| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Calculate frequency density from frequency |
| Difficulty | Moderate -0.8 This is a straightforward S1 statistics question testing standard procedures: frequency density calculation (simple division), linear interpolation for median (routine formula application), coding transformations (direct substitution), and conceptual understanding of mean/SD effects. All parts are textbook exercises requiring recall and basic arithmetic rather than problem-solving or insight. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02j Clean data: missing data, errors |
| Length | Frequency |
| \(20 \leqslant x < 30\) | 2 |
| \(30 \leqslant x < 32\) | 16 |
| \(32 \leqslant x < 36\) | 20 |
| \(36 \leqslant x < 40\) | 8 |
| \(40 \leqslant x < 45\) | 3 |
| \(45 \leqslant x < 50\) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Width \(= 1.25\) [cm] | B1 | No need for units |
| \(18.75 \text{ cm}^2\) for freq of 20 so \(\frac{18.75}{20} \times 16 = 15 \text{ cm}^2\) for freq of 16, or \(w \times h = 15\), or fd \(= 5\) | M1 | For sight of 15 or "their \(w\)" \(\times\) "their \(h\)" \(= 15\) or fd \(= 5\); may be implied by \(h=12\) |
| \([h = 15 \div 1.25\) or \(h = 8 \div 5 \times 7.5 =] 12\) (cm) | A1 | For height \(= 12\); no need for units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Q_2 = [32+]\frac{7}{20} \times 4\) or using \(n+1\) gives \(Q_2 = [32+]\frac{7.5}{20} \times 4\) | M1 | For \(\frac{7}{20}\times 4\) or \(\frac{13}{20}\times 4\) or \(\frac{m-32}{25-18}=\frac{4}{20}\) or \(\frac{36-m}{38-25}=\frac{4}{20}\) oe; allow 25.5 rather than 25 |
| \(= 33.4\) (\(n+1\) gives 33.5) | A1 | 33.4 or if using \((n+1)\): 33.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{y} = \frac{104}{50} [= 2.08]\); \(\sum(w-20) = 10 \times 104 [=1040]\) or \(\sum w = 10 \times 104 + 50 \times 20 [=2040]\) | M1 | For correct method to find mean of \(y\) or \(\sum(w-20)\) or \(\sum w\); \((10\times104+k\) where \(k \neq 20\times50\) is M0) |
| \(\bar{w} = 10 \times \text{"2.08"} + 20 = 40.8\) | A1* | For correct method to find mean of \(w\) leading to 40.8 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\text{Variance of } y =] \frac{233.54}{50} - (\text{"2.08"})^2 \left[= \frac{861}{2500} = 0.3444\right]\) or \(10 \times \text{sd of } y = \text{sd of } w\) | M1 | For correct method to find Variance of \(y\) or writing/using \(10 \times \text{sd of } y = \text{sd of } w\) or correct equation to find \(\sum w^2\) |
| \([\text{Variance of } w =] \text{"0.3444"} \times 100\) or \(\frac{\text{"84954"}}{50} - 40.8^2 \left[= \frac{861}{25} = 34.44\right]\) | M1 | For correct method to find Variance of \(w\) or sd of \(y\) ft their \(\text{Var}(y)\) |
| sd of \(w = \sqrt{\text{"0.3444"} \times 100}\) or \(\sqrt{\text{"34.44"}}\) or \(10 \times \text{"}\frac{\sqrt{861}}{50}\text{"}\) | M1 | For correct method to find sd of \(w\) ft their \(\text{Var}(w)\) |
| \(= 5.868\ldots\) awrt 5.87 | A1 | awrt 5.87; NB exact answer \(\frac{\sqrt{861}}{5}\) scores A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The mean would not change (as 40.8 is the mean) | B1 | No reason needed; allow mean \(= 40.8\) to imply no change |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The standard deviation would decrease (as 40.8 is in the middle so data closer together) | B1 | For sd decreases/be smaller/go down (condone Var decreases); no reason needed |
| Both correct with correct reason for why sd decreases | ddB1 | Both previous B1s awarded; for correct reason for sd decreasing; allow \(\sum(x-\bar{x})^2\) doesn't change and \(n\) increases; allow values would be more concentrated about mean |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Width $= 1.25$ [cm] | B1 | No need for units |
