CAIE S2 2024 November — Question 3 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2024
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeStandard unbiased estimates calculation
DifficultyModerate -0.8 Part (a) is direct application of standard formulas for unbiased estimates (sample mean and s²). Part (b) requires applying CLT to find a value from the sampling distribution of the mean using normal tables, which is routine S2 content. Both parts are straightforward textbook exercises with no problem-solving or novel insight required.
Spec5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance

3 The times, \(T\) minutes, taken by a random sample of 75 students to complete a test were noted. The results were summarised by \(\Sigma t = 230\) and \(\Sigma t ^ { 2 } = 930\).
  1. Calculate unbiased estimates of the population mean and variance of \(T\).
    You should now assume that your estimates from part (a) are the true values of the population mean and variance of \(T\).
  2. The times taken by another random sample of 75 students were noted, and the sample mean, \(\bar { T }\), was found. Find the value of \(a\) such that \(P ( \bar { T } > a ) = 0.234\).

Question 3(a):
AnswerMarks Guidance
AnswerMark Guidance
\(\bar{t} = \frac{230}{75}\) [= 3.0666... or 3.07 (3 sf)] [or 46/15]B1
\(s^2 = \frac{75}{74}(\frac{930}{75} - (\frac{230}{75})^2)\) or \(\frac{1}{74}(930 - 230^2/75)\)M1 Use of correct formula
= 3.0360... or 3.04 (3 sf) or = 337/111A1
Total: 3
Question 3(b):
AnswerMarks Guidance
AnswerMark Guidance
\([\Phi^{-1}(1 - 0.234)] = 0.726\)B1
\(\pm \frac{a - \text{'3.0667'}}{\sqrt{\text{'3.04'}/75}} = \pm\text{'0.726'}\)M1 Ft their 0.726 but must be a z value. Note using 0.766 is M0. Must have \(\sqrt{75}\)
\(a = 3.21\) (3 sf)A1 CWO
Total: 3
## Question 3(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{t} = \frac{230}{75}$ [= 3.0666... or 3.07 (3 sf)] [or 46/15] | B1 | |
| $s^2 = \frac{75}{74}(\frac{930}{75} - (\frac{230}{75})^2)$ or $\frac{1}{74}(930 - 230^2/75)$ | M1 | Use of correct formula |
| = 3.0360... or 3.04 (3 sf) or = 337/111 | A1 | |
| **Total: 3** | | |

## Question 3(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| $[\Phi^{-1}(1 - 0.234)] = 0.726$ | B1 | |
| $\pm \frac{a - \text{'3.0667'}}{\sqrt{\text{'3.04'}/75}} = \pm\text{'0.726'}$ | M1 | Ft their 0.726 but must be a z value. Note using 0.766 is M0. Must have $\sqrt{75}$ |
| $a = 3.21$ (3 sf) | A1 | CWO |
| **Total: 3** | | |
3 The times, $T$ minutes, taken by a random sample of 75 students to complete a test were noted. The results were summarised by $\Sigma t = 230$ and $\Sigma t ^ { 2 } = 930$.
\begin{enumerate}[label=(\alph*)]
\item Calculate unbiased estimates of the population mean and variance of $T$.\\

You should now assume that your estimates from part (a) are the true values of the population mean and variance of $T$.
\item The times taken by another random sample of 75 students were noted, and the sample mean, $\bar { T }$, was found.

Find the value of $a$ such that $P ( \bar { T } > a ) = 0.234$.
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2024 Q3 [6]}}