| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate variance/SD from coded sums |
| Difficulty | Moderate -0.5 This is a standard S1 coding question requiring routine application of formulas for mean and standard deviation transformations. Parts (a)-(b) use direct formula substitution (reversing the coding), while part (c) involves straightforward arithmetic to find the 9th value and recalculate SD. The question is slightly easier than average because it follows a well-practiced template with no conceptual surprises, though it does require careful algebraic manipulation across multiple parts. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([\bar{x}=]\frac{-1.2}{8}[=-0.15]\) and \(\sum b = 21\times 8+2\times(-1.2)[=165.6]\) | M1 | Correct expression for \(\bar{x}\); correct expression for \(\sum b\) |
| \(-0.15=\frac{\bar{b}-21}{2}\) oe and \([\bar{b}=]\frac{165.6}{8}\) | M1 | Using equation \(\bar{x}=\frac{\bar{b}-21}{2}\) where \(-1.2<\bar{x}<1.2\); use of \(\sum b \div n\) where \(\sum b > 18\) |
| \(= 20.7 \text{ cm}\) | A1 | 20.7 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\sigma_x=\sqrt{\frac{5.1}{8}-\left(\frac{-1.2}{8}\right)^2}[=\sqrt{0.615}=0.784\ldots]\) | M1 | Correct method for \(\sigma_x\) or \(\sigma_x^2\), or \(5.1=\frac{\sum b^2-42\times 168.6+8\times 441}{4}\) or \(\sum b^2=3447.6\) |
| \(\sigma_b=2\times 0.784\ldots\) | M1 | For use of \(2\times\) their \(\sigma_x\) (or \(4\times\) their \(\sigma_x^2\)) (adding 21 is M0) or \(\sqrt{\frac{3447.6}{8}-\left(\frac{165.6}{8}\right)^2}\) |
| \(=\text{awrt } \mathbf{1.57} \text{ cm}\) | A1 | awrt 1.57. Allow \(\frac{\sqrt{246}}{10}\). Allow \(s_b=\text{awrt }1.68\) or \(\frac{4\sqrt{246}}{35}\) from \(n-1\) method |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x_9=1.2\rightarrow b_9=1.2\times 2+21\) oe or \(9\times 21-8\times 20.7[=354.6]\) | M1 | Correct equation using \(x_9=1.2\) to enable \(b_9\) to be found e.g. \(1.2=\frac{b-21}{2}\), or correct method to find \(\sum x\) for 9 squirrels ft their 20.7 |
| \(= 23.4 \text{ cm}\) | A1 | 23.4 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\sum x^2=5.1+1.2^2[=6.54]\) \(\Rightarrow \sigma_x=\sqrt{\frac{5.1+1.2^2}{9}-0^2}\) | M1 | For \(5.1+(\pm 1.2)^2[=6.54]\) seen ft their \(x_9\). Condone \(5.1+(\pm 9.6)^2[=97.26]\) |
| \(=\text{awrt } \mathbf{0.852} \text{ cm}\) | A1 | awrt 0.852. Allow \(\frac{\sqrt{654}}{30}\). Allow \(s_x=\text{awrt }0.904\) from \(n-1\) method |
# Question 3:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| $[\bar{x}=]\frac{-1.2}{8}[=-0.15]$ and $\sum b = 21\times 8+2\times(-1.2)[=165.6]$ | M1 | Correct expression for $\bar{x}$; correct expression for $\sum b$ |
| $-0.15=\frac{\bar{b}-21}{2}$ oe and $[\bar{b}=]\frac{165.6}{8}$ | M1 | Using equation $\bar{x}=\frac{\bar{b}-21}{2}$ where $-1.2<\bar{x}<1.2$; use of $\sum b \div n$ where $\sum b > 18$ |
| $= 20.7 \text{ cm}$ | A1 | 20.7 oe |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $\sigma_x=\sqrt{\frac{5.1}{8}-\left(\frac{-1.2}{8}\right)^2}[=\sqrt{0.615}=0.784\ldots]$ | M1 | Correct method for $\sigma_x$ or $\sigma_x^2$, or $5.1=\frac{\sum b^2-42\times 168.6+8\times 441}{4}$ or $\sum b^2=3447.6$ |
| $\sigma_b=2\times 0.784\ldots$ | M1 | For use of $2\times$ their $\sigma_x$ (or $4\times$ their $\sigma_x^2$) (adding 21 is M0) **or** $\sqrt{\frac{3447.6}{8}-\left(\frac{165.6}{8}\right)^2}$ |
| $=\text{awrt } \mathbf{1.57} \text{ cm}$ | A1 | awrt 1.57. Allow $\frac{\sqrt{246}}{10}$. Allow $s_b=\text{awrt }1.68$ or $\frac{4\sqrt{246}}{35}$ from $n-1$ method |
## Part (c)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $x_9=1.2\rightarrow b_9=1.2\times 2+21$ oe **or** $9\times 21-8\times 20.7[=354.6]$ | M1 | Correct equation using $x_9=1.2$ to enable $b_9$ to be found e.g. $1.2=\frac{b-21}{2}$, **or** correct method to find $\sum x$ for 9 squirrels ft their 20.7 |
| $= 23.4 \text{ cm}$ | A1 | 23.4 oe |
## Part (c)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $\sum x^2=5.1+1.2^2[=6.54]$ $\Rightarrow \sigma_x=\sqrt{\frac{5.1+1.2^2}{9}-0^2}$ | M1 | For $5.1+(\pm 1.2)^2[=6.54]$ seen ft their $x_9$. Condone $5.1+(\pm 9.6)^2[=97.26]$ |
| $=\text{awrt } \mathbf{0.852} \text{ cm}$ | A1 | awrt 0.852. Allow $\frac{\sqrt{654}}{30}$. Allow $s_x=\text{awrt }0.904$ from $n-1$ method |
---\begin{enumerate}
\item Morgan is investigating the body length, $b$ centimetres, of squirrels.
\end{enumerate}
A random sample of 8 squirrels is taken and the data for each squirrel is coded using
$$x = \frac { b - 21 } { 2 }$$
The results for the coded data are summarised below
$$\sum x = - 1.2 \quad \sum x ^ { 2 } = 5.1$$
(a) Find the mean of $b$\\
(b) Find the standard deviation of $b$
A 9th squirrel is added to the sample. Given that for all 9 squirrels $\sum x = 0$\\
(c) find\\
(i) the body length of the 9th squirrel,\\
(ii) the standard deviation of $x$ for all 9 squirrels.
\hfill \mbox{\textit{Edexcel S1 2022 Q3 [10]}}