sqrt{5}\sqrt{n} - 297 = 0\) |
| \(\sqrt{n} = \frac{2.4\sqrt{5} \pm \sqrt{(2.4\sqrt{5})^2 + 4 \times 297}}{2}\) \(= 9\sqrt{5}\) or awrt 20.1 | M1 | M1 solving (allow one slip) their 3TQ leading to \(\sqrt{n}=\) or \(n=\) |
| \(n = 405\) only | A1cao | A1 cao with all previous marks scored |
## Question 6:
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $X \sim N\left(\frac{1}{6}n, \frac{5}{36}n\right)$ | M1A1 | M1 Using Normal with mean $\frac{1}{6}n$; A1 mean and variance both correct |
| $P(X < 50) = P\left(Z < \frac{49.5 - \frac{1}{6}n}{\sqrt{\frac{5}{36}n}}\right)$ | M1 dM1 | M1 for $\pm\left(\frac{(48.5 \text{ or } 49 \text{ or } 49.5 \text{ or } 50 \text{ or } 50.5) - \text{their mean}}{\text{their sd}}\right)$; dM1 dep on previous M1 for continuity correction $49 \pm 0.5$ or $50 \pm 0.5$ |
| $\frac{49.5 - \frac{1}{6}n}{\sqrt{\frac{5}{36}n}} = -2.4$ | M1 A1 | M1 setting expression $= z$ value $\|z\| > 2$; A1 correct equation with compatible signs with $z$ value awrt 2.4 |
| $49.5 - \frac{1}{6}n = -2.4\frac{\sqrt{5n}}{6}$ | | Rearranging step |
| $n - 2.4\sqrt{5}\sqrt{n} - 297 = 0$ | M1 A1 | M1 rearranging to get a 3TQ in $\sqrt{n}$ or $n$; A1 correct 3TQ equation e.g. $n - 2.4\sqrt{5}\sqrt{n} - 297 = 0$ |
| $\sqrt{n} = \frac{2.4\sqrt{5} \pm \sqrt{(2.4\sqrt{5})^2 + 4 \times 297}}{2}$ $= 9\sqrt{5}$ or awrt 20.1 | M1 | M1 solving (allow one slip) their 3TQ leading to $\sqrt{n}=$ or $n=$ |
| $n = 405$ only | A1cao | A1 cao with all previous marks scored |
**Total: 10 marks**6. A fair 6 -sided die is thrown $n$ times. The number of sixes, $X$, is recorded. Using a normal approximation, $\mathrm { P } ( X < 50 ) = 0.0082$ correct to 4 decimal places.
Find the value of $n$.\\
(10)\\
\begin{center}
\end{center}
\begin{center}
\end{center}
\begin{center}
\begin{tabular}{|l|l|}
\hline
\hline
END & \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{Edexcel S2 2017 Q6 [10]}}