CAIE Further Paper 4 2022 November — Question 1 6 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2022
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeRecover sample stats from CI
DifficultyChallenging +1.2 This question requires working backwards from a confidence interval to find summary statistics, involving understanding of the t-distribution formula and solving simultaneous equations. While conceptually more demanding than routine confidence interval calculation, it's a standard Further Statistics technique with clear algebraic steps once the setup is recognized.
Spec5.05d Confidence intervals: using normal distribution
1 A basketball club has a large number of players. The heights, \(x \mathrm {~m}\), of a random sample of 10 of these players are measured. A \(90 \%\) confidence interval for the population mean height, \(\mu \mathrm { m }\), of players in this club is calculated. It is assumed that heights are normally distributed. The confidence interval is \(1.78 \leqslant \mu \leqslant 2.02\). Find the values of \(\sum x\) and \(\sum x ^ { 2 }\) for this sample.
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(2.02 = \bar{x} + 1.833\sqrt{\frac{s^2}{10}}\) and \(1.78 = \bar{x} - 1.833\sqrt{\frac{s^2}{10}}\), Add: \(\bar{x} = \frac{2.02 + 1.78}{2} = 1.90\)M1 Add the two equations
\(\sum x = 19\)A1
Subtract: \(2.02 - 1.78 = 2 \times 1.833\sqrt{\frac{s^2}{10}}\)M1 Allow 1.372, 1.383, 1.812 instead of 1.833
\(s^2 = 0.042859\)A1 May be implied. \(s = 0.20702\)
\(s^2 = \frac{1}{9}\left(\sum x^2 - \frac{(\sum x)^2}{10}\right)\)M1
\(\sum x^2 = 9s^2 + \frac{19^2}{10} = 36.5\)A1 36.486
6Using 1.645: maximum M1A1 M0A0 M1A0 3/6 
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2.02 = \bar{x} + 1.833\sqrt{\frac{s^2}{10}}$ and $1.78 = \bar{x} - 1.833\sqrt{\frac{s^2}{10}}$, Add: $\bar{x} = \frac{2.02 + 1.78}{2} = 1.90$ | M1 | Add the two equations |
| $\sum x = 19$ | A1 | |
| Subtract: $2.02 - 1.78 = 2 \times 1.833\sqrt{\frac{s^2}{10}}$ | M1 | Allow 1.372, 1.383, 1.812 instead of 1.833 |
| $s^2 = 0.042859$ | A1 | May be implied. $s = 0.20702$ |
| $s^2 = \frac{1}{9}\left(\sum x^2 - \frac{(\sum x)^2}{10}\right)$ | M1 | |
| $\sum x^2 = 9s^2 + \frac{19^2}{10} = 36.5$ | A1 | 36.486 |
| | **6** | Using 1.645: maximum M1A1 M0A0 M1A0 3/6 |
1 A basketball club has a large number of players. The heights, $x \mathrm {~m}$, of a random sample of 10 of these players are measured. A $90 \%$ confidence interval for the population mean height, $\mu \mathrm { m }$, of players in this club is calculated. It is assumed that heights are normally distributed. The confidence interval is $1.78 \leqslant \mu \leqslant 2.02$.

Find the values of $\sum x$ and $\sum x ^ { 2 }$ for this sample.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q1 [6]}}
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