| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Reverse-engineering from given variance |
| Difficulty | Standard +0.3 This is a straightforward application of variance formula requiring algebraic manipulation. Students must express the mean in terms of a, substitute into the unbiased variance formula (dividing by n-1=4), set equal to 4, and solve a quadratic equation. While it involves multiple steps, each step follows standard procedures taught in S2 with no novel insight required, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{13 + a}{5}\) | B1 | Accept \(\dfrac{2+3+3+5+a}{5}\); do not ignore subsequent working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{5}{4}\!\left(\dfrac{47+a^2}{5} - \left(\dfrac{13+a}{5}\right)^2\right) = 4\) or \(\dfrac{1}{4}\!\left(47 + a^2 - \dfrac{(13+a)^2}{5}\right) = 4\) | M1 | Use of correct formula using *their* value from (a), in terms of \(a\), and equate to 4 |
| \(2a^2 - 13a - 7 = 0\) | A1 | Any correct three-term quadratic equation rearranged to a form ready to solve |
| \(a = 7\) | A1 | Condone the other value of \(a\) \(\left(-\dfrac{1}{2}\right)\) |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{13 + a}{5}$ | B1 | Accept $\dfrac{2+3+3+5+a}{5}$; do not ignore subsequent working |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{5}{4}\!\left(\dfrac{47+a^2}{5} - \left(\dfrac{13+a}{5}\right)^2\right) = 4$ or $\dfrac{1}{4}\!\left(47 + a^2 - \dfrac{(13+a)^2}{5}\right) = 4$ | M1 | Use of correct formula using *their* value from (a), in terms of $a$, and equate to 4 |
| $2a^2 - 13a - 7 = 0$ | A1 | Any correct three-term quadratic equation rearranged to a form ready to solve |
| $a = 7$ | A1 | Condone the other value of $a$ $\left(-\dfrac{1}{2}\right)$ |
---6 A random sample of 5 values of a variable $X$ is given below.
$$\begin{array} { l l l l l }
2 & 3 & 3 & 5 & a
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression, in terms of $a$, for the mean of these values.\\
It is given that an unbiased estimate of the population variance of $X$, using these values, is 4 . It is also given that $a$ is positive.
\item Find and simplify a quadratic equation in terms of $a$ and hence find the value of $a$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q6 [4]}}