| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Standard +0.3 This is a standard S2 binomial distribution question covering routine calculations (parts a-b), a straightforward hypothesis test (part c), and normal approximation to binomial (part d). All techniques are textbook applications with no novel problem-solving required, making it slightly easier than average for A-level maths overall, though typical for S2 module work. |
| Spec | 2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| (a) No. disapproving \(=X \sim B(10, 0.3)\) \(P(X \leq 4) = 0.850\) | B1 M1 A1 | |
| (b) \(P(X \leq 3) - P(X \leq 2) = 0.6496 - 0.3828 = 0.267\) | M1 A1 A1 | |
| (c) No. approving is \(X \sim B(20, p)\) \(H_0 : p = 0 \cdot 7\), \(H_1 : p < 0 \cdot 7\) | B1 B1 | |
| Under \(H_0\), \(P(X \leq 9) = 0.0171 < 5\%\) so reject \(H_0\), i.e. conclude that less than 70% actually do approve | M1 A1 A1 | |
| (d) No. of approvals is \(B(500, 0.45) \approx N(225, 123.75)\), so | M1 A1 | |
| \(P(X < 250) = P(X < 249.5) = P\left(Z < \frac{24.5}{11.12}\right) = P(Z < 2.20) = 0.986\) | M1 A1 A1 | Total: 18 marks |
(a) No. disapproving $=X \sim B(10, 0.3)$ $P(X \leq 4) = 0.850$ | B1 M1 A1 |
(b) $P(X \leq 3) - P(X \leq 2) = 0.6496 - 0.3828 = 0.267$ | M1 A1 A1 |
(c) No. approving is $X \sim B(20, p)$ $H_0 : p = 0 \cdot 7$, $H_1 : p < 0 \cdot 7$ | B1 B1 |
Under $H_0$, $P(X \leq 9) = 0.0171 < 5\%$ so reject $H_0$, i.e. conclude that less than 70% actually do approve | M1 A1 A1 |
(d) No. of approvals is $B(500, 0.45) \approx N(225, 123.75)$, so | M1 A1 |
$P(X < 250) = P(X < 249.5) = P\left(Z < \frac{24.5}{11.12}\right) = P(Z < 2.20) = 0.986$ | M1 A1 A1 | **Total: 18 marks**When a park is redeveloped, it is claimed that 70\% of the local population approve of the new
design. Assuming this to be true, find the probability that, in a group of 10 residents selected
at random,
\begin{enumerate}[label=(\alph*)]
\item 6 or more approve, [3 marks]
\item exactly 7 approve. [3 marks]
\end{enumerate}
A conservation group, however, carries out a survey of 20 people, and finds that only 9
approve.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Use this information to carry out a hypothesis test on the original claim, working at the
5\% significance level. State your conclusion clearly. [5 marks]
\end{enumerate}
If the conservationists are right, and only 45\% approve of the new park,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item use a suitable approximation to the binomial distribution to estimate the probability that in
a larger survey, of 500 people, less than half will approve. [7 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 Q6 [18]}}