| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Find parameter from normal approximation |
| Difficulty | Challenging +1.8 This question requires setting up two simultaneous equations from normal approximation conditions, applying continuity corrections correctly, using inverse normal tables, and solving a system involving both n and p. It goes beyond routine application of normal approximation by requiring parameter recovery rather than just probability calculation, demanding algebraic manipulation and understanding of the relationship between binomial parameters and the approximating normal distribution. |
| Spec | 2.04d Normal approximation to binomial5.05a Sample mean distribution: central limit theorem |
**Question 5**
$\dfrac{60.5 - \mu}{\sigma} = 2.083$, $\quad \dfrac{39.5 - \mu}{\sigma} = 1.417$ — B1M1, A1A1
$\mu = \mathbf{48.0}$, $\sigma = \mathbf{6}$ — M1A1 (Allow 3 sf, can be implied)
$\mu = np$, $\sigma = \sqrt{npq}$ — B1B1
$1 - p = 36 \div 48 = \dfrac{3}{4}$ — M1
$\Rightarrow \quad p = \mathbf{0.25}$, $n = \mathbf{192}$ — A1 ($p \in [0.248, 0.250]$), A1 (192 or 193, must be integer)
**[11]**5 Each year a college has a large fixed number, $n$, of places to fill. The probability, $p$, that a randomly chosen student comes from abroad is constant. Using a suitable normal approximation and applying a continuity correction, it is calculated that the probability of more than 60 students coming from abroad is 0.0187 and the probability of fewer than 40 students coming from abroad is 0.0783 . Find the values of $n$ and $p$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q5 [11]}}