Question 4 - A-Level Maths

Edexcel S1 2024 June — Question 4 13 marks

Exam Board	Edexcel
Module	S1 (Statistics 1)
Year	2024
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear regression
Type	Calculate from raw data
Difficulty	Moderate -0.3 This is a standard S1 linear regression question testing routine calculations (Sdd, Sgg, PMCC, regression equation) with straightforward interpretation parts. All formulas are given in the formula booklet and the question follows a predictable structure with no novel problem-solving required. Slightly easier than average due to being purely procedural, though the multi-part nature and arithmetic involved prevent it from being significantly below average.
Spec	2.02c Scatter diagrams and regression lines 5.08a Pearson correlation: calculate pmcc 5.09a Dependent/independent variables 5.09c Calculate regression line 5.09d Linear coding: effect on regression 5.09e Use regression: for estimation in context

A biologist is studying bears. The biologist records the length, $d \mathrm {~cm}$, and the girth, $g \mathrm {~cm}$, of 8 bears. The biologist summarises the data as follows

$$\begin{gathered} \sum d = 1456.8 \quad \sum g = 713.2 \quad \sum d g = 141978.84 \quad \sum g ^ { 2 } = 72675.98 \\ S _ { d d } = 16769.78 \end{gathered}$$

Calculate the exact value of $S _ { d g }$ and the exact value of $S _ { g g }$
Calculate the value of the product moment correlation coefficient between $d$ and $g$
Show that the equation of the regression line of $g$ on $d$ can be written as $$g = - 42.3 + 0.722 d$$ where the values of the intercept and gradient are given to 3 significant figures.
Give an interpretation, in context, of the gradient of the regression line. Using the equation of the regression line given in part (c)
1. estimate the girth of a bear with a length of 2.5 metres,
2. explain why an estimate for the girth of a bear with a length of 0.5 metres is not reliable. Using the regression line from part (c), the biologist estimates that for each $x \mathrm {~cm}$ increase in the length of a bear there will be a 17.3 cm increase in the girth.
Find the value of $x$

Show mark scheme Show mark scheme source

Question 4:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$S_{dg} = 141978.84 - \frac{1456.8 \times 713.2}{8}$ or $S_{gg} = 72675.98 - \frac{713.2^2}{8}$	M1	For correct expression for $S_{dg}$ or $S_{gg}$
$S_{dg} = 12105.12$	A1	Allow $\frac{302628}{25}$; if exact not seen then SC: M1A0A1 for both awrt 12100 and awrt 9090 if correct methods seen
$S_{gg} = 9094.2$	A1	Allow $\frac{45471}{5}$

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$r = \frac{\text{"12105.12"}}{\sqrt{16769.78 \times \text{"9094.2"}}}$	M1	For valid attempt at $r$ with their $S_{dg}$ not equal to 141978.84
$= 0.9802\ldots$ awrt 0.98	A1	awrt 0.98

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$b = \frac{\text{"12105.12"}}{16769.78} [= 0.7218\ldots]$	M1	For correct method to find $b$; may be implied by 0.7218 or better
$a = \frac{713.2}{8} - \text{"0.7218..."} \times \frac{1456.8}{8} [= -42.297\ldots]$	M1	For correct method to find $a$ ft their $b$; may be implied by $-42.29$ or better
$g = -42.3 + 0.722d$	A1*cso	Both method marks must be awarded with sight of 0.7218 or better or $-42.29$ or better

Part (d):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
For each 1 [cm] increase in length/$d$ the girth/$g$ increases by "0.722…"	B1	For suitable contextual comment mentioning 0.722 (or better); if units stated they must be correct

Part (e)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
138.2 awrt 138	B1	awrt 138; allow 1.38m

Part (e)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
[Unreliable] as get a negative girth	B1	For correct reason e.g. sd $= 45.8$ cm so girth nearly 3sd below mean so likely outlier; allow substitution of 50 leading to $g = -6.2$ and suitable reason; do not allow substitution of 0.5 to imply negative girth

