CAIE Further Paper 4 2023 June — Question 2 6 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2023
SessionJune
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 manipulation of the t-distribution formula and solving simultaneous equations. While the concepts (confidence intervals, t-distribution) are standard Further Maths content, the reverse-engineering aspect and algebraic manipulation of two equations in two unknowns elevates it slightly above routine application, but it remains a structured problem with clear methodology.
Spec5.05d Confidence intervals: using normal distribution

2 Shane is studying the lengths of the tails of male red kangaroos. He takes a random sample of 14 male red kangaroos and measures the length of the tail, \(x \mathrm {~m}\), for each kangaroo. He then calculates a \(90 \%\) confidence interval for the population mean tail length, \(\mu \mathrm { m }\), of male red kangaroos. He assumes that the tail lengths are normally distributed and finds that \(1.11 \leqslant \mu \leqslant 1.14\). Find the values of \(\sum x\) and \(\sum x ^ { 2 }\) for this sample.

Question 2:
AnswerMarks Guidance
AnswerMark Guidance
\(\bar{x} + 1.771\sqrt{\frac{s^2}{14}} = 1.14\) or \(\bar{x} - 1.771\sqrt{\frac{s^2}{14}} = 1.11\)M1 SOI. Allow incorrect \(t\)-value, not \(z\)-value.
\(\bar{x} = \frac{1}{2}(1.14 + 1.11) = 1.125\), \(\sum x = 15.75\)B1 Does not depend on use of a \(t\)-value.
Subtract or substitute: \(\sqrt{\frac{s^2}{14}} = \frac{1}{2}\left(\frac{1.14 - 1.11}{1.771}\right)\)M1 Allow incorrect \(t\)-value, but not a \(z\)-value.
\(s^2 = 0.00100[4]\) or \(s = 0.0316[9]\)A1 \(\frac{450}{448063}\), implied by correct final answer.
\(\sum x^2 = 13s^2 + \frac{(\sum x)^2}{14}\)M1 OE
\(\sum x^2 = 17.7(3)\)A1 CWO
## Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} + 1.771\sqrt{\frac{s^2}{14}} = 1.14$ or $\bar{x} - 1.771\sqrt{\frac{s^2}{14}} = 1.11$ | M1 | SOI. Allow incorrect $t$-value, not $z$-value. |
| $\bar{x} = \frac{1}{2}(1.14 + 1.11) = 1.125$, $\sum x = 15.75$ | B1 | Does not depend on use of a $t$-value. |
| Subtract or substitute: $\sqrt{\frac{s^2}{14}} = \frac{1}{2}\left(\frac{1.14 - 1.11}{1.771}\right)$ | M1 | Allow incorrect $t$-value, but not a $z$-value. |
| $s^2 = 0.00100[4]$ or $s = 0.0316[9]$ | A1 | $\frac{450}{448063}$, implied by correct final answer. |
| $\sum x^2 = 13s^2 + \frac{(\sum x)^2}{14}$ | M1 | OE |
| $\sum x^2 = 17.7(3)$ | A1 | CWO |

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2 Shane is studying the lengths of the tails of male red kangaroos. He takes a random sample of 14 male red kangaroos and measures the length of the tail, $x \mathrm {~m}$, for each kangaroo. He then calculates a $90 \%$ confidence interval for the population mean tail length, $\mu \mathrm { m }$, of male red kangaroos. He assumes that the tail lengths are normally distributed and finds that $1.11 \leqslant \mu \leqslant 1.14$.

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

\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q2 [6]}}