CAIE S2 2023 June — Question 6 3 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2023
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeReverse-engineering from given variance
DifficultyModerate -0.8 This is a straightforward application of the variance formula requiring algebraic manipulation. Students must recall the unbiased variance formula, set up an equation with the unknown 'a', and solve a quadratic. While it involves multiple steps, it's a standard textbook exercise with no conceptual difficulty or novel insight required—easier than average A-level questions.
Spec5.05b Unbiased estimates: of population mean and variance

6 A sample of 5 randomly selected values of a variable \(X\) is as follows: $$\begin{array} { l l l l l } 1 & 2 & 6 & 1 & a \end{array}$$ where \(a > 0\).
Given that an unbiased estimate of the variance of \(X\) calculated from this sample is \(\frac { 11 } { 2 }\), find the value of \(a\).

Question 6:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{5}{4}\left(\frac{1+2^2+6^2+1+a^2}{5} - \left(\frac{1+2+6+1+a}{5}\right)^2\right) = \frac{11}{2}\) or \(\frac{1}{4}\left((42+a^2) - \frac{(10+a)^2}{5}\right) = \frac{11}{2}\)M1* OE attempted or e.g. \(\frac{42+a^2}{5} - \left(\frac{10+a}{5}\right)^2 = \frac{22}{5}\). Allow use of biased i.e. without \(\frac{5}{4}\).
\(4a^2 - 20a + 0 = 0\) or \(a^2 - 5a + 0 = 0\)DM1 Two- or three-term quadratic equation in \(a\), with at least two terms correct.
\(a = 5\)A1 Ignore \(a = 0\), if seen.
Total: 3
## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{5}{4}\left(\frac{1+2^2+6^2+1+a^2}{5} - \left(\frac{1+2+6+1+a}{5}\right)^2\right) = \frac{11}{2}$ or $\frac{1}{4}\left((42+a^2) - \frac{(10+a)^2}{5}\right) = \frac{11}{2}$ | M1* | OE attempted or e.g. $\frac{42+a^2}{5} - \left(\frac{10+a}{5}\right)^2 = \frac{22}{5}$. Allow use of biased i.e. without $\frac{5}{4}$. |
| $4a^2 - 20a + 0 = 0$ or $a^2 - 5a + 0 = 0$ | DM1 | Two- or three-term quadratic equation in $a$, with at least two terms correct. |
| $a = 5$ | A1 | Ignore $a = 0$, if seen. |
| **Total: 3** | | |
6 A sample of 5 randomly selected values of a variable $X$ is as follows:

$$\begin{array} { l l l l l } 
1 & 2 & 6 & 1 & a
\end{array}$$

where $a > 0$.\\
Given that an unbiased estimate of the variance of $X$ calculated from this sample is $\frac { 11 } { 2 }$, find the value of $a$.\\

\hfill \mbox{\textit{CAIE S2 2023 Q6 [3]}}