| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for sampling distribution |
| Difficulty | Moderate -0.8 This is a straightforward application of the Central Limit Theorem with no complications. Part (a) requires stating that the sample mean is approximately N(90, 5²/100) by CLT, and part (b) is a routine normal probability calculation with standardization. The question involves direct recall and standard procedures with no problem-solving insight required, making it easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim N(q^2, \frac{s^2}{n})\) i.e. \(N_3(q, 0.25)\) | M1A1 B1 | Application of central limit theorem as (single large) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\bar{X} > 91) = 1 - \Phi\left(Z < \frac{q1-q}{0.5}\right)\) | M1A1 | Stated. |
| \(= 1 - \Phi(Z \leq 22)\) | ||
| \(= 1 - 0.9772\) | A1 | |
| \(= 0.0228\) | awrt 0.5228 |
## Part (c)
$X \sim N(q^2, \frac{s^2}{n})$ i.e. $N_3(q, 0.25)$ | M1A1 B1 | Application of central limit theorem as (single large)
## Part (u)
$P(\bar{X} > 91) = 1 - \Phi\left(Z < \frac{q1-q}{0.5}\right)$ | M1A1 | Stated.
$= 1 - \Phi(Z \leq 22)$ | |
$= 1 - 0.9772$ | A1 |
$= 0.0228$ | | awrt 0.5228
---A report on the health and nutrition of a population stated that the mean height of three-year old children is 90 cm and the standard deviation is 5 cm. A sample of 100 three-year old children was chosen from the population.
\begin{enumerate}[label=(\alph*)]
\item Write down the approximate distribution of the sample mean height. Give a reason for your answer. [3]
\item Hence find the probability that the sample mean height is at least 91 cm. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2006 Q2 [6]}}