Part (f):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$0.722x = 17.3$	M1	For correct equation; implied by awrt 24
$x = 23.96\ldots$ awrt 24	A1	awrt 24

# Question 4:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_{dg} = 141978.84 - \frac{1456.8 \times 713.2}{8}$ or $S_{gg} = 72675.98 - \frac{713.2^2}{8}$ | M1 | For correct expression for $S_{dg}$ or $S_{gg}$ |
| $S_{dg} = 12105.12$ | A1 | Allow $\frac{302628}{25}$; if exact not seen then SC: M1A0A1 for both awrt 12100 and awrt 9090 if correct methods seen |
| $S_{gg} = 9094.2$ | A1 | Allow $\frac{45471}{5}$ |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \frac{\text{"12105.12"}}{\sqrt{16769.78 \times \text{"9094.2"}}}$ | M1 | For valid attempt at $r$ with their $S_{dg}$ not equal to 141978.84 |
| $= 0.9802\ldots$ awrt **0.98** | A1 | awrt 0.98 |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $b = \frac{\text{"12105.12"}}{16769.78} [= 0.7218\ldots]$ | M1 | For correct method to find $b$; may be implied by 0.7218 or better |
| $a = \frac{713.2}{8} - \text{"0.7218..."} \times \frac{1456.8}{8} [= -42.297\ldots]$ | M1 | For correct method to find $a$ ft their $b$; may be implied by $-42.29$ or better |
| $g = -42.3 + 0.722d$ | A1*cso | Both method marks must be awarded with sight of 0.7218 or better or $-42.29$ or better |

## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For each 1 [cm] increase in length/$d$ the girth/$g$ increases by "0.722…" | B1 | For suitable contextual comment mentioning 0.722 (or better); if units stated they must be correct |

## Part (e)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| 138.2 awrt **138** | B1 | awrt 138; allow 1.38m |

## Part (e)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| [Unreliable] as get a negative girth | B1 | For correct reason e.g. sd $= 45.8$ cm so girth nearly 3sd below mean so likely outlier; allow substitution of 50 leading to $g = -6.2$ and suitable reason; do not allow substitution of 0.5 to imply negative girth |

## Part (f):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.722x = 17.3$ | M1 | For correct equation; implied by awrt 24 |
| $x = 23.96\ldots$ awrt **24** | A1 | awrt 24 |

---

Show LaTeX source

\begin{enumerate}
  \item A biologist is studying bears. The biologist records the length, $d \mathrm {~cm}$, and the girth, $g \mathrm {~cm}$, of 8 bears. The biologist summarises the data as follows
\end{enumerate}

$$\begin{gathered}
\sum d = 1456.8 \quad \sum g = 713.2 \quad \sum d g = 141978.84 \quad \sum g ^ { 2 } = 72675.98 \\
S _ { d d } = 16769.78
\end{gathered}$$

(a) Calculate the exact value of $S _ { d g }$ and the exact value of $S _ { g g }$\\
(b) Calculate the value of the product moment correlation coefficient between $d$ and $g$\\
(c) Show that the equation of the regression line of $g$ on $d$ can be written as

$$g = - 42.3 + 0.722 d$$

where the values of the intercept and gradient are given to 3 significant figures.\\
(d) Give an interpretation, in context, of the gradient of the regression line.

Using the equation of the regression line given in part (c)\\
(e) (i) estimate the girth of a bear with a length of 2.5 metres,\\
(ii) explain why an estimate for the girth of a bear with a length of 0.5 metres is not reliable.

Using the regression line from part (c), the biologist estimates that for each $x \mathrm {~cm}$ increase in the length of a bear there will be a 17.3 cm increase in the girth.\\
(f) Find the value of $x$

\hfill \mbox{\textit{Edexcel S1 2024 Q4 [13]}}

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