| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2021 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate variance from summary statistics |
| Difficulty | Moderate -0.3 This is a multi-part S1 statistics question involving geometric probability, variance calculations from summary statistics, and combining means/standard deviations from different groups. Part (a) requires geometric reasoning but is guided ('show that'), parts (b-c) use standard formulas (variance = Σx²/n - (Σx/n)², combined means/SDs), and part (d) requires insight about π appearing in the corner-covering probability. While multi-step, all techniques are routine S1 content with no novel problem-solving required beyond the geometric setup in (a). |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.03e Model with probability: critiquing assumptions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Centre of disc must land at least 1 cm from each side of the rectangle, i.e. inside a rectangle 3 cm long and 1 cm wide | M1 dM1 | 1st M1 accept suitable diagram showing "winning area" or equivalent in words. 2nd dM1 dep on M1 for dimensions of rectangle (at least 3 or 1 seen) |
| Probability disc lies inside rectangle is \(\frac{3\times1}{5\times3} = \frac{1}{5}\) or \(1-\frac{2(1\times5+1\times1)}{5\times3}\) | A1cso | Complete explanation with evidence seen for both M1 marks (*) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\left[\sigma_x =\right]\sqrt{\frac{295}{15}-\left(\frac{61}{15}\right)^2}\) or \(\sqrt{3.1288...}\) | M1 | Correct expression including \(\sqrt{}\); allow \(\sqrt{3.129}\) or better |
| \(= 1.768866...\) awrt 1.77 | A1 | Ans only 2/2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{y}=3.5 \implies \sum y = 42\), so new \(\sum z = 42+61\ [=103]\) | M1, A1 | 1st M1 for using mean of 3.5 to get sum of 12 students e.g. \(12\times3.5\). 1st A1 for correct sum of \(42+61\) or 103 |
| \(\sigma_y=2 \implies 2^2 = \frac{\sum y^2}{12}-3.5^2\) or \(2=\sqrt{\frac{\sum y^2}{12}-3.5^2}\) | M1 | 2nd M1 for correct equation for \(\sum y^2\) |
| \(\sum y^2 = (2^2+3.5^2)\times12\ [=195]\), so new \(\sum z^2 = (2^2+3.5^2)\times12+295\ [\text{or }490]\) | A1 | 2nd A1 for correct expression for \(\sum z^2\) e.g. \(= 195+295\ [=490]\) |
| New mean \(= \frac{\text{"103"}}{(15+12)} = [3.8148...]\) | dM1 | 3rd dM1 dep on 1st M1 for correct method for finding new mean or awrt 3.81 |
| New standard deviation \(= \sqrt{\frac{\text{"490"}}{(12+15)}-\text{"3.81..."}^2}\ [=1.89613...]\) | dM1 | 4th dM1 dep on 1st and 2nd M1s for correct method for new st. dev. |
| New mean = awrt 3.81, new st. dev = awrt 1.90 | A1 | Both values required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Centre of disc must be within 1 cm of a vertex (so 4 quarter circles) | M1 | M1 for explanation or diagram showing possible region for centre is a full circle |
| So probability of disc covering a vertex is \(\frac{\pi}{15}\) | A1 | 1st A1 for the correct probability. Allow M1A1 for \(\frac{\pi}{15}\) (o.e.) but must be in part (d) |
| So an estimate for \(\pi\) is \(15\times0.2216 = \mathbf{3.324}\) | A1 | 2nd A1 dep on M1 for estimate of 3.324 (accept 3.32 if M1A1 clearly scored) |
| Answer | Marks | Guidance |
|---|---|---|
| As for main scheme | M1dM1, A1cso (3) | As in main scheme |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\sigma_x =\right] \sqrt{\dfrac{295}{300} - \left(\dfrac{61}{300}\right)^2}\) or \(\sqrt{0.941988...}\) | M1 | For a correct expression including \(\sqrt{\phantom{x}}\); allow \(\sqrt{0.942}\) or better |
| \(= 0.9705611...\) awrt \(\mathbf{0.971}\) | A0ft | A0 for awrt 0.971 (first two accuracy ft marks withheld for misread) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{y} = 3.5 \Rightarrow \sum y = 240 \times 3.5 = 840\), so new \(\sum z = 840 + 61 = 901\) | M1, A0ft | 1st M1 for using mean of 3.5 to get sum; 1st A0 for correct sum of \(840 + 61\) or \(901\) |