| $18.75 \text{ cm}^2$ for freq of 20 so $\frac{18.75}{20} \times 16 = 15 \text{ cm}^2$ for freq of 16, or $w \times h = 15$, or fd $= 5$ | M1 | For sight of 15 or "their $w$" $\times$ "their $h$" $= 15$ or fd $= 5$; may be implied by $h=12$ |
| $[h = 15 \div 1.25$ or $h = 8 \div 5 \times 7.5 =] 12$ (cm) | A1 | For height $= 12$; no need for units |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Q_2 = [32+]\frac{7}{20} \times 4$ or using $n+1$ gives $Q_2 = [32+]\frac{7.5}{20} \times 4$ | M1 | For $\frac{7}{20}\times 4$ or $\frac{13}{20}\times 4$ or $\frac{m-32}{25-18}=\frac{4}{20}$ or $\frac{36-m}{38-25}=\frac{4}{20}$ oe; allow 25.5 rather than 25 |
| $= 33.4$ ($n+1$ gives 33.5) | A1 | 33.4 or if using $(n+1)$: 33.5 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{y} = \frac{104}{50} [= 2.08]$; $\sum(w-20) = 10 \times 104 [=1040]$ or $\sum w = 10 \times 104 + 50 \times 20 [=2040]$ | M1 | For correct method to find mean of $y$ or $\sum(w-20)$ or $\sum w$; $(10\times104+k$ where $k \neq 20\times50$ is M0) |
| $\bar{w} = 10 \times \text{"2.08"} + 20 = 40.8$ | A1* | For correct method to find mean of $w$ leading to 40.8 |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\text{Variance of } y =] \frac{233.54}{50} - (\text{"2.08"})^2 \left[= \frac{861}{2500} = 0.3444\right]$ or $10 \times \text{sd of } y = \text{sd of } w$ | M1 | For correct method to find Variance of $y$ or writing/using $10 \times \text{sd of } y = \text{sd of } w$ or correct equation to find $\sum w^2$ |
| $[\text{Variance of } w =] \text{"0.3444"} \times 100$ or $\frac{\text{"84954"}}{50} - 40.8^2 \left[= \frac{861}{25} = 34.44\right]$ | M1 | For correct method to find Variance of $w$ or sd of $y$ ft their $\text{Var}(y)$ |
| sd of $w = \sqrt{\text{"0.3444"} \times 100}$ or $\sqrt{\text{"34.44"}}$ or $10 \times \text{"}\frac{\sqrt{861}}{50}\text{"}$ | M1 | For correct method to find sd of $w$ ft their $\text{Var}(w)$ |
| $= 5.868\ldots$ awrt 5.87 | A1 | awrt 5.87; NB exact answer $\frac{\sqrt{861}}{5}$ scores A0 |
## Part (e)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The mean would not change (as 40.8 is the mean) | B1 | No reason needed; allow mean $= 40.8$ to imply no change |
## Part (e)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The standard deviation would decrease (as 40.8 is in the middle so data closer together) | B1 | For sd decreases/be smaller/go down (condone Var decreases); no reason needed |
| Both correct with correct reason for why sd decreases | ddB1 | Both previous B1s awarded; for correct reason for sd decreasing; allow $\sum(x-\bar{x})^2$ doesn't change and $n$ increases; allow values would be more concentrated about mean |
---\begin{enumerate}
\item The lengths, $x \mathrm {~mm}$, of 50 pebbles are summarised in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | }
\hline
Length & Frequency \\
\hline
$20 \leqslant x < 30$ & 2 \\
\hline
$30 \leqslant x < 32$ & 16 \\
\hline
$32 \leqslant x < 36$ & 20 \\
\hline
$36 \leqslant x < 40$ & 8 \\
\hline
$40 \leqslant x < 45$ & 3 \\
\hline
$45 \leqslant x < 50$ & 1 \\
\hline
\end{tabular}
\end{center}
A histogram is drawn to represent these data.\\
The bar representing the class $32 \leqslant x < 36$ is 2.5 cm wide and 7.5 cm tall.\\
(a) Calculate the width and the height of the bar representing the class $30 \leqslant x < 32$\\
(b) Using linear interpolation, estimate the median of $x$
The weight, $w$ grams, of each of the 50 pebbles is coded using $10 y = w - 20$ These coded data are summarised by
$$\sum y = 104 \quad \sum y ^ { 2 } = 233.54$$
(c) Show that the mean of $w$ is 40.8\\
(d) Calculate the standard deviation of $w$
The weight of a pebble recorded as 40.8 grams is added to the sample.\\
(e) Without carrying out any further calculations, state, giving a reason, what effect this would have on the value of\\
(i) the mean of $w$\\
(ii) the standard deviation of $w$
\hfill \mbox{\textit{Edexcel S1 2024 Q3 [14]}}