| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Find parameter from normal approximation |
| Difficulty | Challenging +1.2 This is a reverse normal approximation problem requiring students to set up two equations from given percentages, find z-scores, apply continuity correction, and solve simultaneous equations involving np and np(1-p). While it requires multiple steps and careful algebraic manipulation, the techniques are standard for Further Maths statistics and the problem structure is methodical rather than requiring novel insight. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem |
For at least one $z$ value. — M1
$\Phi(z) = 0.9599 \Rightarrow z = 1.75$ — A1 (both)
$\Phi(z) = 0.1056 \Rightarrow z = -1.25$ or $\Phi(-z) = 0.8944 \Rightarrow z = -1.25$
Use of $X \sim \text{N}(np,\ npq)$ (wherever it occurs) — M1
At least one of the equations: — M1
$\mu + 1.75\sigma = 55.5$ — A1$\checkmark$ (on a sign)
$\mu - 1.25\sigma = 37.5$ — M1A1
$\mu = 45$, $\sigma = 6$ — A1
$npq = 36$, $np = 45$ — M1
$q = 0.8$, $p = 0.2$ — M1
$n = 225$ — A1 [11]
Special Case: use of 37 and 56 – leading to $n = 420$; scores M1 A0 M1 A1F A1F M1 M1 A0 on the final 8 marks.
**[Total: 11]**10 A box contains a large number, $n$, of identical dice, which are thought to be biased. The probability that one of these dice will show a six in a single roll is $p$. The $n$ dice are rolled many times and the number of sixes obtained in each trial is recorded. In $4.01 \%$ of these trials 56 or more dice showed a six. In $10.56 \%$ of these trials 37 or fewer dice showed a six. Using a suitable normal approximation, find the values of $n$ and $p$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2010 Q10 [11]}}