| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | November |
| Topic | Central limit theorem |
| Type | Justifying CLT for sampling distribution |
| Difficulty | Standard +0.3 This is a straightforward application of CLT to sampling distributions with standard parameter-finding from given mean/variance, followed by routine normal probability calculations. The multi-part structure and need to justify CLT usage adds slight complexity, but all steps follow textbook procedures without requiring novel insight or difficult algebraic manipulation. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
5 The random variable $X$ has a binomial distribution with parameters $n$ and $p$, where $p > 0.5$. A random sample of $4 n$ observations of $X$ is taken and $\bar { X }$ denotes the sample mean. It is given that $\mathrm { E } ( \bar { X } ) = 180$ and $\operatorname { Var } ( \bar { X } ) = 0.0225$.\\
(i) Find
\begin{enumerate}[label=(\alph*)]
\item the values of $p$ and $n$,
\item $\mathrm { P } ( \bar { X } < 179.8 )$,
\item the value of $a$ for which $\mathrm { P } ( 180 - a < \bar { X } < 180 + a ) = 0.99$, giving your answer correct to 2 decimal places.\\
(ii) State how you have used the Central Limit Theorem in part (i).
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q5}}