| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | State distribution of sample mean |
| Difficulty | Moderate -0.8 This is a straightforward application of sampling distribution theory requiring only standard recall: recognizing that the sample mean follows a normal distribution with mean μ and standard deviation σ/√n, then performing a routine normal probability calculation using z-scores. No problem-solving insight or complex manipulation is needed—just direct application of formulas covered in any S3 course. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\bar{T} \sim N(28.5, \frac{7.2^2}{9}) = \sim N(28.5, 6.48)\) | M1 A1 | |
| (b) \(P(25 < \bar{T} < 30) = P(\frac{25-28.5}{\sqrt{6.48}} < Z < \frac{30-28.5}{\sqrt{6.48}})\) | M1 A1 | |
| \(= P(\text{−}1.37 < Z < 0.59) = 0.7224 − (1 − 0.9147) = 0.637\) | M1 A1 | (6) |
**(a)** $\bar{T} \sim N(28.5, \frac{7.2^2}{9}) = \sim N(28.5, 6.48)$ | M1 A1 |
**(b)** $P(25 < \bar{T} < 30) = P(\frac{25-28.5}{\sqrt{6.48}} < Z < \frac{30-28.5}{\sqrt{6.48}})$ | M1 A1 |
$= P(\text{−}1.37 < Z < 0.59) = 0.7224 − (1 − 0.9147) = 0.637$ | M1 A1 | (6)The length of time that registered customers spend on each visit to a supermarket's website is normally distributed with a mean of 28.5 minutes and a standard deviation of 7.2 minutes.
Eight visitors to the site are selected at random and the length of time, $T$ minutes, that each stays is recorded.
\begin{enumerate}[label=(\alph*)]
\item Write down the distribution of $\overline{T}$, the mean time spent at the site by these eight visitors. [2 marks]
\item Find $P(25 < \overline{T} < 30)$. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q2 [6]}}