| \(\sigma_y = 2 \Rightarrow 2^2 = \dfrac{\sum y^2}{240} - 3.5^2\) or \(2 = \sqrt{\dfrac{\sum y^2}{240} - 3.5^2}\) | M1 | 2nd M1 for correct equation for \(\sum y^2\) |
| \(\sum y^2 = (2^2 + 3.5^2) \times 240 = 3900\), so \(\sum z^2 = \sum y^2 + 295 = (2^2 + 3.5^2)\times 240 + 295\) [or 4195] | A1ft | 2nd A1ft for correct expression for \(\sum z^2\), e.g. \(= 3900 + 295\ [= 4195]\) |
| New mean \(= \dfrac{\text{"901"}}{(300+240)} = [1.66851...]\) | dM1 | 3rd dM1 dep on 1st M1 for correct method for finding new mean or awrt 1.67 |
| New standard deviation \(= \sqrt{\dfrac{\text{"4195"}}{(240+300)} - \text{"1.668..."}^2}\ [= 2.2326...]\) | dM1 | 4th dM1 dep on 1st and 2nd M1s for correct method for new st. dev. |
| New mean = awrt \(\mathbf{1.67}\), new st. dev. = awrt \(\mathbf{2.23}\) | A1ft \((7-1=6)\) | 3rd A1ft for both mean \(= 1.67\) and st. dev. \(=\) awrt 2.23 |
| Answer | Marks | Guidance |
|---|---|---|
| Centre of disc must be within 1 cm of a vertex (so 4 quarter circles) | M1 | For explanation or diagram showing possible region for centre is a full circle |
| So probability of disc covering a vertex is \(\dfrac{\pi}{15}\) | A1 | 1st A1 for correct probability; allow M1A1 for \(\dfrac{\pi}{15}\) (o.e.) but must be in part (d) |
| So an estimate for \(\pi\) is \(15 \times 0.2216 = \mathbf{3.324}\) | A1 (3) | 2nd A1 dep on M1 for estimate of 3.324 (accept 3.32 if M1A1 clearly scored); minimum acceptable for 3/3 is \(\pi = 15 \times 0.2216 = 3.324\) |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre of disc must land at least 1 cm from each side of the rectangle, i.e. inside a rectangle 3 cm long and 1 cm wide | M1 dM1 | 1st M1 accept suitable diagram showing "winning area" or equivalent in words. 2nd dM1 dep on M1 for dimensions of rectangle (at least 3 or 1 seen) |
| Probability disc lies inside rectangle is $\frac{3\times1}{5\times3} = \frac{1}{5}$ or $1-\frac{2(1\times5+1\times1)}{5\times3}$ | A1cso | Complete explanation **with evidence seen** for both M1 marks (*) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left[\sigma_x =\right]\sqrt{\frac{295}{15}-\left(\frac{61}{15}\right)^2}$ or $\sqrt{3.1288...}$ | M1 | Correct expression including $\sqrt{}$; allow $\sqrt{3.129}$ or better |
| $= 1.768866...$ awrt **1.77** | A1 | **Ans only 2/2** |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{y}=3.5 \implies \sum y = 42$, so new $\sum z = 42+61\ [=103]$ | M1, A1 | 1st M1 for using mean of 3.5 to get sum of 12 students e.g. $12\times3.5$. 1st A1 for correct sum of $42+61$ or 103 |
| $\sigma_y=2 \implies 2^2 = \frac{\sum y^2}{12}-3.5^2$ or $2=\sqrt{\frac{\sum y^2}{12}-3.5^2}$ | M1 | 2nd M1 for correct equation for $\sum y^2$ |
| $\sum y^2 = (2^2+3.5^2)\times12\ [=195]$, so new $\sum z^2 = (2^2+3.5^2)\times12+295\ [\text{or }490]$ | A1 | 2nd A1 for correct expression for $\sum z^2$ e.g. $= 195+295\ [=490]$ |
| New mean $= \frac{\text{"103"}}{(15+12)} = [3.8148...]$ | dM1 | 3rd dM1 dep on 1st M1 for correct method for finding new mean or awrt 3.81 |
| New standard deviation $= \sqrt{\frac{\text{"490"}}{(12+15)}-\text{"3.81..."}^2}\ [=1.89613...]$ | dM1 | 4th dM1 dep on 1st and 2nd M1s for correct method for new st. dev. |
| New mean = awrt **3.81**, new st. dev = awrt **1.90** | A1 | Both values required |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre of disc must be within 1 cm of a vertex (so 4 quarter circles) | M1 | M1 for explanation or diagram showing possible region for centre is a full circle |
| So probability of disc covering a vertex is $\frac{\pi}{15}$ | A1 | 1st A1 for the correct probability. Allow M1A1 for $\frac{\pi}{15}$ (o.e.) but must be in part (d) |
