| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 12 |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Find minimum/maximum n for probability condition |
| Difficulty | Standard +0.3 This is a standard normal approximation to binomial question with straightforward application of conditions, continuity correction, and inverse normal calculation. Part (iii) requires solving for N using the normal approximation, which is slightly more involved than routine but still follows a standard template. The question is slightly easier than average A-level difficulty as it's methodical application rather than requiring problem-solving insight. |
| Spec | 2.04d Normal approximation to binomial2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER: \(n\) is large (\(\geq 30\)); \(p\) is neither too large nor too small (\(0.1 \leq p \leq 0.9\)) | B1, B1 | |
| OR: \(np > 5\) and \(n(1-p) > 5\) (10 or other suitable value instead of 5 allowed) | B1, B1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= 50 \times \frac{1}{4} = 12.5\); S.D. \(= \sqrt{50 \times \frac{1}{4} \times \frac{3}{4}} = 3.062\) | B1, B1 | |
| \(z = \frac{17.5-12.5}{3.062} = 1.633\) | M1, A1 | |
| \(P(X > 17.5) = 0.0513\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Phi(z) = 0.99 \Rightarrow z = 2.326\) | B1 | |
| \(\frac{0.36N - 0.25N}{\sqrt{N \times \frac{3}{16}}} > 2.326\) (condone equality rather than inequality in working) | M1, A1 | |
| \(\Rightarrow N > \left(\frac{\sqrt{3} \times 2.326}{4 \times 0.11}\right)^2\) (condone equality) | M1 | |
| \(N = 84\) | A1 | 5 |
### (i)
EITHER: $n$ is large ($\geq 30$); $p$ is neither too large nor too small ($0.1 \leq p \leq 0.9$) | B1, B1 |
OR: $np > 5$ and $n(1-p) > 5$ (10 or other suitable value instead of 5 allowed) | B1, B1 | **2**
### (ii)
Mean $= 50 \times \frac{1}{4} = 12.5$; S.D. $= \sqrt{50 \times \frac{1}{4} \times \frac{3}{4}} = 3.062$ | B1, B1 |
$z = \frac{17.5-12.5}{3.062} = 1.633$ | M1, A1 |
$P(X > 17.5) = 0.0513$ | A1 | **5**
### (iii)
$\Phi(z) = 0.99 \Rightarrow z = 2.326$ | B1 |
$\frac{0.36N - 0.25N}{\sqrt{N \times \frac{3}{16}}} > 2.326$ (condone equality rather than inequality in working) | M1, A1 |
$\Rightarrow N > \left(\frac{\sqrt{3} \times 2.326}{4 \times 0.11}\right)^2$ (condone equality) | M1 |
$N = 84$ | A1 | **5**\begin{enumerate}[label=(\roman*)]
\item State briefly the conditions under which the binomial distribution $\text{B}(n, p)$ may be approximated by a normal distribution. [2]
\item A multiple-choice test has $50$ questions. Each question has four possible answers. A student passes the test if answering $36\%$ or more of the questions correctly. Using a suitable distributional approximation, estimate the probability that a student who selects answers to all the questions randomly will pass the test. [5]
\item A test similar to that in part (ii) has $N$ questions instead of $50$ questions. Estimate the least value of $N$ so that the probability that a student gets $36\%$ or more of the questions correct, by selecting answers to all questions randomly, will be less than $0.01$. (A continuity correction is not required in this part of the question.) [5]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q11 [12]}}