| So an estimate for $\pi$ is $15\times0.2216 = \mathbf{3.324}$ | A1 | 2nd A1 dep on M1 for estimate of 3.324 (accept 3.32 if M1A1 clearly scored) |
# Question 6:
## Part (a)
| As for main scheme | M1dM1, A1cso (3) | As in main scheme |
## Part (b) — MR scheme ($n = 300$)
| $\left[\sigma_x =\right] \sqrt{\dfrac{295}{300} - \left(\dfrac{61}{300}\right)^2}$ or $\sqrt{0.941988...}$ | M1 | For a correct expression including $\sqrt{\phantom{x}}$; allow $\sqrt{0.942}$ or better |
| $= 0.9705611...$ awrt $\mathbf{0.971}$ | A0ft | A0 for awrt 0.971 (first two accuracy ft marks withheld for misread) |
## Part (c) — MR scheme ($m = 240$)
| $\bar{y} = 3.5 \Rightarrow \sum y = 240 \times 3.5 = 840$, so new $\sum z = 840 + 61 = 901$ | M1, A0ft | 1st M1 for using mean of 3.5 to get sum; 1st A0 for correct sum of $840 + 61$ or $901$ |
| $\sigma_y = 2 \Rightarrow 2^2 = \dfrac{\sum y^2}{240} - 3.5^2$ or $2 = \sqrt{\dfrac{\sum y^2}{240} - 3.5^2}$ | M1 | 2nd M1 for correct equation for $\sum y^2$ |
| $\sum y^2 = (2^2 + 3.5^2) \times 240 = 3900$, so $\sum z^2 = \sum y^2 + 295 = (2^2 + 3.5^2)\times 240 + 295$ [or 4195] | A1ft | 2nd A1ft for correct expression for $\sum z^2$, e.g. $= 3900 + 295\ [= 4195]$ |
| New mean $= \dfrac{\text{"901"}}{(300+240)} = [1.66851...]$ | dM1 | 3rd dM1 dep on 1st M1 for correct method for finding new mean or awrt 1.67 |
| New standard deviation $= \sqrt{\dfrac{\text{"4195"}}{(240+300)} - \text{"1.668..."}^2}\ [= 2.2326...]$ | dM1 | 4th dM1 dep on 1st and 2nd M1s for correct method for new st. dev. |
| New mean = awrt $\mathbf{1.67}$, new st. dev. = awrt $\mathbf{2.23}$ | A1ft $(7-1=6)$ | 3rd A1ft for both mean $= 1.67$ **and** st. dev. $=$ awrt 2.23 |
## Part (d)
| Centre of disc must be within 1 cm of a vertex (so 4 quarter circles) | M1 | For explanation or diagram showing possible region for centre is a full circle |
| So probability of disc covering a vertex is $\dfrac{\pi}{15}$ | A1 | 1st A1 for correct probability; allow M1A1 for $\dfrac{\pi}{15}$ (o.e.) but must be in part (d) |
| So an estimate for $\pi$ is $15 \times 0.2216 = \mathbf{3.324}$ | A1 (3) | 2nd A1 dep on M1 for estimate of 3.324 (accept 3.32 if M1A1 clearly scored); minimum acceptable for 3/3 is $\pi = 15 \times 0.2216 = 3.324$ |
**[13 marks total]**\begin{enumerate}
\item A disc of radius 1 cm is rolled onto a horizontal grid of rectangles so that the disc is equally likely to land anywhere on the grid. Each rectangle is 5 cm long and 3 cm wide. There are no gaps between the rectangles and the grid is sufficiently large so that no discs roll off the grid.
\end{enumerate}
If the disc lands inside a rectangle without covering any part of the edges of the rectangle then a prize is won.
By considering the possible positions for the centre of the disc,\\
(a) show that the probability of winning a prize on any particular roll is $\frac { 1 } { 5 }$
A group of 15 students each roll the disc onto the grid twenty times and record the number of times, $x$, that each student wins a prize. Their results are summarised as follows
$$\sum x = 61 \quad \sum x ^ { 2 } = 295$$
(b) Find the standard deviation of the number of prizes won per student.
A second group of 12 students each roll the disc onto the grid twenty times and the mean number of prizes won per student is 3.5 with a standard deviation of 2\\
(c) Find the mean and standard deviation of the number of prizes won per student for the whole group of 27 students.
The 27 students also recorded the number of times that the disc covered a corner of a rectangle and estimated the probability to be 0.2216 (to 4 decimal places).\\
(d) Explain how this probability could be used to find an estimate for the value of $\pi$ and state the value of your estimate.
\hfill \mbox{\textit{Edexcel S1 2021 Q6 [15]}}